Compare revisions

20231005/if-07.c

deleted100644 → 0

+0 −6

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void) { int a = 360; printf ("a = %d\n", a); int b; printf
		("Bitte b eingeben: "); scanf ("%d", &b); if (b == 0) { printf
		("Bitte nicht durch 0 teilen.\n"); printf ("Das macht man nicht.\n");
		} else printf ("a / b = %d\n", a / b); return 0; }

20231005/if-08.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		scanf ("%d", &b);
		if (b == 0)
		{
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("Das macht man nicht.\n");
		}
		else
		{
		printf ("a / b = %d\n", a / b);
		}
		return 0;
		}

20231005/init-01.s

deleted100644 → 0

+0 −32

Original line number	Diff line number	Diff line
		.file "init-01.c"
		.text
		.section .rodata.str1.1,"aMS",@progbits,1
		.LC0:
		.string "a = %d\n"
		.LC1:
		.string "b = %d\n"
		.text
		.globl main
		.type main, @function
		main:
		.LFB11:
		.cfi_startproc # #include <stdio.h>
		subq $8, %rsp #
		.cfi_def_cfa_offset 16 # int main (void)
		movl $42, %esi # {
		leaq .LC0(%rip), %rdi # int a = 42;
		movl $0, %eax # int b;
		call printf@PLT # b = 137;
		movl $137, %esi # printf ("a = %d\n", a);
		leaq .LC1(%rip), %rdi # printf ("b = %d\n", b);
		movl $0, %eax # return 0;
		call printf@PLT # }
		movl $0, %eax
		addq $8, %rsp
		.cfi_def_cfa_offset 8
		ret
		.cfi_endproc
		.LFE11:
		.size main, .-main
		.ident "GCC: (Debian 8.3.0-6) 8.3.0"
		.section .note.GNU-stack,"",@progbits

20231005/init-02.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 42;
		int b;
		printf ("Bitte b eingeben: ");
		scanf ("%d", b);
		printf ("a * b + a + b = %d\n", a * b + a + b);
		return 0;
		}

20231005/init-02.s

deleted100644 → 0

+0 −26

Original line number	Diff line number	Diff line
		.file "init-02.c"
		.text
		.section .rodata.str1.1,"aMS",@progbits,1
		.LC0:
		.string "a * b + a + b = %d\n"
		.text
		.globl main
		.type main, @function
		main:
		.LFB11:
		.cfi_startproc
		subq $8, %rsp # #include <stdio.h>
		.cfi_def_cfa_offset 16 #
		movl $5933, %esi # int main (void)
		leaq .LC0(%rip), %rdi # {
		movl $0, %eax # int a = 42;
		call printf@PLT # int b = 137;
		movl $0, %eax # printf ("a * b + a + b = %d\n", a * b + a + b);
		addq $8, %rsp # return 0;
		.cfi_def_cfa_offset 8 # }
		ret
		.cfi_endproc
		.LFE11:
		.size main, .-main
		.ident "GCC: (Debian 8.3.0-6) 8.3.0"
		.section .note.GNU-stack,"",@progbits

20231005/init-03.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 42;
		int b;
		printf ("Bitte b eingeben: ");
		scanf ("%d", b);
		printf ("a * b + a + b = %d\n", a * b + a + b);
		return 0;
		}

20231005/init-04.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 42;
		int b;
		printf ("Bitte b eingeben: ");
		scanf ("%d", &b);
		printf ("a * b + a + b = %d\n", a * b + a + b);
		return 0;
		}

20231005/init-04.s

deleted100644 → 0

+0 −40

Original line number	Diff line number	Diff line
		.file "init-04.c"
		.text
		.section .rodata.str1.1,"aMS",@progbits,1
		.LC0:
		.string "Bitte b eingeben: "
		.LC1:
		.string "%d"
		.LC2:
		.string "a * b + a + b = %d\n"
		.text
		.globl main
		.type main, @function
		main:
		.LFB11:
		.cfi_startproc
		subq $24, %rsp
		.cfi_def_cfa_offset 32 # #include <stdio.h>
		leaq .LC0(%rip), %rdi #
		movl $0, %eax # int main (void)
		call printf@PLT # {
		leaq 12(%rsp), %rsi # int a = 42;
		leaq .LC1(%rip), %rdi # int b;
		movl $0, %eax # printf ("Bitte b eingeben: ");
		call __isoc99_scanf@PLT # scanf ("%d", &b);
		movl 12(%rsp), %eax # printf ("a * b + a + b = %d\n", a * b + a + b);
		leal 1(%rax), %esi # return 0;
		imull $42, %esi, %esi # }
		addl %eax, %esi
		leaq .LC2(%rip), %rdi
		movl $0, %eax
		call printf@PLT
		movl $0, %eax
		addq $24, %rsp
		.cfi_def_cfa_offset 8
		ret
		.cfi_endproc
		.LFE11:
		.size main, .-main
		.ident "GCC: (Debian 8.3.0-6) 8.3.0"
		.section .note.GNU-stack,"",@progbits

20231005/input-02.c

deleted100644 → 0

+0 −10

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		double a;
		printf ("Bitte eine Zahl eingeben: ");
		scanf ("%lf", a);
		printf ("Ihre Antwort war: %lf\n", a);
		return 0;
		}

20231005/input-03.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		double a;
		printf ("a = %lf\n", a);
		printf ("Bitte eine Zahl eingeben: ");
		scanf ("%lf", a);
		printf ("Ihre Antwort war: %lf\n", a);
		return 0;
		}

20231005/input-04.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		double a = &a;
		printf ("a = %lf\n", a);
		printf ("Bitte eine Zahl eingeben: ");
		scanf ("%lf", a);
		printf ("Ihre Antwort war: %lf\n", a);
		return 0;
		}

20231005/input-05.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		double a = (double) (long long unsigned) &a;
		printf ("a = %lf\n", a);
		printf ("Bitte eine Zahl eingeben: ");
		scanf ("%lf", a);
		printf ("Ihre Antwort war: %lf\n", a);
		return 0;
		}

20231005/input-06.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		long long unsigned a = (long long unsigned) &a;
		printf ("a = %llu\n", a);
		printf ("Bitte eine Zahl eingeben: ");
		scanf ("%llu", a);
		printf ("Ihre Antwort war: %llu\n", a);
		return 0;
		}

20231012/functions-01.c

deleted100644 → 0

+0 −4

Original line number	Diff line number	Diff line
		int main (void)
		{
		return 0;
		}

20231012/functions-02.c

deleted100644 → 0

+0 −4

Original line number	Diff line number	Diff line
		int main (void)
		{
		return 42;
		}

20231012/gitlab.png

deleted100644 → 0

−589 B

589 B

20231012/loops-01.c

deleted100644 → 0

+0 −16

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 1;
		do
		{
		if (a <= 10)
		{
		printf ("%d\n", a);
		a++;
		}
		}
		while (a <= 10);
		return 0;
		}

20231012/loops-02.c

deleted100644 → 0

+0 −16

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 11;
		do
		{
		if (a <= 10)
		{
		printf ("%d\n", a);
		a++;
		}
		}
		while (a <= 10);
		return 0;
		}

20231012/side-effects-04.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 3;
		int b = 5;
		printf ("%d\n", a += b);
		printf ("%d\n", a);
		printf ("%d\n", b);
		return 0;
		}

20231012/side-effects-05.c

deleted100644 → 0

+0 −11

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 3;
		int b = 5;
		printf ("%d\n", a + b);
		printf ("%d\n", a);
		printf ("%d\n", b);
		return 0;
		}

20231012/side-effects-11.c

deleted100644 → 0

+0 −9

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 3;
		a = (3,14);
		printf ("%d\n", a);
		return 0;
		}

20231012/side-effects-13.c

deleted100644 → 0

+0 −13

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 137;
		int b = 0;
		if (b == 0)
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("So was macht man nicht.\n");
		else
		printf ("a / b = %d\n", a / b);
		return 0;
		}

20231012/side-effects-15.c

deleted100644 → 0

+0 −13

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 137;
		int b = 0;
		if (b == 0)
		printf ("Bitte nicht durch 0 teilen.\n"),
		printf ("So was macht man nicht.\n");
		else
		printf ("a / b = %d\n", a / b);
		return 0;
		}

20231019/arrays-05c.c

deleted100644 → 0

+0 −15

Original line number	Diff line number	Diff line

		#include <stdio.h>

		int main (void)
		{
		int pflll[8] = { 1, 2, 3, 4, 5, 6, 7, 8 };
		int prime[8] = { 2, 3, 5, 7, 11, 13, 17, 19 };
		printf ("prime:\n");
		for (int p = prime; p; p++)
		printf ("%d\n", *p);
		printf ("pflll:\n");
		for (int q = pflll; q; q++)
		printf ("%d\n", *q);
		return 0;
		}

20231019/files-07.txt

deleted100644 → 0

+0 −40

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <errno.h>
		#include <string.h>

		int main (void)
		{
		FILE *f = fopen ("fhello.txt", "w");
		if (f)
		{
		fprintf (f, "Hello, world!\n");
		fclose (f);
		return 0;
		}
		else
		{
		char *msg = strerror (errno);
		fprintf (stderr, "Fehler: %s\n", msg);
		return errno;
		}
		}
		cassini/home/peter/bo/2023ws/hp/20231019> gcc -Wall -O files-07.c -o files-07
		cassini/home/peter/bo/2023ws/hp/20231019> ./files-07
		Fehler: Permission denied
		cassini/home/peter/bo/2023ws/hp/20231019> echo $?
		13
		cassini/home/peter/bo/2023ws/hp/20231019> rm fhello.txt
		rm: reguläre Datei (schreibgeschützt) 'fhello.txt' entfernen? y
		cassini/home/peter/bo/2023ws/hp/20231019> ./files-07
		cassini/home/peter/bo/2023ws/hp/20231019> echo $?
		0
		cassini/home/peter/bo/2023ws/hp/20231019> ls -l fhello.txt
		-rw-r--r-- 1 peter peter 14 19. Okt 15:56 fhello.txt
		cassini/home/peter/bo/2023ws/hp/20231019> chmod -w fhello.txt
		cassini/home/peter/bo/2023ws/hp/20231019> ls -l fhello.txt
		-r--r--r-- 1 peter peter 14 19. Okt 15:56 fhello.txt
		cassini/home/peter/bo/2023ws/hp/20231019> ./files-07
		Fehler: Permission denied
		cassini/home/peter/bo/2023ws/hp/20231019> echo $?
		13
		cassini/home/peter/bo/2023ws/hp/20231019>

20231019/hp-uebung-20231019.pdf

deleted100644 → 0

−36.7 KiB

File deleted.

20231019/strings-07d.c

deleted100644 → 0

+0 −21

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char hello[] = "Hello, world!\n";

		printf ("%s\n", hello);
		printf ("%zd\n", strlen (hello));

		printf ("%s\n", hello + 17);
		printf ("%zd\n", strlen (hello + 17));

		return 0;

		hello[5] = 0;
		printf ("%s\n", hello);
		printf ("%zd\n", strlen (hello));

		return 0;
		}

20231026/hp-musterloesung-20231026.pdf

deleted100644 → 0

−55.9 KiB

File deleted.

20231026/hp-uebung-20231026.pdf

deleted100644 → 0

−30.1 KiB

File deleted.

20231026/string-ops-03.s

deleted100644 → 0

+0 −57

Original line number	Diff line number	Diff line
		.file "string-ops-03.c"
		.text
		.section .rodata
		.LC0:
		.string "Zacharias"
		.LC1:
		.string "Anton"
		.LC2:
		.string "%d\n"
		.text
		.globl main
		.type main, @function
		main:
		.LFB0:
		.cfi_startproc
		pushq %rbp
		.cfi_def_cfa_offset 16
		.cfi_offset 6, -16
		movq %rsp, %rbp
		.cfi_def_cfa_register 6
		addq $-128, %rsp
		leaq .LC0(%rip), %rax
		movq %rax, -8(%rbp)
		leaq .LC1(%rip), %rax
		movq %rax, -16(%rbp)
		movq -16(%rbp), %rax
		cmpq -8(%rbp), %rax
		setb %al
		movzbl %al, %eax
		movl %eax, %esi
		leaq .LC2(%rip), %rax
		movq %rax, %rdi
		movl $0, %eax
		call printf@PLT
		movq -8(%rbp), %rax
		cmpq -16(%rbp), %rax
		setb %al
		movzbl %al, %eax
		movl %eax, %esi
		leaq .LC2(%rip), %rax
		movq %rax, %rdi
		movl $0, %eax
		call printf@PLT
		movl $1, %esi
		leaq .LC2(%rip), %rax
		movq %rax, %rdi
		movl $0, %eax
		call printf@PLT
		movl $0, %eax
		leave
		.cfi_def_cfa 7, 8
		ret
		.cfi_endproc
		.LFE0:
		.size main, .-main
		.ident "GCC: (Debian 12.2.0-14) 12.2.0"
		.section .note.GNU-stack,"",@progbits

20231026/string-ops-05.c

deleted100644 → 0

+0 −18

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[100] = "Huber ";
		strcat (buffer, anton);
		printf ("%s\n", buffer);

		return 0;
		}

20231026/string-ops-07.c

deleted100644 → 0

+0 −19

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[] = "Huber ";
		strcat (buffer, anton);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-08.c

deleted100644 → 0

+0 −19

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[] = "Huber ";
		strcat (buffer, "Antonius, genannt Toni oder Anton");
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-09.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";
		char *toni = "Antonius, genannt Toni oder Anton";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[] = "Huber ";
		strcat (buffer, toni);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-10.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";
		char *toni = "Antonius!!";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[] = "Huber ";
		strcat (buffer, toni);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-11.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";
		char *toni = "Antonius!";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[] = "Huber ";
		strcat (buffer, toni);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-11.s

deleted100644 → 0

+0 −55

Original line number	Diff line number	Diff line
		.file "string-ops-11.c"
		.text
		.section .rodata.str1.1,"aMS",@progbits,1
		.LC0:
		.string "%d\n"
		.LC1:
		.string "Anton"
		.text
		.globl main
		.type main, @function
		main:
		.LFB11:
		.cfi_startproc
		pushq %rbx
		.cfi_def_cfa_offset 16
		.cfi_offset 3, -16
		subq $16, %rsp
		.cfi_def_cfa_offset 32
		movl $-1, %esi
		leaq .LC0(%rip), %rbx
		movq %rbx, %rdi
		movl $0, %eax
		call printf@PLT
		movl $1, %esi ; #include <stdio.h>
		movq %rbx, %rdi ; #include <string.h>
		movl $0, %eax ;
		call printf@PLT ; int main (void)
		movl $0, %esi ; {
		movq %rbx, %rdi ; char *zacharias = "Zacha";
		movl $0, %eax ; char *anton = "Anton";
		call printf@PLT ; char *toni = "Antonius!";
		movl $1700951368, 9(%rsp) ;
		movl $2126437, 12(%rsp) ; printf ("%d\n", strcmp (anton, zacharias));
		leaq 9(%rsp), %rbx ; printf ("%d\n", strcmp (zacharias, anton));
		movq %rbx, %rdi ; printf ("%d\n", strcmp (anton, anton));
		call strlen@PLT ;
		addq %rbx, %rax ; char buffer[] = "Huber ";
		movabsq $8319671809674079809, %rdx ; strcat (buffer, toni);
		movq %rdx, (%rax) ; printf ("%s\n", buffer);
		movw $33, 8(%rax) ; printf ("%s\n", anton);
		movq %rbx, %rdi ;
		call puts@PLT ; return 0;
		leaq .LC1(%rip), %rdi ; }
		call puts@PLT
		movl $0, %eax
		addq $16, %rsp
		.cfi_def_cfa_offset 16
		popq %rbx
		.cfi_def_cfa_offset 8
		ret
		.cfi_endproc
		.LFE11:
		.size main, .-main
		.ident "GCC: (Debian 12.2.0-14) 12.2.0"
		.section .note.GNU-stack,"",@progbits

20231026/string-ops-12.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";
		char *toni = "Toni";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[] = "Huber ";
		strcat (buffer, toni);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-12.s

deleted100644 → 0

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Original line number	Diff line number	Diff line
		.file "string-ops-12.c"
		.text
		.section .rodata.str1.1,"aMS",@progbits,1
		.LC0:
		.string "%d\n"
		.LC1:
		.string "Anton"
		.text
		.globl main
		.type main, @function
		main:
		.LFB11:
		.cfi_startproc
		pushq %rbx
		.cfi_def_cfa_offset 16
		.cfi_offset 3, -16
		subq $16, %rsp ; #include <stdio.h>
		.cfi_def_cfa_offset 32 ; #include <string.h>
		movl $-1, %esi ;
		leaq .LC0(%rip), %rbx ; int main (void)
		movq %rbx, %rdi ; {
		movl $0, %eax ; char *zacharias = "Zacha";
		call printf@PLT ; char *anton = "Anton";
		movl $1, %esi ; char *toni = "Toni";
		movq %rbx, %rdi ;
		movl $0, %eax ; printf ("%d\n", strcmp (anton, zacharias));
		call printf@PLT ; printf ("%d\n", strcmp (zacharias, anton));
		movl $0, %esi ; printf ("%d\n", strcmp (anton, anton));
		movq %rbx, %rdi ;
		movl $0, %eax ; char buffer[] = "Huber ";
		call printf@PLT ; strcat (buffer, toni);
		movl $1700951368, 9(%rsp) ; printf ("%s\n", buffer);
		movl $2126437, 12(%rsp) ; printf ("%s\n", anton);
		leaq 9(%rsp), %rbx ;
		movq %rbx, %rdi ; return 0;
		call strlen@PLT ; }
		addq %rbx, %rax
		movl $1768845140, (%rax)
		movb $0, 4(%rax)
		movq %rbx, %rdi
		call puts@PLT
		leaq .LC1(%rip), %rdi
		call puts@PLT
		movl $0, %eax
		addq $16, %rsp
		.cfi_def_cfa_offset 16
		popq %rbx
		.cfi_def_cfa_offset 8
		ret
		.cfi_endproc
		.LFE11:
		.size main, .-main
		.ident "GCC: (Debian 12.2.0-14) 12.2.0"
		.section .note.GNU-stack,"",@progbits

20231026/string-ops-13.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";
		char *toni = "Antonius, genannt Toni";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[13] = "Huber ";
		strncat (buffer, toni, 13);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231026/string-ops-14.c

deleted100644 → 0

+0 −20

Original line number	Diff line number	Diff line
		#include <stdio.h>
		#include <string.h>

		int main (void)
		{
		char *zacharias = "Zacha";
		char *anton = "Anton";
		char *toni = "Antonius, genannt Toni";

		printf ("%d\n", strcmp (anton, zacharias));
		printf ("%d\n", strcmp (zacharias, anton));
		printf ("%d\n", strcmp (anton, anton));

		char buffer[14] = "Huber ";
		strncat (buffer, toni, 13);
		printf ("%s\n", buffer);
		printf ("%s\n", anton);

		return 0;
		}

20231102/bc-01.txt

deleted100644 → 0

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Original line number	Diff line number	Diff line
		cassini/home/peter/bo/2023ws/hp/20231102> bc
		bc 1.07.1
		Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006, 2008, 2012-2017 Free Software Foundation, Inc.
		This is free software with ABSOLUTELY NO WARRANTY.
		For details type `warranty'.
		obase=13
		6 * 9
		42
		obase=A
		6 * 9
		54
		obase=13
		6 * 9
		42
		obase=1+1+1+1+1+1+1+1+1+1
		6 * 9
		54

		cassini/home/peter/bo/2023ws/hp/20231102>

20231102/bc-02.txt

deleted100644 → 0

+0 −24

Original line number	Diff line number	Diff line
		cassini/home/peter/bo/2023ws/hp/20231102> bc
		bc 1.07.1
		Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006, 2008, 2012-2017 Free Software Foundation, Inc.
		This is free software with ABSOLUTELY NO WARRANTY.
		For details type `warranty'.
		obase=16
		42
		2A
		ibase=16
		B747 + A380
		15AC7
		obase=10
		5 + 5
		A
		A + A
		14
		18 + 18
		30
		ibase=1+1+1+1+1+1+1+1+1+1
		obase=10
		5 + 5
		10

		cassini/home/peter/bo/2023ws/hp/20231102>

20231102/gtk-01.txt

deleted100644 → 0

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Original line number	Diff line number	Diff line
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -Wall -O gtk-01.c -o gtk-01
		gtk-01.c:1:10: fatal error: gtk/gtk.h: Datei oder Verzeichnis nicht gefunden
		1 \| #include <gtk/gtk.h>
		\| ^~~~~~~~~~~
		compilation terminated.
		cassini/home/peter/bo/2023ws/hp/20231102> locate gtk/gtk.h
		/usr/include/gtk-2.0/gtk/gtk.h
		/usr/include/gtk-3.0/gtk/gtk.h
		/usr/include/gtk-4.0/gtk/gtk.h
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -Wall -I /usr/include/gtk-4.0 -O gtk-01.c -o gtk-01
		In file included from /usr/include/gtk-4.0/gtk/gtk.h:30,
		from gtk-01.c:1:
		/usr/include/gtk-4.0/gtk/css/gtkcss.h:30:10: fatal error: glib.h: Datei oder Verzeichnis nicht gefunden
		30 \| #include <glib.h>
		\| ^~~~~~~~
		compilation terminated.
		cassini/home/peter/bo/2023ws/hp/20231102> locate glib.h
		/home/peter/bo/2012ws/rtech/material/usbstick/fdos/source/command/command/suppl/msglib.h
		/usr/include/jpeglib.h
		/usr/include/dbus-1.0/dbus/dbus-glib.h
		/usr/include/glib-2.0/glib.h
		/usr/include/harfbuzz/hb-glib.h
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -Wall -I /usr/include/gtk-4.0 -I /usr/include/glib-2.0 -O gtk-01.c -o gtk-01
		In file included from /usr/include/glib-2.0/glib/galloca.h:34,
		from /usr/include/glib-2.0/glib.h:32,
		from /usr/include/gtk-4.0/gtk/css/gtkcss.h:30,
		from /usr/include/gtk-4.0/gtk/gtk.h:30,
		from gtk-01.c:1:
		/usr/include/glib-2.0/glib/gtypes.h:34:10: fatal error: glibconfig.h: Datei oder Verzeichnis nicht gefunden
		34 \| #include <glibconfig.h>
		\| ^~~~~~~~~~~~~~
		compilation terminated.
		cassini/home/peter/bo/2023ws/hp/20231102> pkg-config --cflags --libs gtk4
		-I/usr/include/gtk-4.0 -I/usr/include/pango-1.0 -I/usr/include/glib-2.0 -I/usr/lib/x86_64-linux-gnu/glib-2.0/include -I/usr/include/harfbuzz -I/usr/include/freetype2 -I/usr/include/libpng16 -I/usr/include/libmount -I/usr/include/blkid -I/usr/include/fribidi -I/usr/include/cairo -I/usr/include/pixman-1 -I/usr/include/gdk-pixbuf-2.0 -I/usr/include/x86_64-linux-gnu -I/usr/include/graphene-1.0 -I/usr/lib/x86_64-linux-gnu/graphene-1.0/include -mfpmath=sse -msse -msse2 -pthread -lgtk-4 -lpangocairo-1.0 -lpango-1.0 -lharfbuzz -lgdk_pixbuf-2.0 -lcairo-gobject -lcairo -lgraphene-1.0 -lgio-2.0 -lgobject-2.0 -lglib-2.0
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -Wall -O -I/usr/include/gtk-4.0 -I/usr/include/pango-1.0 -I/usr/include/glib-2.0 -I/usr/lib/x86_64-linux-gnu/glib-2.0/include -I/usr/include/harfbuzz -I/usr/include/freetype2 -I/usr/include/libpng16 -I/usr/include/libmount -I/usr/include/blkid -I/usr/include/fribidi -I/usr/include/cairo -I/usr/include/pixman-1 -I/usr/include/gdk-pixbuf-2.0 -I/usr/include/x86_64-linux-gnu -I/usr/include/graphene-1.0 -I/usr/lib/x86_64-linux-gnu/graphene-1.0/include -mfpmath=sse -msse -msse2 -pthread -lgtk-4 -lpangocairo-1.0 -lpango-1.0 -lharfbuzz -lgdk_pixbuf-2.0 -lcairo-gobject -lcairo -lgraphene-1.0 -lgio-2.0 -lgobject-2.0 -lglib-2.0 gtk-01.c -o gtk-01
		/usr/bin/ld: /tmp/cc6S8nDW.o: in function `activate':
		gtk-01.c:(.text+0x7): undefined reference to `gtk_application_window_new'
		/usr/bin/ld: gtk-01.c:(.text+0xf): undefined reference to `gtk_window_get_type'
		/usr/bin/ld: gtk-01.c:(.text+0x1d): undefined reference to `g_type_check_instance_cast'
		/usr/bin/ld: gtk-01.c:(.text+0x2c): undefined reference to `gtk_window_set_title'
		/usr/bin/ld: gtk-01.c:(.text+0x37): undefined reference to `g_type_check_instance_cast'
		/usr/bin/ld: gtk-01.c:(.text+0x3f): undefined reference to `gtk_window_present'
		/usr/bin/ld: /tmp/cc6S8nDW.o: in function `main':
		gtk-01.c:(.text+0x60): undefined reference to `gtk_application_new'
		/usr/bin/ld: gtk-01.c:(.text+0x8a): undefined reference to `g_signal_connect_data'
		/usr/bin/ld: gtk-01.c:(.text+0x8f): undefined reference to `g_application_get_type'
		/usr/bin/ld: gtk-01.c:(.text+0x9a): undefined reference to `g_type_check_instance_cast'
		/usr/bin/ld: gtk-01.c:(.text+0xa7): undefined reference to `g_application_run'
		/usr/bin/ld: gtk-01.c:(.text+0xb1): undefined reference to `g_object_unref'
		collect2: error: ld returned 1 exit status
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -Wall -O gtk-01.c -I/usr/include/gtk-4.0 -I/usr/include/pango-1.0 -I/usr/include/glib-2.0 -I/usr/lib/x86_64-linux-gnu/glib-2.0/include -I/usr/include/harfbuzz -I/usr/include/freetype2 -I/usr/include/libpng16 -I/usr/include/libmount -I/usr/include/blkid -I/usr/include/fribidi -I/usr/include/cairo -I/usr/include/pixman-1 -I/usr/include/gdk-pixbuf-2.0 -I/usr/include/x86_64-linux-gnu -I/usr/include/graphene-1.0 -I/usr/lib/x86_64-linux-gnu/graphene-1.0/include -mfpmath=sse -msse -msse2 -pthread -lgtk-4 -lpangocairo-1.0 -lpango-1.0 -lharfbuzz -lgdk_pixbuf-2.0 -lcairo-gobject -lcairo -lgraphene-1.0 -lgio-2.0 -lgobject-2.0 -lglib-2.0 -o gtk-01
		cassini/home/peter/bo/2023ws/hp/20231102>

20231102/hp-20231102.txt

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Original line number	Diff line number	Diff line
		"Output"
		~~~~~~~~
		AND
		10101100 A
		& 11111011 Bit gezielt löschen & B "Maske"
		-------- ---
		10101000 A, außer dort, wo B Nullen hat
		Dort sind es Nullen.
		OR
		10101100 A
		\| 00000010 Bit gezielt setzen & B "Maske"
		-------- ---
		10101110 A, außer dort, wo B Einsen hat
		Dort sind es Einsen.
		XOR
		10101100 A
		^ 00000010 Bit gezielt umklappen & B "Maske"
		-------- ---
		10101110 A, außer dort, wo B Einsen hat
		Dort wird A umgeklappt.
		00000010
		^ 00001000
		--------
		00001010

		"Input"
		~~~~~~~
		AND
		10101100
		& 00000100
		--------
		00000100 <-- ungleich 0: Bit war gesetzt

		Achtung: Das Ergebnis ist nicht gleich 1!
		(sondern: gleich 4)

20231102/hp-musterloesung-20231102.pdf

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−100 KiB

File deleted.

20231102/hp-musterloesung-20231102.tex

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Original line number	Diff line number	Diff line
		% hp-musterloesung-20231102.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: Zahlensysteme, Ausgabe von Hexadezimalzahlen, Einfügen in Strings

		\documentclass[a4paper]{article}

		\usepackage{pgscript}
		\renewcommand{\breath}{\bigskip}

		\begin{document}

		\section*{Hardwarenahe Programmierung\\
		Musterlösung zu den Übungsaufgaben -- 2.\ November 2023}

		\exercise{Zahlensysteme}

		Wandeln Sie ohne Hilfsmittel

		\begin{minipage}[t]{0.3\textwidth}
		\begin{itemize}
		\item
		nach Dezimal:
		\begin{itemize}
		\item[(a)]
		0010\,0000$_2$
		\item[(b)]
		42$_{16}$
		\item[(c)]
		17$_8$
		\end{itemize}
		\end{itemize}
		\end{minipage}\hfill
		\begin{minipage}[t]{0.3\textwidth}
		\begin{itemize}
		\item
		nach Hexadezimal:
		\begin{itemize}
		\item[(d)]
		0010\,0000$_2$
		\item[(e)]
		42$_{10}$
		\item[(f)]
		192.168.20.254$_{256}$
		\end{itemize}
		\end{itemize}
		\end{minipage}\hfill
		\begin{minipage}[t]{0.3\textwidth}
		\begin{itemize}
		\item
		nach Binär:
		\begin{itemize}
		\item[(g)]
		750$_8$
		\item[(h)]
		42$_{10}$
		\item[(i)]
		AFFE$_{16}$
		\end{itemize}
		\end{itemize}
		\end{minipage}

		\medskip

		Berechnen Sie ohne Hilfsmittel:
		\begin{itemize}
		\item[(j)]
		750$_8$ \& 666$_8$
		\item[(k)]
		A380$_{16}$ + B747$_{16}$
		\item[(l)]
		AFFE$_{16} >> 1$
		\end{itemize}

		Die tiefgestellte Zahl steht für die Basis des Zahlensystems.
		Jede Teilaufgabe zählt 1 Punkt. \addtocounter{points}{12}

		(In der Klausur sind Hilfsmittel zugelassen,
		daher ist dies \emph{keine\/} typische Klausuraufgabe.)

		\solution

		Wandeln Sie ohne Hilfsmittel

		\begin{itemize}
		\item
		nach Dezimal:
		\begin{itemize}
		\item[(a)]
		$0010\,0000_2 = 32_{10}$

		Eine Eins mit fünf Nullen dahinter steht binär für $2^5 = 32$:\\
		mit $1$ anfangen und fünfmal verdoppeln.
		\item[(b)]
		$42_{16} = 4 \cdot 16 + 2 \cdot 1 = 64 + 2 = 66$
		\item[(c)]
		$17_8 = 1 \cdot 8 + 7 \cdot 1 = 8 + 7 = 15$
		\end{itemize}
		Umwandlung von und nach Dezimal ist immer rechenaufwendig.
		Umwandlungen zwischen Binär, Oktal und Hexadezimal gehen ziffernweise
		und sind daher wesentlich einfacher.
		\item
		nach Hexadezimal:
		\begin{itemize}
		\item[(d)]
		$0010\,0000_2 = 20_{16}$

		Umwandlung von Binär nach Hexadezimal geht ziffernweise:\\
		Vier Binärziffern werden zu einer Hex-Ziffer.
		\item[(e)]
		$\rm 42_{10} = 32_{10} + 10_{10} = 20_{16} + A_{16} = 2A_{16}$
		\item[(f)]
		$\rm 192.168.20.254_{256} = C0\,A8\,14\,FE_{16}$

		Umwandlung von der Basis 256 nach Hexadezimal geht ziffernweise:\\
		Eine 256er-Ziffer wird zu zwei Hex-Ziffern.

		Da die 256er-Ziffern dezimal angegeben sind,
		müssen wir viermal Dezimal nach Hexadezimal umwandeln.
		Hierfür bieten sich unterschiedliche Wege an.

		$\rm 192_{10} = 128_{10} + 64_{10} = 1100\,0000_{2} = C0_{16}$

		$\rm 168_{10} = 10_{10} \cdot 16_{10} + 8_{10} = A_{16} \cdot 10_{16} + 8_{16} = A8_{16}$

		$20_{10} = 16_{10} + 4_{10} = 10_{16} + 4_{16} = 14$

		$\rm 254_{10} = 255_{10} - 1_{10} = FF_{16} - 1_{16} = FE_{16}$
		\end{itemize}
		\item
		nach Binär:
		\begin{itemize}
		\item[(g)]
		$750_8 = 111\,101\,000_2$

		Umwandlung von Oktal nach Binär geht ziffernweise:\\
		Eine Oktalziffer wird zu drei Binärziffern.
		\item[(h)]
		$\rm 42_{10} = 2A_{16}$ (siehe oben) $= 0010\,1010_{16}$

		Umwandlung von Hexadezimal nach Binär geht ziffernweise:\\
		Eine Hex-Ziffer wird zu vier Binärziffern.
		\item[(i)]
		$\rm AFFE_{16} = 1010\,1111\,1111\,1110_2$

		Umwandlung von Hexadezimal nach Binär geht ziffernweise:\\
		Eine Hex-Ziffer wird zu vier Binärziffern.
		\end{itemize}
		\end{itemize}

		\medskip

		Berechnen Sie ohne Hilfsmittel:
		\begin{itemize}
		\item[(j)]
		$750_8\,\&\,666_8
		= 111\,101\,000_2\,\&\,110\,110\,110_2
		= 110\,100\,000_2
		= 640_8$

		Binäre Und-Operationen lassen sich am leichtesten
		in binärer Schreibweise durchführen.
		Umwandlung zwischen Oktal und Binär geht ziffernweise:
		Eine Oktalziffer wird zu drei Binärziffern und umgekehrt.

		Mit etwas Übung funktionieren diese Operationen
		auch direkt mit Oktalzahlen im Kopf.

		\item[(k)]
		$\rm\phantom{+}A380_{16}$\\
		$\rm+\kern2ptB747_{16}$\\[-\medskipamount]
		\rule{1.4cm}{0.5pt}\\
		$\rm 15AC7_{16}$
		\begin{picture}(0,0)
		\put(-1.4,0.35){\mbox{\scriptsize\bf 1}}
		\end{picture}

		Mit Hexadezimalzahlen (und Binär- und Oktal- und sonstigen Zahlen)
		kann man genau wie mit Dezimalzahlen schriftlich rechnen.
		Man muß nur daran denken, daß der "`Zehner"'-Überlauf nicht bei
		$10_{10}$ stattfindet, sondern erst bei $10_{16} = 16_{10}$
		(hier: $\rm 8_{16} + 4_{16} = C_{16}$ und
		$\rm 3_{16} + 7_{16} = A_{16}$, aber
		$\rm A_{16} + B_{16} = 10_{10} + 11_{10}
		= 21_{10} = 16_{10} + 5_{10} = 10_{16} + 5_{16} = 15_{16}$).

		\item[(l)]
		$\rm AFFE_{16} >> 1
		= 1010\,1111\,1111\,1110_2 >> 1
		= 0101\,0111\,1111\,1111_2
		= 57FF_{16}$

		Bit-Verschiebungen lassen sich am leichtesten
		in binärer Schreibweise durchführen.
		Umwandlung zwischen Hexadezimal und Binär geht ziffernweise:
		Eine Hex-Ziffer wird zu vier Binärziffern und umgekehrt.

		Mit etwas Übung funktionieren diese Operationen
		auch direkt mit Hexadezimalzahlen im Kopf.

		\end{itemize}

		\exercise{Ausgabe von Hexadezimalzahlen}

		Schreiben Sie eine Funktion \lstinline{void print_hex (uint32_t x)},
		die eine gegebene vorzeichenlose 32-Bit-Ganzzahl \lstinline{x}
		als Hexadezimalzahl ausgibt.
		(Der Datentyp \lstinline{uint32_t} ist mit \lstinline{#include <stdint.h>} verfügbar.)

		Verwenden Sie dafür \emph{nicht\/} \lstinline{printf()} mit
		der Formatspezifikation \lstinline{%x} als fertige Lösung,
		sondern programmieren Sie die nötige Ausgabe selbst.
		(Für Tests ist \lstinline{%x} hingegen erlaubt und sicherlich nützlich.)

		Die Verwendung von \lstinline{printf()}
		mit anderen Formatspezifikationen wie z.\,B.\ \lstinline{%d}
		oder \lstinline{%c} oder \lstinline{%s} ist hingegen zulässig.

		\points{8}

		(Hinweis für die Klausur: Abgabe auf Datenträger ist erlaubt und erwünscht,
		aber nicht zwingend.)

		\solution

		Um die Ziffern von \lstinline{x} zur Basis 16 zu isolieren,
		berechnen wir \lstinline{x % 16} (modulo 16 = Rest bei Division durch 16)
		und dividieren anschließend \lstinline{x} durch 16,
		solange bis \lstinline{x} den Wert 0 erreicht.

		Wenn wir die auf diese Weise ermittelten Ziffern direkt ausgeben,
		sind sie \emph{Little-Endian}, erscheinen also in umgekehrter Reihenfolge.
		Die Datei \gitfile{hp}{2023ws/20231102}{loesung-2-1.c} setzt diesen Zwischenschritt um.

		Die Ausgabe der Ziffern erfolgt in \gitfile{hp}{2023ws/20231102}{loesung-2-1.c}
		über \lstinline{printf ("%d")}
		für die Ziffern 0 bis 9. Für die darüberliegenden Ziffern
		wird der Buchstabe \lstinline{a} um die Ziffer abzüglich 10 inkrementiert
		und der erhaltene Wert mit \lstinline{printf ("%c")} als Zeichen ausgegeben.

		Um die umgekehrte Reihenfolge zu beheben,
		speichern wir die Ziffern von \lstinline{x}
		in einem Array \lstinline{digits[]} zwischen
		und geben sie anschließend in einer zweiten Schleife
		in umgekehrter Reihenfolge aus (siehe \gitfile{hp}{2023ws/20231102}{loesung-2-2.c}).
		Da wir wissen, daß \lstinline{x} eine 32-Bit-Zahl ist
		und daher höchstens 8 Hexadezimalziffern haben kann,
		ist 8 eine sinnvolle Länge für das Ziffern-Array \lstinline{digits[8]}.

		Nun sind die Ziffern in der richtigen Reihenfolge,
		aber wir erhalten zusätzlich zu den eigentlichen Ziffern führende Nullen.
		Da in der Aufgabenstellung nicht von führenden Nullen die Rede war,
		sind diese nicht verboten; \gitfile{hp}{2023ws/20231102}{loesung-2-2.c} ist daher
		eine richtige Lösung der Aufgabe.

		\breath

		Wenn wir die führenden Nullen vermeiden wollen,
		können wir die \lstinline{for}-Schleifen durch \lstinline{while}-Schleifen ersetzen.
		Die erste Schleife zählt hoch, solange \lstinline{x} ungleich 0 ist;
		die zweite zählt von dem erreichten Wert aus wieder herunter
		-- siehe \gitfile{hp}{2023ws/20231102}{loesung-2-3.c}.
		Da wir wissen, daß die Zahl \lstinline{x} höchstens 32 Bit,
		also höchstens 8 Hexadezimalziffern hat,
		wissen wir, daß \lstinline{i} höchstens den Wert 8 erreichen kann,
		das Array also nicht überlaufen wird.

		Man beachte, daß der Array-Index nach der ersten Schleife "`um einen zu hoch"' ist.
		In der zweiten Schleife muß daher \emph{zuerst\/} der Index dekrementiert werden.
		Erst danach darf ein Zugriff auf \lstinline{digit[i]} erfolgen.

		\breath

		Alternativ können wir auch mitschreiben,
		ob bereits eine Ziffer ungleich Null ausgegeben wurde,
		und andernfalls die Ausgabe von Null-Ziffern unterdrücken
		-- siehe \gitfile{hp}{2023ws/20231102}{loesung-2-4.c}.

		\breath

		Weitere Möglichkeiten ergeben sich, wenn man bedenkt,
		daß eine Hexadezimalziffer genau einer Gruppe von vier Binärziffern entspricht.
		Eine Bitverschiebung um 4 nach rechts
		ist daher dasselbe wie eine Division durch 16,
		und eine Und-Verknüpfung mit 15$_{10}$ = f$_{16}$ = 1111$_2$
		ist dasselbe wie die Operation Modulo 16.
		Die Datei \gitfile{hp}{2023ws/20231102}{loesung-2-5.c} ist eine in dieser Weise abgewandelte Variante
		von \gitfile{hp}{2023ws/20231102}{loesung-2-3.c}.

		Mit dieser Methode kann man nicht nur auf die jeweils unterste Ziffer,
		sondern auf alle Ziffern direkt zugreifen.
		Damit ist kein Array als zusätzlicher Speicher mehr nötig.
		Die Datei \gitfile{hp}{2023ws/20231102}{loesung-2-6.c} setzt dies auf einfache Weise um.
		Sie gibt wieder führende Nullen mit aus,
		ist aber trotzdem eine weitere richtige Lösung der Aufgabe.

		Die führenden Nullen ließen sich auf die gleiche Weise vermeiden
		wie in \gitfile{hp}{2023ws/20231102}{loesung-2-4.c}.

		Die Bitverschiebungsmethode hat den Vorteil,
		daß kein zusätzliches Array benötigt wird.
		Auch wird die als Parameter übergebene Zahl \lstinline{x} nicht verändert,
		was bei größeren Zahlen, die über Zeiger übergeben werden, von Vorteil sein kann.
		Demgegenüber steht der Nachteil,
		daß diese Methode nur für eine ganze Anzahl von Bits funktioniert,
		also für Basen, die Zweierpotenzen sind (z.\,B.\ 2, 8, 16, 256).
		Für alle anderen Basen (z.\,B.\ 10) eignet sich nur die Methode
		mit Division und Modulo-Operation.

		\exercise{Einfügen in Strings}

		Wir betrachten das folgende Programm (\gitfile{hp}{2023ws/20231102}{aufgabe-3.c}):
		% \begin{lstlisting}[style=numbered]
		\begin{lstlisting}
		#include <stdio.h>
		#include <string.h>

		void insert_into_string (char src, char *target, int pos)
		{
		int len = strlen (target);
		for (int i = pos; i < len; i++)
		target[i+1] = target[i];
		target[pos] = src;
		}

		int main (void)
		{
		char test[100] = "Hochshule Bochum";
		insert_into_string ('c', test, 5);
		printf ("%s\n", test);
		return 0;
		}
		\end{lstlisting}
		Die Ausgabe des Programms lautet:
		\lstinline[style=terminal]{Hochschhhhhhhhhhh}

		\begin{enumerate}[\quad(a)]
		\item
		Erklären Sie, wie die Ausgabe zustandekommt.
		\points{3}
		% \workspace{12}
		\item
		Schreiben Sie die Funktion \lstinline\|insert_into_string()\| so um,
		daß sie den Buchstaben \lstinline{src} an der Stelle \lstinline{pos}
		in den String \lstinline{target} einfügt.\par
		Die Ausgabe des Programms müßte dann
		\lstinline[style=terminal]{Hochschule Bochum} lauten.
		\points{2}
		% \workspace{13}
		\item
		Was kann passieren, wenn Sie die Zeile
		\lstinline{char test[100] = "Hochshule Bochum";}\\
		durch
		\lstinline{char test[] = "Hochshule Bochum";} ersetzen?
		Begründen Sie Ihre Antwort.
		\points{2}
		% \workspace{10}
		\item
		Was kann passieren, wenn Sie die Zeile
		\lstinline{char test[100] = "Hochshule Bochum";}\\
		durch
		\lstinline{char *test = "Hochshule Bochum";} ersetzen?
		Begründen Sie Ihre Antwort.
		\points{2}
		% \workspace{10}
		% \item
		% Schreiben Sie eine Funktion
		% \lstinline{void insert_into_string_sorted (char src, char *target)},
		% die voraussetzt, daß der String \lstinline{target} alphabetisch sortiert ist
		% und den Buchstaben \lstinline{src} an der alphabetisch richtigen Stelle
		% einfügt. Diese Funktion darf die bereits vorhandene Funktion
		% \lstinline\|insert_into_string()\| aufrufen.\\
		% \points{4}\par
		% Zum Testen eignen sich die folgenden Zeilen im Hauptprogramm:
		% \begin{lstlisting}[gobble=8]
		% char test[100] = "";
		% insert_into_string_sorted ('c', test);
		% insert_into_string_sorted ('a', test);
		% insert_into_string_sorted ('d', test);
		% insert_into_string_sorted ('b', test);
		% \end{lstlisting}
		% Danach sollte \lstinline{test[]} die Zeichenfolge \lstinline{"abcd"} enthalten.
		% \workspace{14}
		% \item
		% Wie schnell (Landau-Symbol in Abhängigkeit von der Länge $n$ des Strings)
		% arbeitet Ihre Funktion
		% \lstinline{void insert_into_string_sorted (char src, char *target)}
		% und warum?
		% \points{1}
		% \workspace{10}
		% \item
		% Beschreiben Sie -- in Worten oder als C-Quelltext --, wie man die Funktion\\
		% \lstinline{void insert_into_string_sorted (char src, char *target)}
		% so gestalten kann,\\
		% daß sie in $\mathcal{O}(\log n)$ arbeitet.
		% \points{3}
		% \workspace{35}
		\end{enumerate}

		\solution

		\begin{enumerate}[\quad(a)]
		\item
		\textbf{Erklären Sie, wie die Ausgabe zustandekommt.}

		In der Schleife wird \emph{zuerst\/} der nächste Buchstabe \lstinline{target[i + 1]}
		gleich dem aktuellen gesetzt
		und \emph{danach\/} der Zähler \lstinline{i} erhöht.
		Dadurch wird im nächsten Schleifendurchlauf der bereits verschobene Buchstabe
		noch weiter geschoben und letztlich alle Buchstaben in \lstinline{target[]}
		durch den an der Stelle \lstinline{pos} ersetzt.

		\item
		\textbf{Schreiben Sie die Funktion \lstinline\|insert_into_string()\| so um,
		daß sie den Buchstben \lstinline{src} an der Stelle \lstinline{pos}
		in den String \lstinline{target} einfügt.}\par
		\textbf{Die Ausgabe des Programms müßte dann
		\lstinline[style=terminal]{Hochschule Bochum} lauten.}

		Um im String "`Platz zu schaffen"', muß man von hinten beginnen,
		also die Schleife umdrehen\\
		(siehe: \gitfile{hp}{2023ws/20231102}{loesung-3.c}):
		\begin{lstlisting}{gobble=8}
		for (int i = len; i >= pos; i--)
		target[i + 1] = target[i];
		\end{lstlisting}

		\item
		\textbf{Was kann passieren, wenn Sie die Zeile
		\lstinline{char test[100] = "Hochshule Bochum";}\\
		durch
		\lstinline{char test[] = "Hochshule Bochum";}
		ersetzen und warum?}

		Die Schreibweise \lstinline{test[]} bedeutet,
		daß der Compiler selbst zählt, wieviel Speicherplatz der String benötigt,
		un dann genau die richtige Menge Speicher reserviert
		(anstatt, wie wir es manuell getan haben, pauschal Platz für 100 Zeichen).

		Wenn wir nun in den String ein zusätzliches Zeichen einfügen,
		ist dafür kein Speicherplatz reserviert worden,
		und wir \textbf{überschreiben} dann Speicher, an dem sich andere Variable befinden,
		was zu einem \textbf{Absturz} führen kann.

		Da wir hier nur ein einziges Zeichen schreiben,
		wird dieser Fehler nicht sofort auffallen.
		Dies ist schlimmer, als wenn das Programm direkt beim ersten Test abstürzt,
		denn dadurch entsteht bei uns der Eindruck, es sei in Ordnung.
		Wenn danach der Fehler in einer Produktivumgebung auftritt,
		kann dadurch Schaden entstehen -- je nach Einsatzgebiet der Software
		u.\,U.\ erheblicher Vermögens-, Sach- und/oder Personenschaden
		(z.\,B.\ Absturz eines Raumflugkörpers).

		\item
		\textbf{Was kann passieren, wenn Sie
		\lstinline{char test[100] = "Hochshule Bochum";}\\
		durch
		\lstinline{char *test = "Hochshule Bochum";}
		ersetzen und warum?}

		In diesem Fall wird der Speicher für den eigentlichen String
		in einem unbenannten, \textbf{nicht schreibbaren} Teil des Speichers reserviert.
		Unser Versuch, dorthin ein zusätzliches Zeichen zu schreiben,
		fürt dann normalerweise zu einem \textbf{Absturz}.

		In manchen Systemen (Betriebssystem, Compiler, \dots)
		ist der Speicherbereich tatsächlich sehr wohl schreibbar.
		In diesem Fall tritt der Absturz nicht immer und nicht immer sofort auf --
		genau wie in Aufgabenteil (c).

		\end{enumerate}

		\end{document}

20231102/hp-uebung-20231102.pdf

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20231102/inttypes-03.txt

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Original line number	Diff line number	Diff line
		2^14 ist nicht "2 hoch 14", sondern "2 xor-verknüpft mit 14".
		Wie rechnet man das aus?

		Binär rechnen:
		14 = 15 - 1 = 1111 - 0001 = 1110
		(dezimal) (binär)

		2 = 10
		(dezimal) (binär)

		0010
		^ 1110
		----
		1100

		Alternative: 2^14 ist dasselbe wie 14^2

		1110
		^ 0010 <-- dieses eine Bit umklappen, die anderen beibehalten
		----
		1100

20231102/libraries-01.txt

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Original line number	Diff line number	Diff line
		cassini/home/peter/bo/2023ws/hp/20231102> ar rs libanswer.a answer.o
		ar: creating libanswer.a
		cassini/home/peter/bo/2023ws/hp/20231102> ls -l libanswer.a
		-rw-r--r-- 1 peter peter 1232 2. Nov 14:45 libanswer.a
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -Wall -O philosophy.c -lanswer -o philosophy /usr/bin/ld: cannot find -lanswer: Datei oder Verzeichnis nicht gefunden
		collect2: error: ld returned 1 exit status
		cassini/home/peter/bo/2023ws/hp/20231102> gcc -L . -Wall -O philosophy.c -lanswer -o philosophy
		cassini/home/peter/bo/2023ws/hp/20231102> ./philosophy
		The answer is 23.
		cassini/home/peter/bo/2023ws/hp/20231102>

20231102/number-systems-01.c

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Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int number[] = { 01, 02, 03, 04, 05, 06, 07, 08, 09, 10 }; // Achtung: Oktalzahlen!
		for (int i = 0; i < 10; i++)
		printf ("%d\n", number[i]);
		return 0;
		}

20231102/number-systems-02.c

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Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int number[] = { 023, 025, 027, // Achtung: Oktalzahlen!
		033, 035, 037,
		123, 125, 127,
		133, 135, 137 };
		for (int i = 0; i < 12; i++)
		printf ("%d\n", number[i]);
		return 0;
		}

20231109/blink-9.s

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Original line number	Diff line number	Diff line
		.file "blink-9.c"
		__SP_H__ = 0x3e
		__SP_L__ = 0x3d
		__SREG__ = 0x3f
		__tmp_reg__ = 0
		__zero_reg__ = 1
		.text
		.global __vector_1
		.type __vector_1, @function
		__vector_1:
		push r1
		push r0
		in r0,__SREG__
		push r0 ; #include <avr/io.h>
		clr __zero_reg__ ; #include <avr/interrupt.h>
		push r24 ; #include <stdint.h>
		/* prologue: Signal */ ;
		/* frame size = 0 */ ; #define F_CPU 16000000l
		/* stack size = 4 */ ; #include <util/delay.h>
		.L__stack_usage = 4 ;
		ldi r24,lo8(1) ; uint8_t key_pressed = 0;
		sts key_pressed,r24 ;
		/* epilogue start / ; ISR (INT0_vect) / PD2 */
		pop r24 ; {
		pop r0 ; key_pressed = 1;
		out __SREG__,r0 ; }
		pop r0 ;
		pop r1 ; int main (void)
		reti ; {
		.size __vector_1, .-__vector_1 ; cli ();
		.section .text.startup,"ax",@progbits ; EICRA = 1 << ISC00 \| 1 << ISC01; /* INT0: steigende Flanke */
		.global main ; EIMSK = 1 << INT0; /* INT0 einschalten */
		.type main, @function ; sei ();
		main: ; DDRD = 0xfb; /* binär: 1111 1011 */
		/* prologue: function / ; PORTD = 0x44; / binär: 0100 0100 */
		/* frame size = 0 */ ; while (1)
		/* stack size = 0 */ ; {
		.L__stack_usage = 0 ; while (!key_pressed)
		/* #APP / ; ; / just wait */
		; 17 "blink-9.c" 1 ; PORTD ^= 0x40;
		cli ; _delay_ms (1); /* Wartezeit zum Entprellen */
		; 0 "" 2 ; key_pressed = 0;
		/* #NOAPP */ ; }
		ldi r24,lo8(3) ; return 0;
		sts 105,r24 ; }
		ldi r24,lo8(1)
		out 0x1d,r24
		/* #APP */
		; 20 "blink-9.c" 1
		sei
		; 0 "" 2
		/* #NOAPP */
		ldi r24,lo8(-5)
		out 0xa,r24
		ldi r24,lo8(68)
		out 0xb,r24
		ldi r25,lo8(64)
		.L4:
		lds r24,key_pressed
		cpse r24,__zero_reg__
		rjmp .L7
		.L6:
		rjmp .L6
		.L7:
		in r24,0xb
		eor r24,r25
		out 0xb,r24
		ldi r30,lo8(3999)
		ldi r31,hi8(3999)
		1: sbiw r30,1
		brne 1b
		rjmp .
		nop
		sts key_pressed,__zero_reg__
		rjmp .L4
		.size main, .-main
		.global key_pressed
		.section .bss
		.type key_pressed, @object
		.size key_pressed, 1
		key_pressed:
		.zero 1
		.ident "GCC: (GNU) 5.4.0"
		.global __do_clear_bss

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		\babel@toc {german}{}\relax
		\beamer@sectionintoc {3}{Bibliotheken}{3}{0}{1}
		\beamer@subsectionintoc {3}{4}{Callbacks}{3}{0}{1}
		\beamer@subsectionintoc {3}{5}{Projekt organisieren: make}{5}{0}{1}
		\beamer@sectionintoc {4}{Hardwarenahe Programmierung}{10}{0}{2}
		\beamer@subsectionintoc {4}{1}{Bit-Operationen}{10}{0}{2}
		\beamer@subsubsectionintoc {4}{1}{1}{Zahlensysteme}{10}{0}{2}
		\beamer@subsubsectionintoc {4}{1}{2}{Bit-Operationen in C}{12}{0}{2}
		\beamer@subsectionintoc {4}{2}{I/O-Ports}{14}{0}{2}
		\beamer@subsectionintoc {4}{3}{Interrupts}{17}{0}{2}
		\beamer@subsectionintoc {4}{4}{volatile-Variable}{19}{0}{2}

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		../common/gitlab.png
		No newline at end of file

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Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		printf ("Hello, world!\n");
		return 0;
		}
		cassini/home/peter/bo/2024ws/hp/20241010> gcc hello-01.c -o hello-01
		cassini/home/peter/bo/2024ws/hp/20241010> ls -l hello-*
		-rwxr-xr-x 1 peter peter 15952 10. Okt 12:02 hello-01
		-rw-r--r-- 1 peter peter 82 10. Okt 12:00 hello-01.c
		-rw-r--r-- 1 peter peter 1261 8. Okt 2023 hello-gtk.c
		cassini/home/peter/bo/2024ws/hp/20241010> ./hello-01
		Hello, world!
		cassini/home/peter/bo/2024ws/hp/20241010>

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		#include <stdio.h>

		int main (void)
		{
		printf ("Die Antwort lautet: %d\n", 42);
		return 0;
		}

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20231005/hp-20231005-fig1.pdf→20241010/hp-20241010-fig1.pdf

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20231005/hp-20231005.tex→20241010/hp-20241010.tex

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Original line number	Diff line number	Diff line
		% hp-20231005.pdf - Lecture Slides on Low-Level Programming
		% Copyright (C) 2012, 2013, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023 Peter Gerwinski
		% hp-20241010.pdf - Lecture Slides on Low-Level Programming
		% Copyright (C) 2012, 2013, 2015-2023, 2024 Peter Gerwinski
		%
		% This document is free software: you can redistribute it and/or
		% modify it either under the terms of the Creative Commons
		@@ -20,7 +20,7 @@
		% Attribution-ShareAlike 3.0 Unported License along with this
		% document. If not, see <http://creativecommons.org/licenses/>.

		% README: Einführung, Einführung in C (bis 2.7: Strukturierte Programmierung)
		% README: Einführung, Einführung in C

		\documentclass[10pt,t]{beamer}

		@@ -41,7 +41,7 @@

		\title{Hardwarenahe Programmierung}
		\author{Prof.\ Dr.\ rer.\ nat.\ Peter Gerwinski}
		\date{5.\ Oktober 2023}
		\date{10.\ Oktober 2024}

		\begin{document}

		@@ -49,11 +49,11 @@

		\date{\begin{picture}(0,0)
		\color{red}
		\put(0.55,1.05){\makebox(0,0)[t]{$\underbrace{\rule{1.45cm}{0pt}}_{%
		\put(0.65,1.05){\makebox(0,0)[t]{$\underbrace{\rule{1.45cm}{0pt}}_{%
		\mbox{\emph{rerum naturalium\/} = der natürlichen Dinge (lat.)}}$}}
		\put(1.55,-3){\makebox(0,0)[bl]{\redurl{https://www.peter.gerwinski.de/physik/}}}
		\end{picture}%
		5.\ Oktober 2023}
		10.\ Oktober 2024}

		\maketitleframe

		@@ -305,11 +305,11 @@

		\end{frame}

		\sectionnonumber{Zu dieser Lehrveranstaltung}
		\nosectionnonumber{Zu dieser Lehrveranstaltung}

		\begin{frame}

		\showsectionnonumber
		\shownosectionnonumber

		\begin{itemize}
		\item
		@@ -385,7 +385,7 @@

		\end{frame}

		\section{Einführung}
		\setcounter{section}{1}
		\section{Einführung in C}
		\subsection{Hello, world!}

20241010/hp-uebung-20241010.pdf

0 → 100644

+37.5 KiB

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20231005/hp-uebung-20231005.tex→20241010/hp-uebung-20241010.tex

+8 −7

Original line number	Diff line number	Diff line
		% hp-uebung-20231005.pdf - Exercises on Low-Level Programming / Applied Computer Sciences
		% Copyright (C) 2013, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023 Peter Gerwinski
		% hp-uebung-20241010.pdf - Exercises on Low-Level Programming / Applied Computer Sciences
		% Copyright (C) 2013, 2015-2023, 2024 Peter Gerwinski
		%
		% This document is free software: you can redistribute it and/or
		% modify it either under the terms of the Creative Commons
		@@ -20,7 +20,7 @@
		% Attribution-ShareAlike 3.0 Unported License along with this
		% document. If not, see <http://creativecommons.org/licenses/>.

		% README: Hello-World-Programme, Schaltjahr ermitteln, Maximum berechnen
		% README: 0: Hello-World-Programme, Schaltjahr ermitteln, Maximum berechnen

		\documentclass[a4paper]{article}

		@@ -33,18 +33,18 @@
		\thispagestyle{empty}

		\section*{Hardwarenahe Programmierung\\
		Übungsaufgaben -- 5.\ Oktober 2023}
		Übungsaufgaben 0 -- 10.\ Oktober 2024}

		\exercise{Hello, world!}

		Unter \url{https://gitlab.cvh-server.de/pgerwinski/hp/tree/2023ws/20231005}
		Unter \url{https://gitlab.cvh-server.de/pgerwinski/hp/tree/2024ws/20241010}
		können Sie
		\href{https://gitlab.cvh-server.de/pgerwinski/hp/raw/2023ws/20231005/hp-uebung-20231005.pdf}%
		{diesen Übungszettel als PDF-Datei}
		herunterladen sowie die folgenden Beispielprogramme:
		\begin{enumerate}[\quad(a)]
		\item \gitfile{hp}{2023ws/20231005}{hello.c}
		\item \gitfile{hp}{2023ws/20231005}{hello-gtk.c}
		\item \gitfile{hp}{2024ws/20241010}{hello.c}
		\item \gitfile{hp}{2024ws/20241010}{hello-gtk.c}
		\end{enumerate}
		Bringen Sie diese Programme auf Ihrem eigenen Rechner
		und/oder auf einem Rechner der Hochschule
		@@ -59,6 +59,7 @@

		\textbf{Hinweis 0:} Diese PDF-Datei enthält Links zu den jeweiligen Dateien.
		Durch Anklicken können Sie diese direkt herunterladen.
		(Falls dies nicht funktionieren sollte, bitte bescheidsagen.)

		\textbf{Hinweis 1:} Für (b) benötigen Sie die Bibliothek GTK+.
		Diese finden Sie -- einschließlich Dokumentation -- als freie Software im Internet.

20231005/if-01.c→20241010/if-01.c

+3 −4

Original line number	Diff line number	Diff line
		@@ -2,10 +2,9 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		printf ("a / b = %d\n", a / b);
		return 0;

20231005/if-10.c→20241010/if-02.c

+4 −5

Original line number	Diff line number	Diff line
		@@ -2,14 +2,13 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b != 0)
		printf ("a / b = %d\n", a / b);
		else
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("Bitte nicht durch 0 dividieren!\n);
		return 0;
		}

20231005/if-02.c→20241010/if-03.c

+14 −0

Original line number	Diff line number	Diff line
		@@ -2,14 +2,13 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b = 0)
		printf ("Bitte nicht durch 0 teilen.\n");
		else
		if (b != 0)
		printf ("a / b = %d\n", a / b);
		else
		printf ("Bitte nicht durch 0 dividieren!\n");
		return 0;
		}

20231005/if-11.c→20241010/if-04.c

+14 −0

Original line number	Diff line number	Diff line
		@@ -2,14 +2,13 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b)
		printf ("a / b = %d\n", a / b);
		else
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("Bitte nicht durch 0 dividieren!\n");
		return 0;
		}

20231005/if-03.c→20241010/if-05.c

+15 −0

Original line number	Diff line number	Diff line
		@@ -2,14 +2,14 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b == 0)
		printf ("Bitte nicht durch 0 teilen.\n");
		else
		if (b)
		printf ("a / b = %d\n", a / b);
		else
		printf ("Bitte nicht durch 0 dividieren!\n");
		printf ("Das macht man nicht!\n");
		return 0;
		}

20231005/if-09.c→20241010/if-06.c

+16 −0

Original line number	Diff line number	Diff line
		@@ -2,17 +2,15 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b == 0) {
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("Das macht man nicht.\n");
		}
		else {
		if (b)
		printf ("a / b = %d\n", a / b);
		else {
		printf ("Bitte nicht durch 0 dividieren!\n");
		printf ("Das macht man nicht!\n");
		}
		return 0;
		}

20231005/if-06.c→20241010/if-07.c

+17 −0

Original line number	Diff line number	Diff line
		@@ -2,17 +2,16 @@

		int main (void)
		{
		int a = 360;
		printf ("a = %d\n", a);
		int b;
		printf ("Bitte b eingeben: ");
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b == 0)
		if (b)
		printf ("a / b = %d\n", a / b);
		else
		{
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("Das macht man nicht.\n");
		printf ("Bitte nicht durch 0 dividieren!\n");
		printf ("Das macht man nicht!\n");
		}
		else
		printf ("a / b = %d\n", a / b);
		return 0;
		}

20241010/if-07a.pas

0 → 100644

+18 −0

Original line number	Diff line number	Diff line
		program if_07a;

		var
		a: Integer = 42;
		b: Integer = 0;

		begin
		WriteLn ('a = 42');
		Write ('Bitte b eingeben: ');
		ReadLn (b);
		if b <> 0 then
		WriteLn ('a / b = ', a / b)
		else
		begin
		WriteLn ('Bitte nicht durch 0 dividieren!');
		WriteLn ('Das macht man nicht!')
		end
		end.

20241010/if-07b.pas

0 → 100644

+18 −0

Original line number	Diff line number	Diff line
		program if_07b;

		var
		a: Integer = 42;
		b: Integer = 0;

		begin
		WriteLn ('a = 42');
		Write ('Bitte b eingeben: ');
		ReadLn (b);
		if b <> 0 then
		WriteLn ('a / b = ', a div b)
		else
		begin
		WriteLn ('Bitte nicht durch 0 dividieren!');
		WriteLn ('Das macht man nicht!')
		end
		end.

20241010/if-07c.c

0 → 100644

+17 −0

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 42;
		int b = 0;
		printf ("a = %d\nBitte b eingeben: ", a);
		scanf ("%d", &b);
		if (b)
		printf ("a / b = %d\n", a / b);
		else
		{
		printf ("Bitte nicht durch 0 dividieren!\n");
		printf ("Das macht man nicht!\n");
		}
		return 0;
		}

20241010/if-07c.pas

0 → 100644

+18 −0

Original line number	Diff line number	Diff line
		program if_07c;

		var
		a: Integer = 42;
		b: Integer = 0;

		begin
		WriteLn ('a = ', a);
		Write ('Bitte b eingeben: ');
		ReadLn (b);
		if b <> 0 then
		WriteLn ('a div b = ', a div b)
		else
		begin
		WriteLn ('Bitte nicht durch 0 dividieren!');
		WriteLn ('Das macht man nicht!')
		end
		end.

20241010/if-08.c

0 → 100644

+15 −0

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a = 42;
		int b = 0;
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b) printf ("a / b = %d\n", a / b); else
		{
		printf ("Bitte nicht durch 0 dividieren!\n");
		printf ("Das macht man nicht!\n");
		}
		return 0;
		}

20241010/if-09.c

0 → 100644

+4 −0

Original line number	Diff line number	Diff line
		#include <stdio.h>
		int main (void) { int a = 42; int b = 0; printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b); if (b) printf ("a / b = %d\n", a / b); else { printf
		("Bitte nicht durch 0 dividieren!\n"); printf ("Das macht man nicht!\n"); } return 0; }

20231012/side-effects-14.c→20241010/if-10.c

+17 −0

Original line number	Diff line number	Diff line
		@@ -2,14 +2,16 @@

		int main (void)
		{
		int a = 137;
		int a = 42;
		int b = 0;
		if (b == 0)
		printf ("a = 42\nBitte b eingeben: ");
		scanf ("%d", &b);
		if (b)
		printf ("a / b = %d\n", a / b);
		else
		{
		printf ("Bitte nicht durch 0 teilen.\n");
		printf ("So was macht man nicht.\n");
		printf ("Bitte nicht durch 0 dividieren!\n");
		printf ("Das macht man nicht!\n");
		}
		else
		printf ("a / b = %d\n", a / b);
		return 0;
		}

20231005/calc-01.c→20241010/input-01.c

+10 −0

Original line number	Diff line number	Diff line
		@@ -3,9 +3,8 @@
		int main (void)
		{
		int a;
		printf ("Bitte eine Zahl eingeben: ");
		printf ("Antwort? ");
		scanf ("%d", &a);
		a = 2 * a;
		printf ("Das Doppelte ist: %d\n", a);
		printf ("Die Antwort lautete: %d\n", a);
		return 0;
		}

20231005/input-01.c→20241010/input-02.c

+2 −2

Original line number	Diff line number	Diff line
		@@ -3,8 +3,8 @@
		int main (void)
		{
		double a;
		printf ("Bitte eine Zahl eingeben: ");
		printf ("Antwort? ");
		scanf ("%lf", &a);
		printf ("Ihre Antwort war: %lf\n", a);
		printf ("Die Antwort lautete: %lf\n", a);
		return 0;
		}

20241010/input-03.c

0 → 100644

+10 −0

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a;
		printf ("Antwort? ");
		scanf ("%d", a);
		printf ("Die Antwort lautete: %d\n", a);
		return 0;
		}

20241010/input-04.c

0 → 100644

+11 −0

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a;
		printf ("Antwort? ");
		scanf ("%d", a);
		a = a + 2;
		printf ("Die Antwort + 2 lautete: %d\n", a);
		return 0;
		}

20241010/input-05.c

0 → 100644

+11 −0

Original line number	Diff line number	Diff line
		#include <stdio.h>

		int main (void)
		{
		int a;
		printf ("Antwort? ");
		scanf ("%d", &a);
		a = a + 2;
		printf ("Die Antwort + 2 lautete: %d\n", a);
		return 0;
		}

20231005/logo-hochschule-bochum-cvh-text-v2.pdf→20241010/logo-hochschule-bochum-cvh-text-v2.pdf

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