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stiti
2024-02-01 13:55:03 +01:00
parent 4fe273c309
commit 113583b37a
228 changed files with 7094 additions and 0 deletions
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17
BUT1/DEV1.1/Entiers/arithmetique.c Normal file
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#include <stdio.h>
#include <stdlib.h>
int main(void)
{
printf("%d\n", 100 / 6);
printf("%d\n", 100 % 6);
printf("%d\n", 0x1A * 015);
printf("%d\n", -3 / 5);
printf("%d\n", -31 / 5);
printf("%d\n", -31 % 5);
printf("%d\n", 100 * (3 / 5));
printf("%d\n", 100 * 3 / 5);
printf("%d\n", 2 - 3 - 5);
printf("%d\n", 2 - (3 - 5));
return EXIT_SUCCESS;
}

10
BUT1/DEV1.1/Entiers/bases.c Normal file
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#include <stdio.h>
#include <stdlib.h>
int main(void)
{
printf("%d\n", 72);
printf("%o\n", 72);
printf("%x\n", 72);
return EXIT_SUCCESS;
}

8
BUT1/DEV1.1/Entiers/multiplication.c Normal file
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#include <stdio.h>
#include <stdlib.h>
int main(void)
{
printf("%d\n", 73 << 4);
return EXIT_SUCCESS;
}

17
BUT1/DEV1.1/Entiers/operations.c Normal file
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#include <stdio.h>
#include <stdlib.h>
int main(void) {
int n = 12;
printf("%d", (n>>7)&1);
printf("%d", (n>>6)&1);
printf("%d", (n>>5)&1);
printf("%d", (n>>4)&1);
printf("%d", (n>>3)&1);
printf("%d", (n>>2)&1);
printf("%d", (n>>1)&1);
printf("%d", n&1);
printf("\n");
return EXIT_SUCCESS;
}

17
BUT1/DEV1.1/Entiers/operations2.c Normal file
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#include <stdio.h>
#include <stdlib.h>
int main(void) {
int n = 12|35;
printf("%d", (n>>7)&1);
printf("%d", (n>>6)&1);
printf("%d", (n>>5)&1);
printf("%d", (n>>4)&1);
printf("%d", (n>>3)&1);
printf("%d", (n>>2)&1);
printf("%d", (n>>1)&1);
printf("%d", n&1);
printf("\n");
return EXIT_SUCCESS;
}

18
BUT1/DEV1.1/Entiers/operations3.c Normal file
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#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int n = 12;
printf("%.32d", (n >> 7) & 1);
printf("%.32d", (n >> 6) & 1);
printf("%.32d", (n >> 5) & 1);
printf("%.32d", (n >> 4) & 1);
printf("%.32d", (n >> 3) & 1);
printf("%.32d", (n >> 2) & 1);
printf("%.32d", (n >> 1) & 1);
printf("%.32d", n & 1);
printf("\n");
return EXIT_SUCCESS;
}

6
BUT1/DEV1.1/Entiers/reponses.txt Normal file
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5) Opérations :
le programme suivant affiche le nombre 0000 1100
On peut en déduire que c'est du binaire
le programme suivant a transformé le nombre 12 écrit en décimal (base 10) en nombre binaire (base 2)
Si on remplace 12 par 35 on pour avoir 00100011