What is printed by the following ANSI C program? #include int main(int argc, char *argv[]) { int x = 1, z[2] = {10, 11}; int *p = NULL; p = &x; *p = 10; p = &z[1]; *(&z[0] + 1) += 3; printf("%d, %d, %d ", x, z[0], z[1]); return 0; }

Question

What is printed by the following ANSI C program?

#include<stdio.h>

int main(int argc, char *argv[])
{
 int x = 1, z[2] = {10, 11};
 int *p = NULL;
 p = &x;
 *p = 10;
 p = &z[1];
 *(&z[0] + 1) += 3;
 printf("%d, %d, %d
", x, z[0], z[1]);
 return 0;
}

Accepted Answer

10, 10, 14. The program modifies the value of 'x' through the pointer 'p', setting it to 10. It then updates 'z[1]' by adding 3 to 'z[0]', which is 10, resulting in 'z[1]' becoming 14. Therefore, the final output is '10, 10, 14'.

What is printed by the following ANSI C program? #include<stdio.h> int main(int argc, char argv[]) { int x = 1, z[2] = {10, 11}; int p = NULL; p = &x; p = 10; p = &z[1]; (&z[0] + 1) += 3; printf("%d, %d, %d ", x, z[0], z[1]); return 0; }

Solution

Related GATE Technical questions

What is printed by the following ANSI C program? #include<stdio.h> int main(int argc, char *argv[]) { int x = 1, z[2] = {10, 11}; int *p = NULL; p = &x; *p = 10; p = &z[1]; *(&z[0] + 1) += 3; printf("%d, %d, %d ", x, z[0], z[1]); return 0; }

Solution

Related GATE Technical questions

What is printed by the following ANSI C program? #include<stdio.h> int main(int argc, char argv[]) { int x = 1, z[2] = {10, 11}; int p = NULL; p = &x; p = 10; p = &z[1]; (&z[0] + 1) += 3; printf("%d, %d, %d ", x, z[0], z[1]); return 0; }