diff --git a/20231116/hp-20231116.pdf b/20231116/hp-20231116.pdf
index 5d8ad8c7c024d23614889e9a42d51a264e8407d6..0b5959d27419f80c0d8be603431b3f1a1130897c 100644
Binary files a/20231116/hp-20231116.pdf and b/20231116/hp-20231116.pdf differ
diff --git a/20231116/hp-20231116.tex b/20231116/hp-20231116.tex
index e10f6c2aa8d43c8ad9c6ec02dbec81360e13cff4..d96bbf3f4148317038c074a825d4b0b7fc10af71 100644
--- a/20231116/hp-20231116.tex
+++ b/20231116/hp-20231116.tex
@@ -20,7 +20,7 @@
% Attribution-ShareAlike 3.0 Unported License along with this
% document. If not, see <http://creativecommons.org/licenses/>.
-% README: Byte-Reihenfolge, Darstellung negativer Zahlen, Darstellung von Gleitkommazahlen, Speicherausrichtung
+% README: Byte-Reihenfolge, Darstellung negativer Zahlen, Darstellung von Gleitkommazahlen
\documentclass[10pt,t]{beamer}
@@ -61,6 +61,7 @@
\item[4.5] Byte-Reihenfolge -- Endianness
\item[4.6] Binärdarstellung negativer Zahlen
\item[4.7] Binärdarstellung von Gleitkommazahlen
+ \color{black}
\item[4.8] Speicherausrichtung -- Alignment
\end{itemize}
\item[\textbf{5}] \textbf{Algorithmen}
@@ -547,6 +548,7 @@
\item[4.5] Byte-Reihenfolge -- Endianness
\item[4.6] Binärdarstellung negativer Zahlen
\item[4.7] Binärdarstellung von Gleitkommazahlen
+ \color{black}
\item[4.8] Speicherausrichtung -- Alignment
\end{itemize}
\item[\textbf{5}] \textbf{Algorithmen}
@@ -871,16 +873,18 @@
\item
Zahlen aufsummieren:\\
vorher sortieren, mit der kleinsten Zahl beginnen
- \item
- Ableitungen bilden:\\
- Beim Bilden von Differenzquotienten\\
- verliert man notwendigerweise an Präzision!\\
- \textarrow\ Die Differenzen sehr sorgfältig auswählen.\\
- \textarrow\ Am besten gar nicht ableiten, sondern integrieren.
+% \item
+% Ableitungen bilden:\\
+% Beim Bilden von Differenzquotienten\\
+% verliert man notwendigerweise an Präzision!\\
+% \textarrow\ Die Differenzen sehr sorgfältig auswählen.\\
+% \textarrow\ Am besten gar nicht ableiten, sondern integrieren.
\end{itemize}
\end{frame}
+\iffalse
+
\subsection{Speicherausrichtung -- Alignment}
\begin{frame}[fragile]
@@ -947,4 +951,6 @@
\end{frame}
+\fi
+
\end{document}
diff --git a/20231116/hp-musterloesung-20231116.pdf b/20231116/hp-musterloesung-20231116.pdf
index e1a8b074f3e986af2da73bb1c7491cc74eb1a684..429cab8899551fd68e5c426323d1bdff75083470 100644
Binary files a/20231116/hp-musterloesung-20231116.pdf and b/20231116/hp-musterloesung-20231116.pdf differ
diff --git a/20231116/hp-musterloesung-20231116.tex b/20231116/hp-musterloesung-20231116.tex
index 5e52b6fb67e9b3ece80d2034c1a923a7b9d77b1b..e3bcfb3ae99b4b11d79ebbb91714b75f532d496d 100644
--- a/20231116/hp-musterloesung-20231116.tex
+++ b/20231116/hp-musterloesung-20231116.tex
@@ -1,4 +1,4 @@
-% hp-musterloesung-20231109.pdf - Solutions to the Exercises on Low-Level Programming / Applied Computer Sciences
+% hp-musterloesung-20231116.pdf - Solutions to the Exercises on Low-Level Programming / Applied Computer Sciences
% Copyright (C) 2013, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023 Peter Gerwinski
%
% This document is free software: you can redistribute it and/or
@@ -33,7 +33,7 @@
\begin{document}
\section*{Hardwarenahe Programmierung\\
- Musterlösung zu den Übungsaufgaben -- 9.\ November 2023}
+ Musterlösung zu den Übungsaufgaben -- 16.\ November 2023}
\exercise{Trickprogrammierung}
diff --git a/20231116/sum-01.c b/20231116/sum-01.c
new file mode 100644
index 0000000000000000000000000000000000000000..0362b2f512138c77dde1ae1cd5e113dba56ed301
--- /dev/null
+++ b/20231116/sum-01.c
@@ -0,0 +1,15 @@
+#include <stdio.h>
+#include <math.h>
+
+/* 1/1² + 1/2² + 1/3² + ... = pi²/6 */
+
+int main (void)
+{
+ float pi = M_PI;
+ printf ("%0.9f\n", pi * pi / 6.0);
+ float S = 0.0;
+ for (float x = 1; x <= 100; x++)
+ S += 1.0 / (x * x);
+ printf ("%0.9f\n", S);
+ return 0;
+}
diff --git a/20231116/sum-02.c b/20231116/sum-02.c
new file mode 100644
index 0000000000000000000000000000000000000000..c0fabe8394e2c1cb6619a0ba817ae4a7b2b67670
--- /dev/null
+++ b/20231116/sum-02.c
@@ -0,0 +1,15 @@
+#include <stdio.h>
+#include <math.h>
+
+/* 1/1² + 1/2² + 1/3² + ... = pi²/6 */
+
+int main (void)
+{
+ float pi = M_PI;
+ printf ("%0.9f\n", pi * pi / 6.0);
+ float S = 0.0;
+ for (float x = 1; x <= 1000000; x++)
+ S += 1.0 / (x * x);
+ printf ("%0.9f\n", S);
+ return 0;
+}
diff --git a/20231116/sum-03.c b/20231116/sum-03.c
new file mode 100644
index 0000000000000000000000000000000000000000..472d8b7cb1fd74ac102e4a662ce7d27dfce1eb8a
--- /dev/null
+++ b/20231116/sum-03.c
@@ -0,0 +1,15 @@
+#include <stdio.h>
+#include <math.h>
+
+/* 1/1² + 1/2² + 1/3² + ... = pi²/6 */
+
+int main (void)
+{
+ float pi = M_PI;
+ printf ("%0.9f\n", pi * pi / 6.0);
+ float S = 0.0;
+ for (float x = 1; x <= 10000000; x++) /* ca. 1/20 s Rechenzeit */
+ S += 1.0 / (x * x);
+ printf ("%0.9f\n", S);
+ return 0;
+}
diff --git a/20231116/sum-04.c b/20231116/sum-04.c
new file mode 100644
index 0000000000000000000000000000000000000000..c49d13aa687b7019d2b306e9f10f30da478d56f1
--- /dev/null
+++ b/20231116/sum-04.c
@@ -0,0 +1,15 @@
+#include <stdio.h>
+#include <math.h>
+
+/* 1/1² + 1/2² + 1/3² + ... = pi²/6 */
+
+int main (void)
+{
+ float pi = M_PI;
+ printf ("%0.9f\n", pi * pi / 6.0);
+ float S = 0.0;
+ for (float x = 1; x <= 100000000; x++) /* erwarte: ca. 1/2 s Rechenzeit */
+ S += 1.0 / (x * x);
+ printf ("%0.9f\n", S);
+ return 0;
+}
diff --git a/20231116/sum-05.c b/20231116/sum-05.c
new file mode 100644
index 0000000000000000000000000000000000000000..8612524abcc79fa720193972d61b623ff750e443
--- /dev/null
+++ b/20231116/sum-05.c
@@ -0,0 +1,14 @@
+#include <stdio.h>
+
+int main (void)
+{
+ float x = 10000000;
+ printf ("%f\n", x);
+ x++;
+ printf ("%f\n", x);
+ x = 100000000;
+ printf ("%f\n", x);
+ x++;
+ printf ("%f\n", x);
+ return 0;
+}
diff --git a/20231116/sum-06.c b/20231116/sum-06.c
new file mode 100644
index 0000000000000000000000000000000000000000..477e087168ae5d43158edbf04261292b51ebb16d
--- /dev/null
+++ b/20231116/sum-06.c
@@ -0,0 +1,15 @@
+#include <stdio.h>
+#include <math.h>
+
+/* 1/1² + 1/2² + 1/3² + ... = pi²/6 */
+
+int main (void)
+{
+ float pi = M_PI;
+ printf ("%0.9f\n", pi * pi / 6.0);
+ float S = 0.0;
+ for (int i = 1; i <= 100000000; i++) /* erwarte: ca. 1/2 s Rechenzeit */
+ S += 1.0 / (i * i);
+ printf ("%0.9f\n", S);
+ return 0;
+}
diff --git a/20231116/sum-07.c b/20231116/sum-07.c
new file mode 100644
index 0000000000000000000000000000000000000000..af482431387bbb2b7b4e17f8e5c658c759824322
--- /dev/null
+++ b/20231116/sum-07.c
@@ -0,0 +1,8 @@
+#include <stdio.h>
+
+int main (void)
+{
+ int i = 65536;
+ printf ("%d\n", i * i);
+ return 0;
+}
diff --git a/20231116/sum-09.c b/20231116/sum-09.c
new file mode 100644
index 0000000000000000000000000000000000000000..e2c732f2d878c76989bccbfd30f1ec3b38997e28
--- /dev/null
+++ b/20231116/sum-09.c
@@ -0,0 +1,25 @@
+#include <stdio.h>
+#include <math.h>
+
+/* 1/1² + 1/2² + 1/3² + ... = pi²/6 */
+
+int main (void)
+{
+ float pi = M_PI;
+ printf ("%0.9f\n", pi * pi / 6.0);
+ float S = 0.0;
+ for (int i = 1; i <= 100000000; i++)
+ {
+ float x = i;
+ S += 1.0 / (x * x); /* kleine Zahl zu großer addieren: verschwindet */
+ }
+ printf ("%0.9f\n", S);
+ S = 0.0;
+ for (int i = 100000000; i >= 1; i--) /* Zuerst die kleinen Zahlen addieren, dann die großen. */
+ {
+ float x = i;
+ S += 1.0 / (x * x);
+ }
+ printf ("%0.9f\n", S);
+ return 0;
+}
diff --git a/20231123/aufgabe-2.c b/20231123/aufgabe-2.c
new file mode 100644
index 0000000000000000000000000000000000000000..47595ef0658e94d76a42263e82200f94895cdeea
--- /dev/null
+++ b/20231123/aufgabe-2.c
@@ -0,0 +1,23 @@
+#include <stdio.h>
+#include <string.h>
+
+typedef struct
+{
+ char first_name[10];
+ char family_name[20];
+ char day, month;
+ int year;
+} person;
+
+int main (void)
+{
+ person sls;
+ sls.day = 26;
+ sls.month = 7;
+ sls.year = 1951;
+ strcpy (sls.first_name, "Sabine");
+ strcpy (sls.family_name, "Leutheusser-Schnarrenberger");
+ printf ("%s %s wurde am %d.%d.%d geboren.\n",
+ sls.first_name, sls.family_name, sls.day, sls.month, sls.year);
+ return 0;
+}
diff --git a/20231123/aufgabe-3.c b/20231123/aufgabe-3.c
new file mode 100644
index 0000000000000000000000000000000000000000..5b0cb23fdd5ee15a4403808c18d2104ed49caf3f
--- /dev/null
+++ b/20231123/aufgabe-3.c
@@ -0,0 +1,62 @@
+#include <gtk/gtk.h>
+
+#define WIDTH 320
+#define HEIGHT 240
+
+double t = 0.0;
+double dt = 0.2;
+
+int r = 5;
+
+double x = 10;
+double y = 200;
+double vx = 20;
+double vy = -60;
+double g = 9.81;
+
+gboolean draw (GtkWidget *widget, cairo_t *c, gpointer data)
+{
+ GdkRGBA blue = { 0.0, 0.5, 1.0, 1.0 };
+
+ gdk_cairo_set_source_rgba (c, &blue);
+ cairo_arc (c, x, y, r, 0, 2 * G_PI);
+ cairo_fill (c);
+
+ return FALSE;
+}
+
+gboolean timer (GtkWidget *widget)
+{
+ t += dt;
+ x += vx * dt;
+ y += vy * dt;
+ vx = vx;
+ vy = 0.5 * g * (t * t);
+ if (y + r >= HEIGHT)
+ vy = -vy * 0.9;
+ if (x + r >= WIDTH)
+ vx = -vx * 0.9;
+ if (x - r <= 0)
+ vx = -vx * 0.9;
+ gtk_widget_queue_draw_area (widget, 0, 0, WIDTH, HEIGHT);
+ g_timeout_add (50, (GSourceFunc) timer, widget);
+ return FALSE;
+}
+
+int main (int argc, char **argv)
+{
+ gtk_init (&argc, &argv);
+
+ GtkWidget *window = gtk_window_new (GTK_WINDOW_TOPLEVEL);
+ gtk_widget_show (window);
+ gtk_window_set_title (GTK_WINDOW (window), "Hello");
+ g_signal_connect (window, "destroy", G_CALLBACK (gtk_main_quit), NULL);
+
+ GtkWidget *drawing_area = gtk_drawing_area_new ();
+ gtk_widget_show (drawing_area);
+ gtk_container_add (GTK_CONTAINER (window), drawing_area);
+ gtk_widget_set_size_request (drawing_area, WIDTH, HEIGHT);
+
+ gtk_main ();
+ return 0;
+}
diff --git a/20231123/hp-20231123.pdf b/20231123/hp-20231123.pdf
new file mode 100644
index 0000000000000000000000000000000000000000..721a2fe874f61111db02c6f2e18d0b937123a3ec
Binary files /dev/null and b/20231123/hp-20231123.pdf differ
diff --git a/20231123/hp-20231123.tex b/20231123/hp-20231123.tex
new file mode 100644
index 0000000000000000000000000000000000000000..1975ab620ce0aa95de87e6b30f9be90c46c309e7
--- /dev/null
+++ b/20231123/hp-20231123.tex
@@ -0,0 +1,883 @@
+% hp-20231123.pdf - Lecture Slides on Low-Level Programming
+% Copyright (C) 2012, 2013, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023 Peter Gerwinski
+%
+% This document is free software: you can redistribute it and/or
+% modify it either under the terms of the Creative Commons
+% Attribution-ShareAlike 3.0 License, or under the terms of the
+% GNU General Public License as published by the Free Software
+% Foundation, either version 3 of the License, or (at your option)
+% any later version.
+%
+% This document is distributed in the hope that it will be useful,
+% but WITHOUT ANY WARRANTY; without even the implied warranty of
+% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
+% GNU General Public License for more details.
+%
+% You should have received a copy of the GNU General Public License
+% along with this document. If not, see <http://www.gnu.org/licenses/>.
+%
+% You should have received a copy of the Creative Commons
+% Attribution-ShareAlike 3.0 Unported License along with this
+% document. If not, see <http://creativecommons.org/licenses/>.
+
+% README: Speicherausrichtung, Algorithmen: Differentialgleichungen
+
+\documentclass[10pt,t]{beamer}
+
+\usepackage{pgslides}
+\usepackage{tikz}
+
+\newcommand{\redurl}[1]{\href{#1}{\color{red}\nolinkurl{#1}}}
+
+\title{Hardwarenahe Programmierung}
+\author{Prof.\ Dr.\ rer.\ nat.\ Peter Gerwinski}
+\date{23.\ November 2023}
+
+\begin{document}
+
+\maketitleframe
+
+\title{Hardwarenahe Programmierung}
+
+\nosectionnonumber{\inserttitle}
+
+\begin{frame}
+
+ \shownosectionnonumber
+
+ \begin{itemize}
+ \item[\textbf{1}] \textbf{Einführung}
+ \hfill\makebox(0,0)[br]{\raisebox{2.25ex}{\url{https://gitlab.cvh-server.de/pgerwinski/hp}}}
+ \item[\textbf{2}] \textbf{Einführung in C}
+ \item[\textbf{3}] \textbf{Bibliotheken}
+ \item[\textbf{4}] \textbf{Hardwarenahe Programmierung}
+ \begin{itemize}
+ \item[4.1] Bit-Operationen
+ \item[4.2] I/O-Ports
+ \item[4.3] Interrupts
+ \item[4.4] volatile-Variable
+ \color{medgreen}
+ \item[4.5] Byte-Reihenfolge -- Endianness
+ \item[4.6] Binärdarstellung negativer Zahlen
+ \item[4.7] Binärdarstellung von Gleitkommazahlen
+ \color{red}
+ \item[4.8] Speicherausrichtung -- Alignment
+ \end{itemize}
+ \item[\textbf{5}] \textbf{Algorithmen}
+ \item[\textbf{6}] \textbf{Objektorientierte Programmierung}
+ \item[\textbf{7}] \textbf{Datenstrukturen}
+ \end{itemize}
+
+\end{frame}
+
+\setcounter{section}{3}
+\section{Hardwarenahe Programmierung}
+\setcounter{subsection}{4}
+\subsection{Byte-Reihenfolge -- Endianness}
+\subsubsection{Konzept}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+ \showsubsubsection
+
+ Eine Zahl geht über mehrere Speicherzellen.\\
+ Beispiel: 16-Bit-Zahl in 2 8-Bit-Speicherzellen
+
+ \smallskip
+
+ Welche Bits liegen wo?
+
+% \pause
+ \bigskip
+
+ $1027 = 1024 + 2 + 1 = 0000\,0100\,0000\,0011_2 = 0403_{16}$
+
+% \pause
+ \bigskip
+ Speicherzellen:
+
+ \medskip
+ \begin{tabular}{|c|c|l}\cline{1-2}
+ \raisebox{-0.25ex}{04} & \raisebox{-0.25ex}{03} & \strut Big-Endian "`großes Ende zuerst"' \\\cline{1-2}
+ \multicolumn{2}{c}{} & für Menschen leichter lesbar \\
+ \multicolumn{3}{c}{} \\[-5pt]\cline{1-2}
+ \raisebox{-0.25ex}{03} & \raisebox{-0.25ex}{04} & \strut Little-Endian "`kleines Ende zuerst"' \\\cline{1-2}
+ \multicolumn{2}{c}{} & bei Additionen effizienter
+ \end{tabular}
+
+% \pause
+ \medskip
+ \textarrow\ Geschmackssache
+% \pause
+ \\
+ \quad\textbf{\dots\ außer bei Datenaustausch!}
+
+% \pause
+% \bigskip
+%
+% Aber: nicht verwechseln! \qquad $0304_{16} = 772$
+
+\end{frame}
+
+\begin{frame}
+
+ \showsubsection
+ \showsubsubsection
+
+ Eine Zahl geht über mehrere Speicherzellen.\\
+ Beispiel: 16-Bit-Zahl in 2 8-Bit-Speicherzellen
+
+ \smallskip
+
+ Welche Bits liegen wo?
+
+ \medskip
+
+ \textarrow\ Geschmackssache\\
+ \textbf{\dots\ außer bei Datenaustausch!}
+
+ \begin{itemize}
+ \item
+ Dateiformate
+ \item
+ Datenübertragung
+ \end{itemize}
+
+\end{frame}
+
+\subsubsection{Dateiformate}
+
+\begin{frame}
+
+ \showsubsection
+ \showsubsubsection
+
+ Audio-Formate: Reihenfolge der Bytes in 16- und 32-Bit-Zahlen
+ \begin{itemize}
+ \item
+ RIFF-WAVE-Dateien (\file{.wav}): Little-Endian
+ \item
+ Au-Dateien (\file{.au}): Big-Endian
+% \pause
+ \item
+ ältere AIFF-Dateien (\file{.aiff}): Big-Endian
+ \item
+ neuere AIFF-Dateien (\file{.aiff}): Little-Endian
+ \end{itemize}
+
+% \pause
+ \bigskip
+
+ Grafik-Formate: Reihenfolge der Bits in den Bytes
+ \begin{itemize}
+ \item
+ PBM-Dateien: Big-Endian\only<1->{, MSB first}
+ \item
+ XBM-Dateien: Little-Endian\only<1->{, LSB first}
+ \end{itemize}
+ \only<1->{MSB/LSB = most/least significant bit}
+
+\end{frame}
+
+\subsubsection{Datenübertragung}
+
+\begin{frame}
+
+ \showsubsection
+ \showsubsubsection
+
+ \begin{itemize}
+ \item
+ RS-232 (serielle Schnittstelle): LSB first
+ \item
+ I$^2$C: MSB first
+ \item
+ USB: beides
+% \pause
+ \medskip
+ \item
+ Ethernet: LSB first
+ \item
+ TCP/IP (Internet): Big-Endian
+ \end{itemize}
+
+\end{frame}
+
+\subsection{Binärdarstellung negativer Zahlen}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+ Speicher ist begrenzt!\\
+ \textarrow\ feste Anzahl von Bits
+
+ \medskip
+
+ 8-Bit-Zahlen ohne Vorzeichen: \lstinline{uint8_t}\\
+ \textarrow\ Zahlenwerte von \lstinline{0x00} bis \lstinline{0xff} = 0 bis 255\\
+% \pause
+ \textarrow\ 255 + 1 = 0
+
+% \pause
+ \medskip
+
+ 8-Bit-Zahlen mit Vorzeichen: \lstinline{int8_t}\\
+ \lstinline{0xff} = 255 ist die "`natürliche"' Schreibweise für $-1$.\\
+% \pause
+ \textarrow\ Zweierkomplement
+
+% \pause
+ \medskip
+
+ Oberstes Bit = 1: negativ\\
+ Oberstes Bit = 0: positiv\\
+ \textarrow\ 127 + 1 = $-128$
+
+\end{frame}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+ Speicher ist begrenzt!\\
+ \textarrow\ feste Anzahl von Bits
+
+ \medskip
+
+ 16-Bit-Zahlen ohne Vorzeichen:
+ \lstinline{uint16_t}\hfill\lstinline{uint8_t}\\
+ \textarrow\ Zahlenwerte von \lstinline{0x0000} bis \lstinline{0xffff}
+ = 0 bis 65535\hfill 0 bis 255\\
+ \textarrow\ 65535 + 1 = 0\hfill 255 + 1 = 0
+
+ \medskip
+
+ 16-Bit-Zahlen mit Vorzeichen:
+ \lstinline{int16_t}\hfill\lstinline{int8_t}\\
+ \lstinline{0xffff} = 66535 ist die "`natürliche"' Schreibweise für $-1$.\hfill
+ \lstinline{0xff} = 255 = $-1$\\
+ \textarrow\ Zweierkomplement
+
+ \medskip
+
+ Oberstes Bit = 1: negativ\\
+ Oberstes Bit = 0: positiv\\
+ \textarrow\ 32767 + 1 = $-32768$
+
+ \bigskip
+ Literatur: \url{http://xkcd.com/571/}
+
+\end{frame}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+ Frage: \emph{Für welche Zahl steht der Speicherinhalt\,
+ \raisebox{2pt}{%
+ \tabcolsep0.25em
+ \begin{tabular}{|c|c|}\hline
+ \rule{0pt}{11pt}a3 & 90 \\\hline
+ \end{tabular}}
+ (hexadezimal)?}
+
+% \pause
+ \smallskip
+ Antwort: \emph{Das kommt darauf an.} ;--)
+
+% \pause
+ \medskip
+ Little-Endian:
+
+ \smallskip
+
+ \begin{tabular}{lrl}
+ als \lstinline,int8_t,: & $-93$ & (nur erstes Byte)\\
+ als \lstinline,uint8_t,: & $163$ & (nur erstes Byte)\\
+ als \lstinline,int16_t,: & $-28509$\\
+ als \lstinline,uint16_t,: & $37027$\\
+ \lstinline,int32_t, oder größer: & $37027$
+ & (zusätzliche Bytes mit Nullen aufgefüllt)
+ \end{tabular}
+
+% \pause
+ \medskip
+ Big-Endian:
+
+ \smallskip
+
+ \begin{tabular}{lrl}
+ als \lstinline,int8_t,: & $-93$ & (nur erstes Byte)\\
+ als \lstinline,uint8_t,: & $163$ & (nur erstes Byte)\\
+ als \lstinline,int16_t,: & $-23664$\\
+ als \lstinline,uint16_t,: & $41872$\\ als \lstinline,int32_t,: & $-1550843904$ & (zusätzliche Bytes\\
+ als \lstinline,uint32_t,: & $2744123392$ & mit Nullen aufgefüllt)\\
+ als \lstinline,int64_t,: & $-6660823848880963584$\\
+ als \lstinline,uint64_t,: & $11785920224828588032$\\
+ \end{tabular}
+
+ \vspace*{-1cm}
+
+\end{frame}
+
+\subsection{Binärdarstellung von Gleitkommazahlen}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+% (Diese Seite wurde unbewußt leer gelassen.)
+
+ Beispiel für Gleitkommazahl: $2{,}351\cdot10^5$ (oder: $2.351\times10^5$)
+
+ \smallskip
+
+ Bezeichnungen: $\text{Mantisse} \cdot 10^{\text{Exponent}}$
+
+ \smallskip
+
+ C-Schreibweise: \lstinline{2.351e5} (oder: \lstinline{2.351E5})
+
+% \pause
+ \bigskip
+
+ Wie speichert man Gleitkommazahlen?
+
+ \smallskip
+
+ $m$-Bit-Zahl, davon
+ \begin{itemize}
+ \item
+ $e$ Bits für den Exponenten (einschließlich Vorzeichen),
+ \item
+ $1$ Bit für das Vorzeichen der Mantisse,
+ \item
+ $m - e - 1$ Bits für die Mantisse.
+ \end{itemize}
+
+% \pause
+ \begin{picture}(0,0)
+ \color{red}
+ \put(1.95,0.65){\makebox(0,0){\tikz{\draw(0,0)--(0.5,0.25);}}}
+ \put(1.95,0.65){\makebox(0,0){\tikz{\draw(0,0.25)--(0.5,0);}}}
+ \end{picture}%
+ {\color{red}Trick: Mantisse als \newterm{normalisierte Zahl\/} abspeichern}
+
+% \pause
+ \bigskip
+ Vorteil gegenüber ganzen Zahlen:\\
+ größerer Wertebereich bei vergleichbarem Speicherplatzbedarf
+
+ \medskip
+
+ Nachteil gegenüber ganzen Zahlen: Rundungsfehler\\
+ \textcolor{red}{\textarrow\
+ \textbf{ungeeignet} für Anwendungen, bei denen es auf jedes Bit ankommt\\
+ \phantom{\textarrow\ }(z.\,B.\ Verschlüsselung)}
+ \vspace*{-1cm}
+
+\end{frame}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+ Problem beim Arbeiten mit Gleitkommazahlen: Auslöschung von Ziffern
+ \begin{itemize}
+ \item
+ Zahlen aufsummieren:\\
+ vorher sortieren, mit der kleinsten Zahl beginnen
+ \pause
+ \item
+ Ableitungen bilden:\\
+ Beim Bilden von Differenzquotienten\\
+ verliert man notwendigerweise an Präzision!\\
+ \textarrow\ Die Differenzen sehr sorgfältig auswählen.\\
+ \textarrow\ Am besten gar nicht ableiten, sondern integrieren.
+ \end{itemize}
+
+\end{frame}
+
+\subsection{Speicherausrichtung -- Alignment}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+ \begin{lstlisting}
+ #include <stdint.h>
+
+ uint8_t a;
+ uint16_t b;
+ uint8_t c;
+ \end{lstlisting}
+
+ \pause
+ \bigskip
+
+ Speicheradresse durch 2 teilbar -- "`16-Bit-Alignment"'
+ \begin{itemize}
+ \item
+ 2-Byte-Operation: effizienter
+ \pause
+ \item
+ \dots\ oder sogar nur dann erlaubt
+ \pause
+ \arrowitem
+ Compiler optimiert Speicherausrichtung
+ \end{itemize}
+
+ \medskip
+
+ \pause
+ \begin{minipage}{3cm}
+ \begin{lstlisting}[gobble=6]
+ ¡uint8_t a;
+ uint8_t dummy;
+ uint16_t b;
+ uint8_t c;¿
+ \end{lstlisting}
+ \end{minipage}
+ \pause
+ \begin{minipage}{3cm}
+ \begin{lstlisting}[gobble=6]
+ ¡uint8_t a;
+ uint8_t c;
+ uint16_t b;¿
+ \end{lstlisting}
+ \end{minipage}
+
+ \pause
+ \vspace{-1.75cm}
+ \strut\hfill
+ \begin{minipage}{6.5cm}
+ Fazit:
+ \begin{itemize}
+ \item
+ \textbf{Adressen von Variablen\\
+ sind systemabhängig}
+ \item
+ Bei Definition von Datenformaten\\
+ Alignment beachten \textarrow\ effizienter
+ \end{itemize}
+ \end{minipage}
+
+\end{frame}
+
+\section{Algorithmen}
+\subsection{Differentialgleichungen}
+
+\begin{frame}[fragile]
+
+ \showsection
+ \showsubsection
+
+ \textbf{Beispiel 1: Gleichmäßig beschleunigte Bewegung}
+
+ \strut\hfill
+ \begin{minipage}{2.5cm}
+ \vspace*{0.6cm}
+ \begin{align*}
+ x'(t) &= v_x(t) \\[0.65cm]
+ y'(t) &= v_y(t) \\[0.75cm]
+ v_x'(t) &= 0 \\[0.65cm]
+ v_y'(t) &= -g
+ \end{align*}
+ \vspace*{0.0cm}
+ \end{minipage}%
+ \only<1>{\hspace*{9.49cm}}\strut
+ \only<2->{\hfill$\Rightarrow$\hfill}%
+ \begin{onlyenv}<2-8>
+ \begin{minipage}{8.3cm}
+ \begin{align*}
+ x(t) &= \int v_x(t)\,dt
+ \visible<4->{= \int v_{0x}\,dt}
+ \visible<5->{= x_0 + v_{0x}\cdot t}\\[\medskipamount]
+ y(t) &= \int v_y(t)\,dt
+ \visible<7->{= \int v_{0y} - g\cdot t\,dt}
+ \visible<8->{= y_0 + v_{0y}\cdot t
+ - {\textstyle\frac12}gt^2}\\[\bigskipamount]
+ v_x(t) &= \int 0\,dt
+ \visible<3->{= v_{0x}} \\[\medskipamount]
+ v_y(t) &= \int -g\,dt
+ \visible<6->{= v_{0y} - g\cdot t}
+ \end{align*}
+ \end{minipage}%
+ \end{onlyenv}%
+ \begin{onlyenv}<9->
+ \begin{minipage}{3.5cm}
+ \vspace*{0.5cm}
+ \begin{lstlisting}[gobble=8,xleftmargin=0.5em]
+ ¡x += vx * dt;¿
+ \end{lstlisting}
+ \vspace{0.75cm}
+ \begin{lstlisting}[gobble=8,xleftmargin=0.5em]
+ ¡y += vy * dt;¿
+ \end{lstlisting}
+ \vspace{0.90cm}
+ \begin{lstlisting}[gobble=8,xleftmargin=0.5em]
+ ¡vx += 0 * dt;¿
+ \end{lstlisting}
+ \vspace{0.75cm}
+ \begin{lstlisting}[gobble=8,xleftmargin=0.5em]
+ ¡vy += -g * dt;¿
+ \end{lstlisting}
+ \end{minipage}%
+ \begin{minipage}{5.13cm}
+% Siehe: \file{gtk-13.c}
+ \strut
+ \end{minipage}
+ \end{onlyenv}%
+ \hfill\strut
+
+\end{frame}
+
+\begin{frame}[fragile]
+ \showsection
+ \showsubsection
+
+ \textbf{Beispiel 1: Gleichmäßig beschleunigte Bewegung}
+
+ \medskip
+
+ \textbf{Beispiel 2: Mathematisches Pendel}
+
+ \vspace*{-2\bigskipamount}
+
+ \begin{picture}(0,0)
+ \put(8,-6.5){\includegraphics{pendulum.pdf}}
+ \end{picture}
+
+ \begin{eqnarray*}
+ \varphi'(t) &=& \omega(t) \\[\smallskipamount]
+ \omega'(t) &=& -\frac{g}{l}\cdot\sin\varphi(t)\hspace*{7.1cm}
+ \end{eqnarray*}
+ \vspace*{-1.5\medskipamount}
+ \begin{itemize}
+ \item
+ Von Hand (analytisch):\\
+ Lösung raten (Ansatz), Parameter berechnen
+ \item
+ Mit Computer (numerisch):\\
+ Eulersches Polygonzugverfahren
+ \end{itemize}
+ \smallskip
+ \begin{lstlisting}[gobble=0]
+ phi += dt * omega;
+ omega += - dt * g / l * sin (phi);
+ \end{lstlisting}
+
+ \pause
+ \bigskip
+
+ \textbf{Beispiel 3: Weltraum-Simulation}
+
+ Praktikumsaufgabe
+ \vspace*{-1cm}
+
+\end{frame}
+
+\iffalse
+
+\nosectionnonumber{\inserttitle}
+
+\begin{frame}
+
+ \shownosectionnonumber
+
+ \begin{itemize}
+ \item[\textbf{1}] \textbf{Einführung}
+ \hfill\makebox(0,0)[br]{\raisebox{2.25ex}{\url{https://gitlab.cvh-server.de/pgerwinski/hp}}}
+ \item[\textbf{2}] \textbf{Einführung in C}
+ \item[\textbf{4}] \textbf{Hardwarenahe Programmierung}
+ \begin{itemize}
+ \item[4.1] Bit-Operationen
+ \item[4.2] I/O-Ports
+ \item[4.3] Interrupts
+ \item[4.4] volatile-Variable
+ \color{medgreen}
+ \item[4.5] Byte-Reihenfolge -- Endianness
+ \item[4.6] Binärdarstellung negativer Zahlen
+ \item[4.7] Speicherausrichtung -- Alignment
+ \end{itemize}
+ \item[\textbf{5}] \textbf{Algorithmen}
+ \begin{itemize}
+ \color{medgreen}
+ \item[5.1] Differentialgleichungen
+ \color{red}
+ \item[5.2] Rekursion
+ \color{black}
+ \item[5.3] Aufwandsabschätzungen
+ \end{itemize}
+ \item[\textbf{6}] \textbf{Objektorientierte Programmierung}
+ \item[\textbf{7}] \textbf{Datenstrukturen}
+ \end{itemize}
+
+\end{frame}
+
+\setcounter{section}{4}
+\section{Algorithmen}
+\setcounter{subsection}{1}
+\subsection{Rekursion}
+
+\begin{frame}[fragile]
+
+ \showsubsection
+
+ Vollständige Induktion:
+ \vspace*{-0.725cm}
+ \begin{displaymath}
+ \hspace*{4cm}
+ \left.
+ \begin{array}{r}
+ \mbox{Aussage gilt für $n = 1$}\\[2pt]
+ \mbox{Schluß von $n - 1$ auf $n$}
+ \end{array}
+ \right\}
+ \mbox{Aussage gilt für alle $n\in\mathbb{N}$}
+ \end{displaymath}
+ \vspace*{-0.5cm}
+
+ \pause
+
+ Türme von Hanoi
+
+ \begin{onlyenv}<2>
+ \begin{center}
+ \includegraphics[width=12.2cm]{Tower_of_Hanoi.jpeg}
+ \end{center}
+ \end{onlyenv}
+
+ \begin{onlyenv}<3->
+ \begin{itemize}
+ \item
+ 64 Scheiben, 3 Plätze,
+ \only<3-4>{\hfill\makebox(0,0)[rt]{\includegraphics[width=6cm]{Tower_of_Hanoi.jpeg}}}\\
+ immer 1 Scheibe verschieben
+ \item
+ Ziel: Turm verschieben
+ \item
+ Es dürfen nur kleinere Scheiben\\
+ auf größeren liegen.
+ \bigskip
+ \pause
+ \pause
+ \item
+ $n = 1$ Scheibe: fertig
+ \item
+ Wenn $n - 1$ Scheiben verschiebbar:\\
+ schiebe $n - 1$ Scheiben auf Hilfsplatz,\\
+ verschiebe die darunterliegende,\\
+ hole $n - 1$ Scheiben von Hilfsplatz
+ \end{itemize}
+ \begin{onlyenv}<5>
+ \vspace{-4.3cm}
+ \begin{lstlisting}[gobble=8,xleftmargin=6.4cm]
+ void move (int from, int to, int disks)
+ {
+ if (disks == 1)
+ move_one_disk (from, to);
+ else
+ {
+ int help = 0 + 1 + 2 - from - to;
+ move (from, help, disks - 1);
+ move (from, to, 1);
+ move (help, to, disks - 1);
+ }
+ }
+ \end{lstlisting}
+ \end{onlyenv}
+% \begin{onlyenv}<6->
+% \vspace{-5.0cm}
+% \hspace*{7.4cm}\begin{minipage}[t]{5cm}
+% 32 Scheiben:
+% \begin{lstlisting}[gobble=10,style=terminal]
+% $ ¡time ./hanoi-9b¿
+% ...
+% real 0m30,672s
+% user 0m30,662s
+% sys 0m0,008s
+% \end{lstlisting}
+% \pause[7]
+% \begin{itemize}
+% \arrowitem
+% etwas über 1 Minute\\
+% für 64 Scheiben
+% \end{itemize}
+% \pause
+% \vspace*{-0.5cm}
+% \begin{picture}(0,0)
+% \color{red}
+% \put(0,0){\makebox(0,0)[bl]{\tikz[line width=1pt]{\draw(0,0)--(4,0.8);}}}
+% \put(0,0.8){\makebox(0,0)[tl]{\tikz[line width=1pt]{\draw(0,0)--(4,-0.8);}}}
+% \end{picture}
+%
+% Für jede zusätzliche Scheibe\\verdoppelt sich die Rechenzeit!
+% % 30.672 * 2^32 / 3600 / 24 / 365.25 = 4174.43775518138261464750
+% \begin{itemize}
+% \arrowitem
+% $\frac{30,672\,\text{s}\,\cdot\,2^{32}}{3600\,\cdot\,24\,\cdot\,365,25} \approx 4174$
+% Jahre\\[\smallskipamount]
+% für 64 Scheiben
+% \end{itemize}
+% \end{minipage}
+% \end{onlyenv}
+ \end{onlyenv}
+
+\end{frame}
+
+\subsection{Aufwandsabschätzungen \protect\color{gray}-- Komplexitätsanalyse}
+
+\begin{frame}[fragile]
+
+% \newcommand{\w}{\hspace*{0.75pt}}
+
+ \showsubsection
+
+ \begin{picture}(0,0)
+ \put(7.6,-0.5){%
+ \begin{minipage}[t]{5.3cm}
+% \vspace*{-1.0cm}\includegraphics{landau-symbols.pdf}
+ \vspace*{-1.0cm}\alt<16->{\includegraphics{landau-symbols-3.pdf}}%
+ {\alt<15->{\includegraphics{landau-symbols-2.pdf}}%
+ {\includegraphics{landau-symbols.pdf}}}
+ \small
+ \begin{description}\itemsep0pt\leftskip-0.5cm
+ \item[$n$:] Eingabedaten
+ \item[$g(n)$:] Rechenzeit
+ \end{description}
+ \end{minipage}}
+ \end{picture}
+
+ \vspace*{-\bigskipamount}
+
+ Wann ist ein Programm "`schnell"'?
+
+ \medskip
+
+ \begin{onlyenv}<1-2>
+ Türme von Hanoi: $\mathcal{O}(2^n)$
+ \par\medskip
+ Für jede zusätzliche Scheibe\\verdoppelt sich die Rechenzeit!
+ \begin{itemize}
+ \arrowitem
+ $\frac{30,672\,\text{s}\,\cdot\,2^{32}}{3600\,\cdot\,24\,\cdot\,365,25} \approx 4174$
+ Jahre\\[\smallskipamount]
+ für 64 Scheiben
+ \end{itemize}
+
+ \bigskip
+ \end{onlyenv}
+
+ \begin{onlyenv}<2->
+ Faustregel:\\Schachtelung der Schleifen zählen\\
+ $k$ Schleifen ineinander \textarrow\ $\mathcal{O}(n^k)$
+
+ \bigskip
+ \end{onlyenv}
+
+ \begin{onlyenv}<3-13>
+ \textbf{Beispiel: Sortieralgorithmen}
+
+ \smallskip
+
+ Anzahl der Vergleiche bei $n$ Strings
+ \begin{itemize}
+ \item
+ Maximum suchen \pause[4]mit Schummeln\pause: $\mathcal{O}(1)$
+ \pause
+ \item
+ Maximum suchen\pause: $\mathcal{O}(n)$
+ \pause
+ \item
+ Selection-Sort\pause: $\mathcal{O}(n^2)$
+ \pause
+ \item
+ Bubble-Sort\pause: $\mathcal{O}(n)$ bis $\mathcal{O}(n^2)$
+ \pause
+ \item
+ Quicksort\pause: $\mathcal{O}(n\log n)$ bis $\mathcal{O}(n^2)$
+ \end{itemize}
+
+ \end{onlyenv}
+
+ \begin{onlyenv}<14>
+ \textbf{Wie schnell ist RSA-Verschlüsselung?}
+
+ \smallskip
+
+ \begin{math}
+ c = m^e\,\%\,N
+ \end{math}
+ \quad
+ ("`$\%$"' = "`modulo"')
+
+ \medskip
+
+ \begin{lstlisting}[gobble=6,xleftmargin=2em]
+ int c = 1;
+ for (int i = 0; i < e; i++)
+ c = (c * m) % N;
+ \end{lstlisting}
+
+ \smallskip
+
+ \begin{itemize}
+ \item
+ $\mathcal{O}(e)$ Iterationen
+% \item
+% wenn $n$ die Anzahl der Binärziffern (Bits) von $e$ ist:
+% $\mathcal{O}(2^n)$ Iterationen
+ \item
+ mit Trick:
+ $\mathcal{O}(\log e)$ Iterationen ($\log e$ = Anzahl der Ziffern von $e$)
+ \end{itemize}
+
+ \smallskip
+
+ Jede Iteration enthält eine Multiplikation und eine Division.\\
+ Aufwand dafür: $\mathcal{O}(\log e)$\\
+ \textarrow\ Gesamtaufwand: $\mathcal{O}\bigl((\log e)^2\bigr)$
+
+ \end{onlyenv}
+
+ \begin{onlyenv}<15->
+
+ \textbf{Wie schnell ist RSA?}\\
+
+ \smallskip
+
+ ($n$ = typische beteiligte Zahl, z.\,B. $e,p,q$)
+
+ \begin{itemize}
+ \item
+ Ver- und Entschlüsselung (Exponentiation):\\
+ \strut\hbox to 3.5cm{\color{red}$\mathcal{O}\!\left((\log n)^2\right)$\hss}
+ \only<16->{{\color{magenta}$\mathcal{O}(n^2)$}}
+ \item
+ Schlüsselerzeugung (Berechnung von $d$):\\
+ \strut\hbox to 3.5cm{\color{red}$\mathcal{O}\!\left((\log n)^2\right)$\hss}
+ \only<16->{{\color{magenta}$\mathcal{O}(n^2)$}}
+ \item
+ Verschlüsselung brechen (Primfaktorzerlegung):\\
+ \strut\hbox to 3.5cm{\color{red}$\mathcal{O}\bigl(2^{\sqrt{\log n\,\cdot\,\log\log n}}\bigr)$\hss}
+ \only<16->{{\color{magenta}$\mathcal{O}\bigl(2^{\sqrt{n\log n}}\bigr)$}}
+ \end{itemize}
+
+ \vspace{0cm plus 1filll}
+
+ \textbf{Die Sicherheit von RSA beruht darauf,
+ daß das Brechen der Verschlüsselung aufwendiger ist als
+ \boldmath$\mathcal{O}\bigl((\log n)^k\bigr)$ (für beliebiges $k$).}
+
+ \vspace*{0.65cm}
+
+ \end{onlyenv}
+
+\end{frame}
+
+\fi
+
+\end{document}
diff --git a/20231123/hp-musterloesung-20231123.pdf b/20231123/hp-musterloesung-20231123.pdf
new file mode 100644
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diff --git a/20231123/hp-musterloesung-20231123.tex b/20231123/hp-musterloesung-20231123.tex
new file mode 100644
index 0000000000000000000000000000000000000000..810480f0f431e547c1f4cc9f472b1337f2598e0b
--- /dev/null
+++ b/20231123/hp-musterloesung-20231123.tex
@@ -0,0 +1,394 @@
+% hp-musterloesung-20231116.pdf - Solutions to the Exercises on Low-Level Programming / Applied Computer Sciences
+% Copyright (C) 2013, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023 Peter Gerwinski
+%
+% This document is free software: you can redistribute it and/or
+% modify it either under the terms of the Creative Commons
+% Attribution-ShareAlike 3.0 License, or under the terms of the
+% GNU General Public License as published by the Free Software
+% Foundation, either version 3 of the License, or (at your option)
+% any later version.
+%
+% This document is distributed in the hope that it will be useful,
+% but WITHOUT ANY WARRANTY; without even the implied warranty of
+% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
+% GNU General Public License for more details.
+%
+% You should have received a copy of the GNU General Public License
+% along with this document. If not, see <http://www.gnu.org/licenses/>.
+%
+% You should have received a copy of the Creative Commons
+% Attribution-ShareAlike 3.0 Unported License along with this
+% document. If not, see <http://creativecommons.org/licenses/>.
+
+% README: Kondensator, Personen-Datenbank, Hexdumps
+
+\documentclass[a4paper]{article}
+
+\usepackage{pgscript}
+\usepackage{gnuplot-lua-tikz}
+
+\begin{document}
+
+ \section*{Hardwarenahe Programmierung\\
+ Musterlösung zu den Übungsaufgaben -- 16.\ November 2023}
+
+ \exercise{Kondensator}
+
+ Ein Kondensator der Kapazität $C = 100\,\mu{\rm F}$
+ ist auf die Spannung $U_0 = 5\,{\rm V}$ aufgeladen
+ und wird über einen Widerstand $R = 33\,{\rm k}\Omega$ entladen.
+
+ \begin{enumerate}[(a)]
+ \item
+ Schreiben Sie ein C-Programm, das
+ den zeitlichen Spannungsverlauf in einer Tabelle darstellt.
+ \points{5}
+ \item
+ Schreiben Sie ein C-Programm, das ermittelt,
+ wie lange es dauert, bis die Spannung unter $0.1\,{\rm V}$ gefallen ist.
+ \points{4}
+ \item
+ Vergleichen Sie die berechneten Werte mit der exakten theoretischen Entladekurve:
+ \begin{math}
+ U(t) = U_0 \cdot e^{-\frac{t}{RC}}
+ \end{math}\\
+ \points{3}
+ \end{enumerate}
+
+ Hinweise:
+ \begin{itemize}
+ \item
+ Für die Simulation zerlegen wir den Entladevorgang in kurze Zeitintervalle $dt$.
+ Innerhalb jedes Zeitintervalls betrachten wir den Strom $I$ als konstant
+ und berechnen, wieviel Ladung $Q$ innerhalb des Zeitintervalls
+ aus dem Kondensator herausfließt.
+ Aus der neuen Ladung berechnen wir die Spannung am Ende des Zeitintervalls.
+ \item
+ Für den Vergleich mit der exakten theoretischen Entladekurve
+ benötigen Sie die Exponentialfunktion \lstinline{exp()}.
+ Diese finden Sie in der Mathematik-Bibliothek:
+ \lstinline{#include <math.h>} im Quelltext,
+ beim \lstinline[style=cmd]{gcc}-Aufruf \lstinline[style=cmd]{-lm} mit angeben.
+ \item
+ $Q = C \cdot U$,\quad $U = R \cdot I$,\quad $I = \frac{dQ}{dt}$
+ \end{itemize}
+
+ \solution
+
+ \begin{itemize}
+ \item
+ \textbf{Schreiben Sie ein C-Programm, das
+ den zeitlichen Spannungsverlauf in einer Tabelle darstellt.}
+
+ In dem Programm \gitfile{hp}{2023ws/20231123}{loesung-1a.c}
+ arbeiten wir, dem ersten Hinweis folgend,
+ mit einem Zeitintervall von \lstinline{dt = 0.01}.
+ Mit dieser Schrittweite lassen wir uns eine Tabelle ausgeben,
+ die jeweils die Zeit und durch die Simulation berechnete Spannung ausgibt.
+
+ Wir simulieren, wie die Ladung $Q = C \cdot U$ des Kondensators
+ im Laufe der Zeit abfließt.
+ Dazu berechnen wir in jedem Zeitschritt zunächst den Strom $I = U / R$,
+ der aus dem Kondensator fließt.
+ Dieser Strom bewirkt, daß innerhalb des Zeitintervalls $dt$
+ die Ladung $dQ = I \cdot dt$ aus dem Kondensator abfließt.
+ Am Ende des Zeitintervalls berechnen wir die zur neuen Ladung $Q$
+ gehörende neue Spannung $U = Q / C$.
+
+ Für eine einfache Ausgabe der Tabelle
+ verwenden wir die Formatspezifikationen \lstinline{%10.3lf} für drei
+ bzw.\ \lstinline{%15.8lf} für acht Nachkommastellen Genauigkeit.
+ Damit schreiben wir jeweils eine \emph{lange Fließkommazahl\/} (\lstinline{%lf})
+ rechtsbündig in ein Feld der Breite 10 bzw.\ 15
+ und lassen uns 3 bzw.\ 8 Nachkommastellen ausgeben.
+
+ Wir compilieren das Programm mit:
+ \lstinline[style=cmd]{gcc -Wall -O loesung-1a.c -o loesung-1a}
+
+ \item
+ \textbf{Schreiben Sie ein C-Programm, das ermittelt,
+ wie lange es dauert, bis die Spannung unter \boldmath $0.1\,{\rm V}$ gefallen ist.}
+
+ Wir ändern das Programm \gitfile{hp}{2023ws/20231123}{loesung-1a.c} so ab,
+ daß zum einen die Schleife abbricht, sobald die Spannung
+ den Wert $0.1\,{\rm V}$ unterschreitet (\gitfile{hp}{2023ws/20231123}{loesung-1b.c}),
+ und daß zum anderen nicht jedesmal eine Zeile für die Tabelle ausgegeben wird,
+ sondern erst am Ende die Zeit (und die Spannung).
+
+ Der Ausgabe entnehmen wir, daß die Spannung bei etwa $t = 12.90\,{\rm s}$
+ den Wert $0.1\,{\rm V}$ unterschreitet.
+
+ \item
+ \textbf{Vergleichen Sie die berechneten Werte
+ mit der exakten theoretischen Entladekurve:\\[0.5\smallskipamount]
+ \boldmath
+ \begin{math}
+ U(t) = U_0 \cdot e^{-\frac{t}{RC}}
+ \end{math}}
+
+ Wir ändern das Programm \gitfile{hp}{2023ws/20231123}{loesung-1a.c} so ab,
+ daß es zusätzlich zur Zeit und zur simulierten Spannung
+ die exakte Spannung $U_0 \cdot e^{-\frac{t}{RC}}$
+ gemäß der theoretischen Entladekurve ausgibt (\gitfile{hp}{2023ws/20231123}{loesung-1c.c}),
+
+ Da dieses Programm die Exponentialfunktion verwendet, müssen wir nun
+ beim Compilieren zusätzlich \lstinline[style=cmd]{-lm}\hspace{1pt}
+ für das Einbinden der Mathematik-Bibliothek angeben.
+
+ Der erweiterten Tabelle können wir entnehmen,
+ daß die durch die Simulation berechnete Spannung
+ mit der Spannung $U_0 \cdot e^{-\frac{t}{RC}}$
+ gemäß der theoretischen Entladekurve
+ bis auf wenige Prozent übereinstimmt.
+ Dies ist für viele praktische Anwendungen ausreichend,
+ wenn auch nicht für Präzisionsmessungen.
+
+ Wenn Sie die Ausgabe des Programms, z.\,B.\ mit
+ \lstinline[style=cmd]{./loesung-1c > loesung-1c.dat},
+ in einer Datei \gitfile{hp}{2023ws/20231123}{loesung-1c.dat} speichern,
+ können Sie sich die beiden Kurven graphisch darstellen lassen,
+ z.\,B.\ mit \file{gnuplot} und dem folgenden Befehl:
+ \begin{lstlisting}[style=cmd,gobble=8]
+ plot "loesung-1c.dat" using 1:2 with lines title "Simulation",
+ "loesung-1c.dat" using 1:3 with lines title "Theorie"
+ \end{lstlisting}
+ \vspace*{-\bigskipamount}
+ \begin{center}
+ \input{loesung-1c.tikz}
+ \end{center}
+ Der Unterschied zwischen der simulierten und der theoretischen Entladungskurve
+ ist mit bloßem Auge nicht sichtbar.
+ \end{itemize}
+
+ \exercise{Personen-Datenbank}
+
+ Wir betrachten das folgende Programm (\gitfile{hp}{2023ws/20231123}{aufgabe-2.c}):
+ \begin{lstlisting}
+ #include <stdio.h>
+ #include <string.h>
+
+ typedef struct
+ {
+ char first_name[10];
+ char family_name[20];
+ char day, month;
+ int year;
+ } person;
+
+ int main (void)
+ {
+ person sls;
+ sls.day = 26;
+ sls.month = 7;
+ sls.year = 1951;
+ strcpy (sls.first_name, "Sabine");
+ strcpy (sls.family_name, "Leutheusser-Schnarrenberger");
+ printf ("%s %s wurde am %d.%d.%d geboren.\n",
+ sls.first_name, sls.family_name, sls.day, sls.month, sls.year);
+ return 0;
+ }
+ \end{lstlisting}
+
+ Die Standard-Funktion \lstinline{strcpy()} bewirkt ein Kopieren eines Strings
+ von rechts nach links, hier also z.\,B.\ die Zuweisung der String-Konstanten
+ \lstinline{"Sabine"} an die String-Variable \lstinline{sls.first_name[]}.
+
+ Das Programm wird für einen 32-Bit-Rechner compiliert und ausgeführt.\\
+ (Die \lstinline[style=cmd]{gcc}-Option \lstinline[style=cmd]{-m32} sorgt dafür,
+ daß \lstinline[style=cmd]{gcc} Code für einen 32-Bit-Prozessor erzeugt.)
+
+ \begin{lstlisting}[style=terminal]
+ $ ¡gcc -Wall -O -m32 aufgabe-2.c -o aufgabe-2¿
+ $ ¡./aufgabe-2¿
+ Sabine Leutheusser-Schnarrenberger wurde am 110.98.1701278309 geboren.
+ Speicherzugriffsfehler
+ \end{lstlisting}
+
+ \begin{enumerate}[\quad(a)]
+ \item
+ Erklären Sie die Ausgabe des Programms einschließlich der Zahlenwerte.
+ \points{4}
+ \item
+ Welche Endianness hat der verwendete Rechner?
+ Begründen Sie Ihre Antwort.
+ \points{1}
+ \item
+ Wie sähe die Ausgabe auf einem Rechner mit entgegengesetzter Endianness aus?
+ \points{2}
+ \item
+ Erklären Sie den Speicherzugriffsfehler.
+ (Es kann sein, daß sich der Fehler auf Ihrem Rechner nicht bemerkbar macht.
+ Er ist aber trotzdem vorhanden.)
+ \points{2}
+ \end{enumerate}
+
+ \goodbreak
+
+ \solution
+
+ \begin{enumerate}[\quad(a)]
+ \item
+ \textbf{Erklären Sie die Ausgabe des Programms einschließlich der Zahlenwerte.}
+
+ Der String \lstinline{"Leutheusser-Schnarrenberger"}
+ hat 27 Zeichen und daher mehr als die in der Variablen
+ \lstinline{sls.family_name} vorgesehenen 20 Zeichen.
+ Das \lstinline{"nberger"} paßt nicht mehr in die String-Variable.
+
+ Die Zuweisung \lstinline{strcpy (sls.family_name, "Leutheusser-Schnarrenberger")}
+ überschreibt daher 8 Speicherzellen außerhalb der String-Variablen
+ \lstinline{sls.family_name} mit dem String \lstinline{"nberger"}
+ (7 Buchstaben zzgl.\ String-Ende-Symbol) -- und damit insbesondere die Variablen
+ \lstinline{sls.day}, \lstinline{sls.month} und \lstinline{sls.year}.
+
+ Die überschriebenen Speicherzellen sehen demnach folgendermaßen aus:
+ \begin{center}
+ \begin{picture}(8,1.5)(0,-0.5)
+ \put(0,0){\line(1,0){8}}
+ \put(0,1){\line(1,0){8}}
+ \multiput(0,0)(1,0){9}{\line(0,1){1}}
+ \put(0.4,0.38){\lstinline{n}}
+ \put(1.4,0.38){\lstinline{b}}
+ \put(2.4,0.38){\lstinline{e}}
+ \put(3.4,0.38){\lstinline{r}}
+ \put(4.4,0.38){\lstinline{g}}
+ \put(5.4,0.38){\lstinline{e}}
+ \put(6.4,0.38){\lstinline{r}}
+ \put(7.5,0.5){\makebox(0,0){\lstinline{0x00}}}
+ \put(0.5,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{0.95cm}{0pt}}_{\mbox{\lstinline{day}}}$}}
+ \put(1.5,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{0.95cm}{0pt}}_{\mbox{\lstinline{month}}}$}}
+ \put(4.0,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{3.95cm}{0pt}}_{\mbox{\lstinline{year}}}$}}
+ \put(7.0,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{1.95cm}{0pt}}_{\mbox{?}}$}}
+ \end{picture}
+ \end{center}
+ ("`?"' steht für zwei Speicherzellen, von denen wir nicht wissen,
+ wofür sie genutzt werden.)
+
+ Wenn wir die ASCII-Zeichen in Hexadezimalzahlen umrechnen, entspricht dies:
+ \begin{center}
+ \begin{picture}(7,1.5)(0,-0.5)
+ \put(0,0){\line(1,0){8}}
+ \put(0,1){\line(1,0){8}}
+ \multiput(0,0)(1,0){9}{\line(0,1){1}}
+ \put(0.5,0.5){\makebox(0,0){\lstinline{0x6e}}}
+ \put(1.5,0.5){\makebox(0,0){\lstinline{0x62}}}
+ \put(2.5,0.5){\makebox(0,0){\lstinline{0x65}}}
+ \put(3.5,0.5){\makebox(0,0){\lstinline{0x72}}}
+ \put(4.5,0.5){\makebox(0,0){\lstinline{0x67}}}
+ \put(5.5,0.5){\makebox(0,0){\lstinline{0x65}}}
+ \put(6.5,0.5){\makebox(0,0){\lstinline{0x72}}}
+ \put(7.5,0.5){\makebox(0,0){\lstinline{0x00}}}
+ \put(0.5,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{0.95cm}{0pt}}_{\mbox{\lstinline{day}}}$}}
+ \put(1.5,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{0.95cm}{0pt}}_{\mbox{\lstinline{month}}}$}}
+ \put(4.0,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{3.95cm}{0pt}}_{\mbox{\lstinline{year}}}$}}
+ \put(7.0,-0.1){\makebox(0,0)[t]{$\underbrace{\rule{1.95cm}{0pt}}_{\mbox{?}}$}}
+ \end{picture}
+ \end{center}
+ Dies entspricht bereits genau den Werten \lstinline{110} und \lstinline{98}
+ für die Variablen \lstinline{sls.day} bzw.\ \lstinline{sls.month}.
+
+ Für die Variable \lstinline{sls.year} müssen wir ihre vier Speicherzellen
+ unter der Berücksichtigung der Endianness des Rechners zusammenziehen.
+ Für Big-Endian ergibt dies \lstinline{0x65726765 == 1701996389}.
+ Für Little-Endian ergibt sich der Wert \lstinline{0x65677265 == 1701278309},
+ der auch in der Ausgabe des Programms auftaucht.
+
+ \item
+ \textbf{Welche Endianness hat der verwendete Rechner?
+ Begründen Sie Ihre Antwort.}
+
+ Wie in (a) begründet, ergibt sich die Ausgabe von
+ \lstinline[style=terminal]{1701278309} für das Jahr
+ aus dem Speicherformat Little-Endian.
+
+ \item
+ \textbf{Wie sähe die Ausgabe auf einem Rechner mit entgegengesetzter Endianness aus?}
+
+ Wie in (a) begründet, ergäbe sich aus dem Speicherformat Big-Endian
+ die Ausgabe von \lstinline[style=terminal]{1701996389} für das Jahr.
+
+ \item
+ \textbf{Erklären Sie den Speicherzugriffsfehler.
+ (Es kann sein, daß sich der Fehler auf Ihrem Rechner nicht bemerkbar macht.
+ Er ist aber trotzdem vorhanden.)}
+
+ Die zwei in (a) mit "`?"' bezeichneten Speicherzellen
+ wurden ebenfalls überschrieben.
+ Dies ist in der Ausgabe zunächst nicht sichtbar,
+ bewirkt aber später den Speicherzugriffsfehler.
+
+ (Tatsächlich handelt es sich bei den überschriebenen Speicherzellen
+ um einen Teil der Rücksprungadresse, die \lstinline{main()} verwendet,
+ um mit \lstinline{return 0} an das Betriebssystem zurückzugeben.)
+
+ \end{enumerate}
+
+ \textbf{Hinweis 1:}
+ Um auf einen solchen Lösungsweg zu kommen, wird empfohlen,
+ "`geheimnisvolle"' Zahlen nach hexadezimal umzurechnen
+ und in Speicherzellen (Zweiergruppen von Hex-Ziffern) zu zerlegen.
+ Oft erkennt man dann direkt ASCII-Zeichen:
+ Großbuchstaben beginnen mit der Hex-Ziffer \lstinline{4} oder \lstinline{5},
+ Kleinbuchstaben mit \lstinline{6} oder \lstinline{7}.
+
+ \textbf{Hinweis 2:}
+ Um derartige Programmierfehler in der Praxis von vorneherein zu vermeiden,
+ wird empfohlen, anstelle von \lstinline{strcpy()}
+ grundsätzlich die Funktion \lstinline{strncpy()} zu verwenden.
+ Diese erwartet einen zusätzlichen Parameter,
+ der die maximal zulässige Länge des Strings enthält.
+ Ohne einen derartigen expliziten Parameter kann die Funktion nicht wissen,
+ wie lang die Variable ist, in der der String gespeichert werden soll.
+
+ \exercise{Hexdumps}
+
+ Das folgende Programm (\gitfile{hp}{2023ws/20231123}{aufgabe-4.c}) liest
+ einen String ein und gibt die ASCII-Werte der Buchstaben hexadezimal aus.
+ (Anders als z.\,B.\ \lstinline{scanf()}
+ akzeptiert die Funktion \lstinline{fgets()} zum Lesen von Strings auch Leerzeichen,
+ und sie vermeidet Pufferüberläufe.)
+ \begin{lstlisting}[style=numbered]
+ #include <stdio.h>
+
+ int main (void)
+ {
+ char buffer[100];
+ fgets (buffer, 100, stdin);
+ for (char *p = buffer; *p; p++)
+ printf ("%02x", *p);
+ printf ("\n");
+ }
+ \end{lstlisting}
+ Beispiel: Bei der Eingabe von \lstinline[style=cmd]{Dies ist ein Test.}
+ erscheint die Ausgabe\\
+ \lstinline[style=terminal]{44696573206973742065696e20546573742e0a}.
+
+ Schreiben Sie ein Programm, das diese Umwandlung in umgekehrter Richtung vornimmt,
+ also z.\,B.\ bei Eingabe von \lstinline[style=cmd]{44696573206973742065696e20546573742e0a}
+ wieder \lstinline[style=terminal]{Dies ist ein Test.} ausgibt.
+
+ \points{6}
+
+ Hinweis für die Klausur:
+ Abgabe in digitaler Form ist erwünscht, aber nicht zwingend.
+
+ \solution
+
+ Siehe \gitfile{hp}{2023ws/20231123}{loesung-3.c}.
+
+ Das Programm macht mehrfach davon Gebrauch,
+ daß in C Zeichen und Zahlen äquivalent sind.
+ Wenn z.\,B.\ die \lstinline{char}-Variable \lstinline{c}
+ den Wert \lstinline{'3'} (Ziffer 3) enthält,
+ dann hat der Ausdruck \lstinline{c - '0'} den Wert \lstinline{3} (Zahlenwert 3).
+ Hierfür ist es insbesondere nicht nötig, vorauszusetzen,
+ daß wir den ASCII-Zeichensatz verwenden und \lstinline{'0'}
+ den Wert \lstinline{48} hat.
+
+ Bei Eingabe von \lstinline[style=cmd]{44696573206973742065696e20546573742e0a}
+ gibt das Programm zusätzlich eine Leerzeile aus.
+ Die liegt daran, daß das \lstinline[style=cmd]{0a} am Ende
+ bereits eine Zeilenschaltung enthält und das Programm mit
+ \lstinline{printf ("\n")} eine zusätzliche Zeilenschaltung ausgibt.
+
+\end{document}
diff --git a/20231123/hp-uebung-20231123.pdf b/20231123/hp-uebung-20231123.pdf
new file mode 100644
index 0000000000000000000000000000000000000000..6beb192a0a95fd8864f56b5dac454eb0f06a3e18
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diff --git a/20231123/hp-uebung-20231123.tex b/20231123/hp-uebung-20231123.tex
new file mode 100644
index 0000000000000000000000000000000000000000..5ec4c23a85d0cc054e38ed08b2d28df9af5104cc
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+++ b/20231123/hp-uebung-20231123.tex
@@ -0,0 +1,188 @@
+% hp-uebung-20231123.pdf - Exercises on Low-Level Programming / Applied Computer Sciences
+% Copyright (C) 2013, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023 Peter Gerwinski
+%
+% This document is free software: you can redistribute it and/or
+% modify it either under the terms of the Creative Commons
+% Attribution-ShareAlike 3.0 License, or under the terms of the
+% GNU General Public License as published by the Free Software
+% Foundation, either version 3 of the License, or (at your option)
+% any later version.
+%
+% This document is distributed in the hope that it will be useful,
+% but WITHOUT ANY WARRANTY; without even the implied warranty of
+% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
+% GNU General Public License for more details.
+%
+% You should have received a copy of the GNU General Public License
+% along with this document. If not, see <http://www.gnu.org/licenses/>.
+%
+% You should have received a copy of the Creative Commons
+% Attribution-ShareAlike 3.0 Unported License along with this
+% document. If not, see <http://creativecommons.org/licenses/>.
+
+% README: Kondensator, Personen-Datenbank, Hexdumps
+
+\documentclass[a4paper]{article}
+
+\usepackage{pgscript}
+\usepackage{gensymb}
+
+\newcommand{\ItwoC}{I\raisebox{0.5ex}{\footnotesize 2}C}
+\newcommand{\ITWOC}{I\raisebox{0.5ex}{\normalsize 2}C}
+
+\begin{document}
+
+ \thispagestyle{empty}
+
+ \section*{Hardwarenahe Programmierung\\
+ Übungsaufgaben -- 23.\ November 2023}
+
+ Diese Übung enthält Punkteangaben wie in einer Klausur.
+ Um zu "`bestehen"', müssen Sie innerhalb von 80 Minuten
+ unter Verwendung ausschließlich zugelassener Hilfsmittel
+ 14 Punkte (von insgesamt \totalpoints) erreichen.
+
+ \exercise{Kondensator}
+
+ Ein Kondensator der Kapazität $C = 100\,\mu{\rm F}$
+ ist auf die Spannung $U_0 = 5\,{\rm V}$ aufgeladen
+ und wird über einen Widerstand $R = 33\,{\rm k}\Omega$ entladen.
+
+ \begin{enumerate}[(a)]
+ \item
+ Schreiben Sie ein C-Programm, das
+ den zeitlichen Spannungsverlauf in einer Tabelle darstellt.
+ \points{5}
+ \item
+ Schreiben Sie ein C-Programm, das ermittelt,
+ wie lange es dauert, bis die Spannung unter $0.1\,{\rm V}$ gefallen ist.
+ \points{4}
+ \item
+ Vergleichen Sie die berechneten Werte mit der exakten theoretischen Entladekurve:
+ \begin{math}
+ U(t) = U_0 \cdot e^{-\frac{t}{RC}}
+ \end{math}\\
+ \points{3}
+ \end{enumerate}
+
+ Hinweise:
+ \begin{itemize}
+ \item
+ Für die Simulation zerlegen wir den Entladevorgang in kurze Zeitintervalle $dt$.
+ Innerhalb jedes Zeitintervalls betrachten wir den Strom $I$ als konstant
+ und berechnen, wieviel Ladung $Q$ innerhalb des Zeitintervalls
+ aus dem Kondensator herausfließt.
+ Aus der neuen Ladung berechnen wir die Spannung am Ende des Zeitintervalls.
+ \item
+ Für den Vergleich mit der exakten theoretischen Entladekurve
+ benötigen Sie die Exponentialfunktion \lstinline{exp()}.
+ Diese finden Sie in der Mathematik-Bibliothek:
+ \lstinline{#include <math.h>} im Quelltext,
+ beim \lstinline[style=cmd]{gcc}-Aufruf \lstinline[style=cmd]{-lm} mit angeben.
+ \item
+ $Q = C \cdot U$,\quad $U = R \cdot I$,\quad $I = \frac{dQ}{dt}$
+ \end{itemize}
+
+ \exercise{Personen-Datenbank}
+
+ Wir betrachten das folgende Programm (\gitfile{hp}{2023ws/20231123}{aufgabe-2.c}):
+ \begin{lstlisting}
+ #include <stdio.h>
+ #include <string.h>
+
+ typedef struct
+ {
+ char first_name[10];
+ char family_name[20];
+ char day, month;
+ int year;
+ } person;
+
+ int main (void)
+ {
+ person sls;
+ sls.day = 26;
+ sls.month = 7;
+ sls.year = 1951;
+ strcpy (sls.first_name, "Sabine");
+ strcpy (sls.family_name, "Leutheusser-Schnarrenberger");
+ printf ("%s %s wurde am %d.%d.%d geboren.\n",
+ sls.first_name, sls.family_name, sls.day, sls.month, sls.year);
+ return 0;
+ }
+ \end{lstlisting}
+
+ Die Standard-Funktion \lstinline{strcpy()} bewirkt ein Kopieren eines Strings
+ von rechts nach links, hier also z.\,B.\ die Zuweisung der String-Konstanten
+ \lstinline{"Sabine"} an die String-Variable \lstinline{sls.first_name[]}.
+
+ Das Programm wird für einen 32-Bit-Rechner compiliert und ausgeführt.\\
+ (Die \lstinline[style=cmd]{gcc}-Option \lstinline[style=cmd]{-m32} sorgt dafür,
+ daß \lstinline[style=cmd]{gcc} Code für einen 32-Bit-Prozessor erzeugt.)
+
+ \begin{lstlisting}[style=terminal]
+ $ ¡gcc -Wall -O -m32 aufgabe-2.c -o aufgabe-2¿
+ $ ¡./aufgabe-2¿
+ Sabine Leutheusser-Schnarrenberger wurde am 110.98.1701278309 geboren.
+ Speicherzugriffsfehler
+ \end{lstlisting}
+
+ \begin{enumerate}[\quad(a)]
+ \item
+ Erklären Sie die Ausgabe des Programms einschließlich der Zahlenwerte.
+ \points{4}
+ \item
+ Welche Endianness hat der verwendete Rechner?
+ Begründen Sie Ihre Antwort.
+ \points{1}
+ \item
+ Wie sähe die Ausgabe auf einem Rechner mit entgegengesetzter Endianness aus?
+ \points{2}
+ \item
+ Erklären Sie den Speicherzugriffsfehler.
+ (Es kann sein, daß sich der Fehler auf Ihrem Rechner nicht bemerkbar macht.
+ Er ist aber trotzdem vorhanden.)
+ \points{2}
+ \end{enumerate}
+
+ \exercise{Hexdumps}
+
+ Das folgende Programm (\gitfile{hp}{2023ws/20231123}{aufgabe-4.c}) liest
+ einen String ein und gibt die ASCII-Werte der Buchstaben hexadezimal aus.
+ (Anders als z.\,B.\ \lstinline{scanf()}
+ akzeptiert die Funktion \lstinline{fgets()} zum Lesen von Strings auch Leerzeichen,
+ und sie vermeidet Pufferüberläufe.)
+ \begin{lstlisting}[style=numbered]
+ #include <stdio.h>
+
+ int main (void)
+ {
+ char buffer[100];
+ fgets (buffer, 100, stdin);
+ for (char *p = buffer; *p; p++)
+ printf ("%02x", *p);
+ printf ("\n");
+ }
+ \end{lstlisting}
+ Beispiel: Bei der Eingabe von \lstinline[style=cmd]{Dies ist ein Test.}
+ erscheint die Ausgabe\\
+ \lstinline[style=terminal]{44696573206973742065696e20546573742e0a}.
+
+ Schreiben Sie ein Programm, das diese Umwandlung in umgekehrter Richtung vornimmt,
+ also z.\,B.\ bei Eingabe von \lstinline[style=cmd]{44696573206973742065696e20546573742e0a}
+ wieder \lstinline[style=terminal]{Dies ist ein Test.} ausgibt.
+
+ \points{6}
+
+ Hinweis für die Klausur:
+ Abgabe in digitaler Form ist erwünscht, aber nicht zwingend.
+
+ \begin{flushright}
+ \textit{Viel Erfolg!}
+ \end{flushright}
+
+ \makeatletter
+ \immediate\write\@mainaux{\string\gdef\string\totalpoints{\arabic{points}}}
+ \makeatother
+
+\end{document}
diff --git a/20231123/loesung-1a.c b/20231123/loesung-1a.c
new file mode 100644
index 0000000000000000000000000000000000000000..f355706a6ff6f95bd72897123a985fb2b7ead7c3
--- /dev/null
+++ b/20231123/loesung-1a.c
@@ -0,0 +1,22 @@
+#include <stdio.h>
+#include <math.h>
+
+int main (void)
+{
+ double C = 0.0001;
+ double U0 = 5.0;
+ double U = U0;
+ double R = 33000.0;
+ double t = 0.0;
+ double dt = 0.01;
+ double Q = C * U;
+ while (U > 0.09)
+ {
+ printf ("%10.3lf%15.8lf\n", t, U);
+ double I = U / R;
+ Q -= I * dt;
+ U = Q / C;
+ t += dt;
+ }
+ return 0;
+}
diff --git a/20231123/loesung-1b.c b/20231123/loesung-1b.c
new file mode 100644
index 0000000000000000000000000000000000000000..f5fc22022ef10a1b986bbd11fba8298a7547792d
--- /dev/null
+++ b/20231123/loesung-1b.c
@@ -0,0 +1,22 @@
+#include <stdio.h>
+#include <math.h>
+
+int main (void)
+{
+ double C = 0.0001;
+ double U0 = 5.0;
+ double U = U0;
+ double R = 33000.0;
+ double t = 0.0;
+ double dt = 0.01;
+ double Q = C * U;
+ while (U >= 0.1)
+ {
+ double I = U / R;
+ Q -= I * dt;
+ U = Q / C;
+ t += dt;
+ }
+ printf ("%10.3lf%15.8lf\n", t, U);
+ return 0;
+}
diff --git a/20231123/loesung-1c.c b/20231123/loesung-1c.c
new file mode 100644
index 0000000000000000000000000000000000000000..71802e036dd56534470028b61f646d297d1d8840
--- /dev/null
+++ b/20231123/loesung-1c.c
@@ -0,0 +1,22 @@
+#include <stdio.h>
+#include <math.h>
+
+int main (void)
+{
+ double C = 0.0001;
+ double U0 = 5.0;
+ double U = U0;
+ double R = 33000.0;
+ double t = 0.0;
+ double dt = 0.01;
+ double Q = C * U;
+ while (U > 0.09)
+ {
+ printf ("%10.3lf%15.8lf%15.8lf\n", t, U, U0 * exp (-t / (R * C)));
+ double I = U / R;
+ Q -= I * dt;
+ U = Q / C;
+ t += dt;
+ }
+ return 0;
+}
diff --git a/20231123/loesung-1c.dat b/20231123/loesung-1c.dat
new file mode 100644
index 0000000000000000000000000000000000000000..4df706295bde25a74e025c5335f43617d93129a8
--- /dev/null
+++ b/20231123/loesung-1c.dat
@@ -0,0 +1,1324 @@
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+ --(8.877,0.983)--(8.885,0.983)--(8.893,0.981)--(8.901,0.979)--(8.909,0.979)--(8.916,0.978)%
+ --(8.924,0.976)--(8.932,0.975)--(8.940,0.975)--(8.948,0.973)--(8.955,0.972)--(8.963,0.972)%
+ --(8.971,0.970)--(8.979,0.969)--(8.987,0.969)--(8.995,0.967)--(9.002,0.965)--(9.010,0.965)%
+ --(9.018,0.964)--(9.026,0.962)--(9.034,0.962)--(9.041,0.961)--(9.049,0.959)--(9.057,0.959)%
+ --(9.065,0.958)--(9.073,0.956)--(9.080,0.956)--(9.088,0.955)--(9.096,0.953)--(9.104,0.953)%
+ --(9.112,0.951)--(9.120,0.950)--(9.127,0.950)--(9.135,0.948)--(9.143,0.947)--(9.151,0.947)%
+ --(9.159,0.945)--(9.166,0.944)--(9.174,0.944)--(9.182,0.942)--(9.190,0.941)--(9.198,0.941)%
+ --(9.205,0.939)--(9.213,0.939)--(9.221,0.937)--(9.229,0.936)--(9.237,0.936)--(9.244,0.934)%
+ --(9.252,0.933)--(9.260,0.933)--(9.268,0.931)--(9.276,0.931)--(9.284,0.930)--(9.291,0.928)%
+ --(9.299,0.928)--(9.307,0.927)--(9.315,0.927)--(9.323,0.925)--(9.330,0.923)--(9.338,0.923)%
+ --(9.346,0.922)--(9.354,0.922)--(9.362,0.920)--(9.369,0.919)--(9.377,0.919)--(9.385,0.917)%
+ --(9.393,0.917)--(9.401,0.916)--(9.409,0.914)--(9.416,0.914)--(9.424,0.913)--(9.432,0.913)%
+ --(9.440,0.911)--(9.448,0.911)--(9.455,0.910)--(9.463,0.908)--(9.471,0.908)--(9.479,0.906)%
+ --(9.487,0.906)--(9.494,0.905)--(9.502,0.905)--(9.510,0.903)--(9.518,0.902)--(9.526,0.902)%
+ --(9.533,0.900)--(9.541,0.900)--(9.549,0.899)--(9.557,0.899)--(9.565,0.897)--(9.573,0.897)%
+ --(9.580,0.896)--(9.588,0.894)--(9.596,0.894)--(9.604,0.892)--(9.612,0.892)--(9.619,0.891)%
+ --(9.627,0.891)--(9.635,0.889)--(9.643,0.889)--(9.651,0.888)--(9.658,0.888)--(9.666,0.886)%
+ --(9.674,0.886)--(9.682,0.885)--(9.690,0.885)--(9.698,0.883)--(9.705,0.882)--(9.713,0.882)%
+ --(9.721,0.880)--(9.729,0.880)--(9.737,0.878)--(9.744,0.878)--(9.752,0.877)--(9.760,0.877)%
+ --(9.768,0.875)--(9.776,0.875)--(9.783,0.874)--(9.791,0.874)--(9.799,0.872)--(9.807,0.872)%
+ --(9.815,0.871)--(9.822,0.871)--(9.830,0.869)--(9.838,0.869)--(9.846,0.868)--(9.854,0.868)%
+ --(9.862,0.866)--(9.869,0.866)--(9.877,0.864)--(9.885,0.864)--(9.893,0.863)--(9.901,0.863)%
+ --(9.908,0.861)--(9.916,0.861)--(9.924,0.861)--(9.932,0.860)--(9.940,0.860)--(9.947,0.858)%
+ --(9.955,0.858)--(9.963,0.857)--(9.971,0.857)--(9.979,0.855)--(9.987,0.855)--(9.994,0.854)%
+ --(10.002,0.854)--(10.010,0.852)--(10.018,0.852)--(10.026,0.851)--(10.033,0.851)--(10.041,0.851)%
+ --(10.049,0.849)--(10.057,0.849)--(10.065,0.847)--(10.072,0.847)--(10.080,0.846)--(10.088,0.846)%
+ --(10.096,0.844)--(10.104,0.844)--(10.111,0.843)--(10.119,0.843)--(10.127,0.843)--(10.135,0.841)%
+ --(10.143,0.841)--(10.151,0.840)--(10.158,0.840)--(10.166,0.838)--(10.174,0.838)--(10.182,0.838)%
+ --(10.190,0.837)--(10.197,0.837)--(10.205,0.835)--(10.213,0.835)--(10.221,0.833)--(10.229,0.833)%
+ --(10.236,0.833)--(10.244,0.832)--(10.252,0.832)--(10.260,0.830)--(10.268,0.830)--(10.276,0.829)%
+ --(10.283,0.829)--(10.291,0.829)--(10.299,0.827)--(10.307,0.827)--(10.315,0.826)--(10.322,0.826)%
+ --(10.330,0.826)--(10.338,0.824)--(10.346,0.824)--(10.354,0.823)--(10.361,0.823)--(10.369,0.823)%
+ --(10.377,0.821)--(10.385,0.821)--(10.393,0.819)--(10.400,0.819)--(10.408,0.819)--(10.416,0.818)%
+ --(10.424,0.818)--(10.432,0.816)--(10.440,0.816)--(10.447,0.816)--(10.455,0.815)--(10.463,0.815)%
+ --(10.471,0.813)--(10.479,0.813)--(10.486,0.813)--(10.494,0.812)--(10.502,0.812)--(10.510,0.812)%
+ --(10.518,0.810)--(10.525,0.810)--(10.533,0.809)--(10.541,0.809)--(10.549,0.809)--(10.557,0.807)%
+ --(10.565,0.807)--(10.572,0.805)--(10.580,0.805)--(10.588,0.805)--(10.596,0.804)--(10.604,0.804)%
+ --(10.611,0.804)--(10.619,0.802)--(10.627,0.802)--(10.635,0.802)--(10.643,0.801)--(10.650,0.801)%
+ --(10.658,0.799)--(10.666,0.799)--(10.674,0.799)--(10.682,0.798)--(10.689,0.798)--(10.697,0.798)%
+ --(10.705,0.796)--(10.713,0.796)--(10.721,0.796)--(10.729,0.795)--(10.736,0.795)--(10.744,0.795)%
+ --(10.752,0.793)--(10.760,0.793)--(10.768,0.793)--(10.775,0.791)--(10.783,0.791)--(10.791,0.791)%
+ --(10.799,0.790)--(10.807,0.790)--(10.814,0.790)--(10.822,0.788)--(10.830,0.788)--(10.838,0.788)%
+ --(10.846,0.787)--(10.854,0.787)--(10.861,0.787)--(10.869,0.785)--(10.877,0.785)--(10.885,0.785)%
+ --(10.893,0.784)--(10.900,0.784)--(10.908,0.784)--(10.916,0.782)--(10.924,0.782)--(10.932,0.782)%
+ --(10.939,0.781)--(10.947,0.781)--(10.955,0.781)--(10.963,0.779)--(10.971,0.779)--(10.978,0.779)%
+ --(10.986,0.778)--(10.994,0.778)--(11.002,0.778)--(11.010,0.776)--(11.018,0.776)--(11.025,0.776)%
+ --(11.033,0.774)--(11.041,0.774)--(11.049,0.774)--(11.057,0.774)--(11.064,0.773)--(11.072,0.773)%
+ --(11.080,0.773)--(11.088,0.771)--(11.096,0.771)--(11.103,0.771)--(11.111,0.770)--(11.119,0.770)%
+ --(11.127,0.770)--(11.135,0.768)--(11.142,0.768)--(11.150,0.768)--(11.158,0.768)--(11.166,0.767)%
+ --(11.174,0.767)--(11.182,0.767)--(11.189,0.765)--(11.197,0.765)--(11.205,0.765)--(11.213,0.765)%
+ --(11.221,0.764)--(11.228,0.764)--(11.236,0.764)--(11.244,0.762)--(11.252,0.762)--(11.260,0.762)%
+ --(11.267,0.762)--(11.275,0.760)--(11.283,0.760)--(11.291,0.760)--(11.299,0.759)--(11.307,0.759)%
+ --(11.314,0.759)--(11.322,0.759)--(11.330,0.757)--(11.338,0.757)--(11.346,0.757);
+\gpcolor{color=gp lt color border}
+\draw[gp path] (1.012,8.381)--(1.012,0.616)--(11.947,0.616)--(11.947,8.381)--cycle;
+%% coordinates of the plot area
+\gpdefrectangularnode{gp plot 1}{\pgfpoint{1.012cm}{0.616cm}}{\pgfpoint{11.947cm}{8.381cm}}
+\end{tikzpicture}
+%% gnuplot variables
diff --git a/20231123/loesung-3.c b/20231123/loesung-3.c
new file mode 100644
index 0000000000000000000000000000000000000000..872058ac9fecdcb59ac1104ca67841cb3dc974a9
--- /dev/null
+++ b/20231123/loesung-3.c
@@ -0,0 +1,28 @@
+#include <stdio.h>
+
+int read_hex (char c)
+{
+ if (c >= '0' && c <= '9')
+ return c - '0';
+ else if (c >= 'A' && c <= 'F')
+ return c - 'A' + 10;
+ else if (c >= 'a' && c <= 'f')
+ return c - 'a' + 10;
+ else
+ {
+ fprintf (stderr, "invalid hex digit '%c'\n", c);
+ return 0;
+ }
+}
+
+int main (void)
+{
+ char buffer[100];
+ fgets (buffer, 100, stdin);
+ for (char *p = buffer; p[0] && p[1]; p += 2)
+ {
+ char c = 16 * read_hex (p[0]) + read_hex (p[1]);
+ printf ("%c", c);
+ }
+ printf ("\n");
+}
diff --git a/20231123/logo-hochschule-bochum-cvh-text-v2.pdf b/20231123/logo-hochschule-bochum-cvh-text-v2.pdf
new file mode 120000
index 0000000000000000000000000000000000000000..4aa99b8f81061aca6dcaf43eed2d9efef40555f8
--- /dev/null
+++ b/20231123/logo-hochschule-bochum-cvh-text-v2.pdf
@@ -0,0 +1 @@
+../common/logo-hochschule-bochum-cvh-text-v2.pdf
\ No newline at end of file
diff --git a/20231123/logo-hochschule-bochum.pdf b/20231123/logo-hochschule-bochum.pdf
new file mode 120000
index 0000000000000000000000000000000000000000..b6b9491e370e499c9276918182cdb82cb311bcd1
--- /dev/null
+++ b/20231123/logo-hochschule-bochum.pdf
@@ -0,0 +1 @@
+../common/logo-hochschule-bochum.pdf
\ No newline at end of file
diff --git a/20231123/pendulum.pdf b/20231123/pendulum.pdf
new file mode 120000
index 0000000000000000000000000000000000000000..7d1d87305cdb8840a248ff2207538d758464f452
--- /dev/null
+++ b/20231123/pendulum.pdf
@@ -0,0 +1 @@
+../common/pendulum.pdf
\ No newline at end of file
diff --git a/20231123/pgscript.sty b/20231123/pgscript.sty
new file mode 120000
index 0000000000000000000000000000000000000000..95c888478c99ea7fda0fd11ccf669ae91be7178b
--- /dev/null
+++ b/20231123/pgscript.sty
@@ -0,0 +1 @@
+../common/pgscript.sty
\ No newline at end of file
diff --git a/20231123/pgslides.sty b/20231123/pgslides.sty
new file mode 120000
index 0000000000000000000000000000000000000000..5be1416f4216f076aa268901f52a15d775e43f64
--- /dev/null
+++ b/20231123/pgslides.sty
@@ -0,0 +1 @@
+../common/pgslides.sty
\ No newline at end of file
diff --git a/README.md b/README.md
index 492885ffb1e1a03c045faef9bc843c1335247f98..e381558054eaaca6545c8ce2c2028d3b9cd25a55 100644
--- a/README.md
+++ b/README.md
@@ -23,7 +23,8 @@ Vortragsfolien und Beispiele:
* [26.10.2023: Einführung in C: String-Operationen; Bibliotheken](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231026/hp-20231026.pdf) [**(Beispiele)**](https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231026/)
* [02.11.2023: Bibliotheken](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231102/hp-20231102.pdf) [**(Beispiele)**](https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231102/)
* [09.11.2023: Hardwarenahe Programmierung](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231109/hp-20231109.pdf) [**(Beispiele)**](https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231109/)
- * [16.11.2023: Byte-Reihenfolge, Darstellung negativer Zahlen, Darstellung von Gleitkommazahlen, Speicherausrichtung](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231116/hp-20231116.pdf) [**(Beispiele)**](https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231116/)
+ * [16.11.2023: Byte-Reihenfolge, Darstellung negativer Zahlen, Darstellung von Gleitkommazahlen](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231116/hp-20231116.pdf) [**(Beispiele)**](https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231116/)
+ * [23.11.2023: Speicherausrichtung, Algorithmen: Differentialgleichungen](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231123/hp-20231123.pdf) [**(Beispiele)**](https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231123/)
* [alle in 1 Datei](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/hp-slides-2023ws.pdf)
Übungsaufgaben:
@@ -35,6 +36,7 @@ Vortragsfolien und Beispiele:
* [02.11.2023: Zahlensysteme, Ausgabe von Hexadezimalzahlen, Einfügen in Strings](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231102/hp-uebung-20231102.pdf)
* [09.11.2023: Text-Grafik-Bibliothek, Mikrocontroller, LED-Blinkmuster](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231109/hp-uebung-20231109.pdf)
* [16.11.2023: Trickprogrammierung, Thermometer-Baustein an I²C-Bus, Speicherformate von Zahlen](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231116/hp-uebung-20231116.pdf)
+ * [23.11.2023: Kondensator, Personen-Datenbank, Hexdumps](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231123/hp-uebung-20231123.pdf)
Musterlösungen:
---------------
@@ -44,6 +46,7 @@ Musterlösungen:
* [02.11.2023: Zahlensysteme, Ausgabe von Hexadezimalzahlen, Einfügen in Strings](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231102/hp-musterloesung-20231102.pdf)
* [09.11.2023: Text-Grafik-Bibliothek, Mikrocontroller, LED-Blinkmuster](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231109/hp-musterloesung-20231109.pdf)
* [16.11.2023: Trickprogrammierung, Thermometer-Baustein an I²C-Bus, Speicherformate von Zahlen](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231116/hp-musterloesung-20231116.pdf)
+ * [23.11.2023: Kondensator, Personen-Datenbank, Hexdumps](https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231123/hp-musterloesung-20231123.pdf)
Praktikumsunterlagen:
---------------------
diff --git a/hp-slides-2023ws.pdf b/hp-slides-2023ws.pdf
index f4b1aafb1a74f044ff673ee7270568b722418c96..4194a11bffce9b374c9a4ac056895cf9b987e760 100644
Binary files a/hp-slides-2023ws.pdf and b/hp-slides-2023ws.pdf differ
diff --git a/hp-slides-2023ws.tex b/hp-slides-2023ws.tex
index f7cd9bd26bcb913256ca6d6966191fa8a8735bb9..f577a24e6499716afb7c5732b60640c4a6bc1ff7 100644
--- a/hp-slides-2023ws.tex
+++ b/hp-slides-2023ws.tex
@@ -23,6 +23,8 @@
\includepdf[pages=-]{20231102/hp-20231102.pdf}
\pdfbookmark[1]{09.11.2023: Hardwarenahe Programmierung}{20231109}
\includepdf[pages=-]{20231109/hp-20231109.pdf}
- \pdfbookmark[1]{16.11.2023: Byte-Reihenfolge, Darstellung negativer Zahlen, Darstellung von Gleitkommazahlen, Speicherausrichtung}{20231116}
+ \pdfbookmark[1]{16.11.2023: Byte-Reihenfolge, Darstellung negativer Zahlen, Darstellung von Gleitkommazahlen}{20231116}
\includepdf[pages=-]{20231116/hp-20231116.pdf}
+ \pdfbookmark[1]{23.11.2023: Speicherausrichtung}{20231123}
+ \includepdf[pages=-]{20231123/hp-20231123.pdf}
\end{document}