char *p; p
Can you combine the following two statements into one? char *p; p = (char*) malloc(100);
char *p; p = (char*) malloc(100);
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1) What will be the output of the following C code?#include <stdio.h> int main() { char *p = NULL; char *q = 0; if (p) printf(" p "); else printf("nullp"); if (q) printf("q\n"); else printf(" nullq\n"); }
#include <stdio.h> int main() { char *p = NULL; char *q = 0; if (p) printf(" p "); else printf("nullp"); if (q) printf("q\n"); else printf(" nullq\n"); }
2) Which is an indirection operator among the following?
3) What would be the equivalent pointer expression for referring the array element a[i][j][k][l]
4) What will be the output of the following C code? #include <stdio.h> int main() { int i = 10; void *p = &i; printf("%d\n", (int)*p); return 0; }
#include <stdio.h> int main() { int i = 10; void *p = &i; printf("%d\n", (int)*p); return 0; }
5) What will be the output of the program ? #include<stdio.h> int main() { static char *s[] = {"black", "white", "pink", "violet"}; char **ptr[] = {s+3, s+2, s+1, s}, ***p; p = ptr; ++p; printf("%s", **p+1); return 0; }
#include<stdio.h> int main() { static char *s[] = {"black", "white", "pink", "violet"}; char **ptr[] = {s+3, s+2, s+1, s}, ***p; p = ptr; ++p; printf("%s", **p+1); return 0; }
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