Q1: If the size of pointer is 4 bytes then What will be the output of the program ?

#include<stdio.h>

int main()
{
    char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"};
    printf("%d, %d", sizeof(str), strlen(str[0]));
    return 0;
}

ANS:C - 24, 5

Step 1: char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; The variable str is declared as an pointer to the array of 6 strings. 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).