#include<stdio.h> int main() { int i; printf("%d\n", scanf("%d", &i)); return 0; }
printf("%d\n", scanf("%d", &i)); The scanf function returns the value 1(one).
Therefore, the output of the program is '1'.
#include<stdio.h> int X=40; int main() { int X=20; printf("%d\n", X); return 0; }
/* myprog.c */ #include<stdio.h> #include<stdlib.h> int main(int argc, char **argv) { int i; for(i=1; i<=3; i++) printf("%u\n", &argv[i]); return 0; }If the first value printed by the above program is 65517, what will be the rest of output?
#include<stdio.h> #include<stdlib.h> int main() { int *p; p = (int *)malloc(20); /* Assume p has address of 1314 */ free(p); printf("%u", p); return 0; }
#include<stdio.h> int main() { int arr[2][2][2] = {10, 2, 3, 4, 5, 6, 7, 8}; int *p, *q; p = &arr[1][1][1]; q = (int*) arr; printf("%d, %d\n", *p, *q); return 0; }
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