What will be output of the following program? Assume that you are running this program in little-endian processor.
#include<stdio.h> int main() {
short a = 320;
char * ptr;
ptr = (char * ) & a;
printf("%d", * ptr);
return 0;
} |
(A) 1
(B) 320
(C) 64
(D) Compilation error
Answer: (C)
Explanation: Big Endian Byte Order: The most significant byte of the data is placed at the byte with the lowest address. The rest of the data is placed in order in the next three bytes in memory.
Little Endian Byte Order: The least significant byte of the data is placed at the byte with the lowest address. The rest of the data is placed in order in the next three bytes in memory.
a= 320, 2-byte representation of a:
But as we are using little-endian format here, so it would actually stored in memory like :
ptr which is a character pointer, on dereferencing will output a single byte, i.e the first byte: (01000000)2 which is equal to 64.
So, the correct option is (C).