#include<stdio.h> int main() { char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; printf("%d, %d", sizeof(str), strlen(str[0])); return 0; }
Step 2: printf("%d, %d", sizeof(str), strlen(str[0]));
sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints '24'
strlen(str[0])); becomes strlen(Frogs)). Hence it prints '5';
Hence the output of the program is 24, 5
Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).
#include<stdio.h> int main() { char huge *near *far *ptr1; char near *far *huge *ptr2; char far *huge *near *ptr3; printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3)); return 0; }
#include<stdio.h> int main() { struct node { int data; struct node *link; }; struct node *p, *q; p = (struct node *) malloc(sizeof(struct node)); q = (struct node *) malloc(sizeof(struct node)); printf("%d, %d\n", sizeof(p), sizeof(q)); return 0; }
After sometime the stack memory will be filled completely. Hence stack overflow error will occur.
#include<stdio.h> int main() { int arr[] = {12, 13, 14, 15, 16}; printf("%d, %d, %d\n", sizeof(arr), sizeof(*arr), sizeof(arr[0])); return 0; }
void *cmp();
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