What is the total size in bytes of the structure on a 32-bit machine (int=4B, short=2B, 4B alignment enforced)?
struct Data {
short s;
int i;
short t;
};s takes 2B + 2B padding (4B). 'i' takes 4B (8B). 't' takes 2B + 2B trailing padding to align to 4B, yielding 12B.
What will be printed by the following code?
#include <stdio.h>
union U {
int x;
char c[4];
};
int main(void) {
union U u;
u.x = 0;
u.c[0] = 5;
u.c[1] = 2;
printf("%d", u.c[0] + u.c[1]);
return 0;
}u.c[0] = 5 and u.c[1] = 2. Sum = 5 + 2 = 7.
How many total bytes are allocated on the stack for the array declaration 'struct Node arr[5];' if sizeof(struct Node) is 16 bytes?
Total memory allocated = 5 elements * 16 bytes/element = 80 bytes.
What is the output of the C program?
#include <stdio.h>
struct Box {
int h, w, l;
};
int volume(struct Box b) {
return b.h * b.w * b.l;
}
int main(void) {
struct Box b = {2, 3, 4};
printf("%d", volume(b));
return 0;
}volume(b) calculates 2 * 3 * 4 = 24.
What is the printed value of total?
#include <stdio.h>
struct Num {
int val;
};
int main(void) {
struct Num a = {10}, b = {20}, c = {30};
struct Num arr[3] = {a, b, c};
int total = 0;
for(int i=0; i<3; i++) total += arr[i].val;
printf("%d", total);
return 0;
}total = 10 + 20 + 30 = 60.