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  1. If the size of pointer is 4 bytes then What will be the output of the program ?

    #include 

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

    2. A.

    22, 4
    B.

    25, 5
    C.

    24, 5
    D.

    20, 2

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    Workspace

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    Answer : Option C

    Explanation :

    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).

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