What would be the equivalent pointer expression for referring the array element a[i][j][k][l]
What is (void*)0?
What will be the output of the program?
#include<stdio.h> int main() { int arr[2][2][= {8}; int *p, *q; p = &arr[1][1][1]; q = (int*) arr; printf("%d, %d\n", *p, *q); return }
What will be the output of the program ?
#include<stdio.h> #include<string.h> int main() { int i, n; char *x="Alice"; n = strlen(x); *x = x[n]; for(i=i<=n; i++) { printf("%s ", x); x++; } printf("\n", x); return}
A pointer is
Can you combine the following two statements into one?
char *p; p = (char*) malloc(100);
What will be the output of the program assuming that the array begins at the locationand size of an integer is 4 bytes?
#include<stdio.h> int main() { int a[3][= {}; printf("%u, %u, %u\n", a[0]+*(a[0]+1), *(*(a+0)+1)); return }
#include<stdio.h> int main() { int i, a[] = {10}; change(a, 5); for(i=i<=i++) printf("%d, ", a[i]); return} void change(int *b, int n) { int i; for(i=i<n; i++) *(b+= *(b+i)+}
#include<stdio.h> void fun(void *p); int i; int main() { void *vptr; vptr = &i; fun(vptr); return } void fun(void *p) { int **q; q = (int**)&p; printf("%d\n", **q); }
The operator used to get value at address stored in a pointer variable is
In which header file is the NULL macro defined?
#include<stdio.h> int main() { int arr[= {4}; char *p; p = arr; p = (char*)((int*)(p)); printf("%d, ", *p); p = (int*)(p+1); printf("%d", *p); return }
If the size of integer is 4bytes, What will be the output of the program?
#include<stdio.h> int main() { int arr[] = {16}; printf("%d, %d, %d\n", sizeof(arr), sizeof(*arr), sizeof(arr[0])); return}
#include<stdio.h> int main() { char *str; str = "%s"; printf(str, "K\n"); return}
How many bytes are occupied by near, far and huge pointers (DOS)?
#include<stdio.h> int main() { char *str; str = "%d\n"; str++; str++; printf(str-300); return}
#include<stdio.h> int *check(static int, static int); int main() { int *c; c = check(20); printf("%d\n", c); return} int *check(static int i, static int j) { int *p, *q; p = &i; q = &j; if(i >= return (p); else return (q); }
#include<stdio.h> int main() { char str[] = "peace"; char *s = str; printf("%s\n", s++ +3); return}
#include<stdio.h> int main() { static char *s[] = {"black", "white", "pink", "violet"}; char **ptr[] = {s+s+s+s}, ***p; p = ptr; ++p; printf("%s", **p+1); return}
Point out the compile time error in the program given below.
#include<stdio.h> int main() { int *x; *x= return}
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