Question
#include<stdio.h>
int main()
{
char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"};
printf("%d, %d", sizeof(str), strlen(str[0]));
return 0;
}
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;
}
Answer: Option C
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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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