Question

This program below moves the last (junior) and the penultimate bytes variable i type int. I’m trying to understand why the programmer wrote this

i = (i & LEADING_TWO_BYTES_MASK) | ((i & PENULTIMATE_BYTE_MASK) >> 8) | ((i & LAST_BYTE_MASK) << 8);

Can anyone explain to me in plain English whats going on in the program below.

#include <stdio.h>
#include <cstdlib>

#define LAST_BYTE_MASK 255 //11111111
#define PENULTIMATE_BYTE_MASK 65280 //1111111100000000
#define LEADING_TWO_BYTES_MASK 4294901760 //11111111111111110000000000000000


int main(){
    unsigned int i = 0;
    printf("i = ");
    scanf("%d", &i);
    i = (i & LEADING_TWO_BYTES_MASK) | ((i & PENULTIMATE_BYTE_MASK) >> 8) | ((i & LAST_BYTE_MASK) << 8);
    printf("i = %d", i);
    system("pause");
}

Answer 1

The expression is indeed a bit convoluted but in essence the author does this:

// Mask out relevant bytes
unsigned higher_order_bytes = i & LEADING_TWO_BYTES_MASK;
unsigned first_byte = i & LAST_BYTE_MASK;
unsigned second_byte = i & PENULTIMATE_BYTE_MASK;

// Switch positions:
unsigned first_to_second = first_byte << 8;
unsigned second_to_first = second_byte >> 8;

// Concatenate back together:
unsigned result = higher_order_bytes | first_to_second | second_to_first;

Incidentally, defining the masks using hexadecimal notation is more readable than using decimal. Furthermore, using #define here is misguided. Both C and C++ have const:

unsigned const LEADING_TWO_BYTES_MASK = 0xFFFF0000;
unsigned const PENULTIMATE_BYTE_MASK = 0xFF00;
unsigned const LAST_BYTE_MASK = 0xFF;

To understand this code you need to know what &, | and bit shifts are doing on the bit level.