Vorbereitung 30.3.2023

20230323/ad-20230323.tex

-% ad-20230322.pdf - Lecture Slides on Algorithms and Data Structures in C/C++
+% ad-20230323.pdf - Lecture Slides on Algorithms and Data Structures in C/C++
 % Copyright (C) 2018, 2019, 2020, 2021, 2022, 2023  Peter Gerwinski
 %
 % This document is free software: you can redistribute it and/or
@@ -20,7 +20,7 @@
 % Attribution-ShareAlike 3.0 Unported License along with this
 % document.  If not, see <http://creativecommons.org/licenses/>.
 
-% README: Einführung
+% README: Einführung in C++
 
 \documentclass[10pt,t]{beamer}
 
@@ -37,7 +37,7 @@
 
 \title{Algorithmen und Datenstrukturen in C/C++}
 \author{Prof.\ Dr.\ rer.\ nat.\ Peter Gerwinski}
-\date{22.\ März 2023}
+\date{23.\ März 2023}
 
 \begin{document}
 
@@ -1034,6 +1034,8 @@
   \end{lstlisting}
 \end{frame}
 
+\iffalse
+
 \subsection{Objekte}
 
 \begin{frame}[fragile]
@@ -1085,4 +1087,6 @@
   \end{itemize}
 \end{frame}
 
+\fi
+
 \end{document}

20230323/ad-20230323.txt

+Lösung zur Übungsaufgabe, 23.03.2023, 15:17:55
+~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+(b) Trivial.
+
+(a) Wir setzen (b) als gelöst voraus und bringen die doppelt verkettette Liste
+    in die umgekehrte Reihenfolge (lists-19.c).
+
+(c) O(n)
+
+(d) O(1)

20230323/const-01.c

+#include <stdio.h>
+
+const int n = 5;
+int prime[n] = { 2, 3, 5, 7, 11 };
+
+int main (void)
+{
+  for (int i = 0; i < 5; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-02.c

+#include <stdio.h>
+
+int main (void)
+{
+  const int n = 5;
+  int prime[n] = { 2, 3, 5, 7, 11 };
+  for (int i = 0; i < 5; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-03.c

+#include <stdio.h>
+
+int main (void)
+{
+  const int n = 5;
+  int prime[n];
+  prime[0] = 2;
+  prime[1] = 3;
+  prime[2] = 5;
+  prime[3] = 7;
+  prime[4] = 11;
+  for (int i = 0; i < 5; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-04.c

+#include <stdio.h>
+
+const int n = 5;
+int prime[n];
+
+int main (void)
+{
+  prime[0] = 2;
+  prime[1] = 3;
+  prime[2] = 5;
+  prime[3] = 7;
+  prime[4] = 11;
+  for (int i = 0; i < 5; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-05.c

+#include <stdio.h>
+
+#define n 5
+int prime[n];
+
+int main (void)
+{
+  prime[0] = 2;
+  prime[1] = 3;
+  prime[2] = 5;
+  prime[3] = 7;
+  prime[4] = 11;
+  for (int i = 0; i < 5; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-06.c

+#include <stdio.h>
+
+const int n = 6;
+int prime[n] = { 2, 3, 5, 7, 11, 13 };
+
+int main (void)
+{
+  for (int i = 0; i < n; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-07.c

+#include <stdio.h>
+
+int fak (int n)
+{
+  int f = 1;
+  for (int i = 2; i <= n; i++)
+    f *= i;
+  return f;
+}
+
+const int n = fak (3);
+int prime[n] = { 2, 3, 5, 7, 11, 13 };
+
+int main (void)
+{
+  for (int i = 0; i < n; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-08.c

+#include <stdio.h>
+
+constexpr int fak (int n)
+{
+  int f = 1;
+  for (int i = 2; i <= n; i++)
+    f *= i;
+  return f;
+}
+
+const int n = fak (3);
+int prime[n] = { 2, 3, 5, 7, 11, 13 };
+
+int main (void)
+{
+  for (int i = 0; i < n; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-09.c

+#include <stdio.h>
+
+constexpr int fak (int n)
+{
+  if (n <= 1)
+    return 1;
+  else
+    return n * fak (n - 1);
+}
+
+const int n = fak (3);
+int prime[n] = { 2, 3, 5, 7, 11, 13 };
+
+int main (void)
+{
+  for (int i = 0; i < n; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-10.c

+#include <stdio.h>
+
+constexpr int fak (int n)
+{
+  return n <= 1 ? 1 : n * fak (n - 1);
+}
+
+const int n = fak (3);
+int prime[n] = { 2, 3, 5, 7, 11, 13 };
+
+int main (void)
+{
+  for (int i = 0; i < n; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/const-11.c

+#include <stdio.h>
+
+int eins = 1;
+
+constexpr int fak (int n)
+{
+  return n <= eins ? eins : n * fak (n - 1);
+}
+
+const int n = fak (3);
+int prime[n] = { 2, 3, 5, 7, 11, 13 };
+
+int main (void)
+{
+  for (int i = 0; i < n; i++)
+    printf ("%d\n", prime[i]);
+  return 0;
+}

20230323/dynmem-01.c

+#include <stdio.h>
+#include <stdlib.h>
+
+int main (void)
+{
+  int *p = malloc (1);
+  *p = 42;
+  printf ("The answer is: %d\n", *p);
+  free (p);
+  return 0;
+}

20230323/dynmem-02.c

+#include <stdio.h>
+#include <stdlib.h>
+
+int main (void)
+{
+  int *p = malloc (sizeof (int));
+  *p = 42;
+  printf ("The answer is: %d\n", *p);
+  free (p);
+  return 0;
+}

20230323/dynmem-03.c

+#include <stdio.h>
+#include <stdlib.h>
+
+int main (void)
+{
+  int *p = new int;
+  *p = 42;
+  printf ("The answer is: %d\n", *p);
+  delete p;
+  return 0;
+}

20230323/dynmem-04.c

+#include <stdio.h>
+#include <stdlib.h>
+
+int main (void)
+{
+  int *p = new sizeof (int);
+  *p = 42;
+  printf ("The answer is: %d\n", *p);
+  delete p;
+  return 0;
+}

20230323/functions-01.c

+#include <stdio.h>
+
+void print (const char *s)
+{
+  printf ("%s", s);
+}
+
+void print (int i)
+{
+  printf ("%d", i);
+}
+
+int main ()
+{
+  print ("Hello, world!\n");
+  print (42);
+}

20230323/functions-02.c

+#include <stdio.h>
+
+void print (const char *s = "\n")
+{
+  printf ("%s", s);
+}
+
+void print (int i)
+{
+  printf ("%d", i);
+}
+
+int main ()
+{
+  print ("Hello, world!\n");
+  print (42);
+  print ();
+}