C Program: Convert to Binary and Change Endian-ness

[lolgasm]

Hello all!

This is my C program that is supposed to:
1. prompt user for integer
2. display integer in decimal
3. display binary value of integer
4. flip from little-endian to big-endian and display binary
5. display NEW big-endian in decimal
6. quit on user input 'q'.

First question: this program does not display the binary values properly.. where did I mess up?

Also:
Besides alternatives to scanf or printf, or using function calls to convert to binary, where can I improve my code? I'm looking for efficiency, easy implementation etc..

Thanks!!

#include <stdio.h>


int main(void)
{
    
    while (1)
    {
        short i;
        short j;
        char input[40];
        
        printf("Enter a 32-bit integer ('q' to quit): ");
        scanf("%s", input);
   
        if (input[0] == 'q')
            return 0;
            
        int number = atoi(input);


        char little[32];
        
        for( i = 31; i >= 0; i--) 
        {
            if( (1 << i) & number)
                little[31 - i] = '1';
            else
                little[31 - i] = '0';                   
        }
    
        char big[32];
        
        for ( j = 4; j > 0; j--)
        {
            for ( i = 8; i > 0; i--)
            {
                big[(8 * j) - i] = little[(8 * (5 - j)) - i];
            }
        }
        
        int flipped = 0;
        
        for( i = 31; i >= 0; i--) 
        {
            if ( big[i] = '1')
                flipped = flipped + 2^i;                 
        }
        
        printf("\n32-bit integer: %d\nLittle-endian binary: %s\nBig-endian binary: %s\nNew 32-bit integer: %d\n\n",number,little,big,flipped);
    
    }
    return 0;
}

[iMalc]

So, what are you inputting, what is it outputting, and what were you expecting it to output?

[oogabooga]

Why are i and j declared as short? Unless you are trying to save space (say in a large array) you should always use int, even if the numbers will be small.

You probably want == instead of = here

if ( big[i] = '1')


What do you think this means? Hint, it doesn't mean 2 to the power of i !!!

2^i

Use instead what you did earlier:

(1 << i)


Overall, though, you seem to be misunderstanding little and big endian. They don't have every bit in reverse order, just the bytes. So for two bytes, if this was big endian

11110000 00001111

then this would be little endian

00001111 11110000

[lolgasm]

input:
5

output:

32-bit integer: 5
Little-endian binary: 00000000000000000000000000000101ÿÿ
Big-endian binary:11111111111111111111111111111111000000000000000000 00000000000101ÿÿ
New 32-bit integer: 64

EXPECTED output:

32-bit integer: 5
Little-endian binary: 00000000000000000000000000000101
Big-endian binary: 00000101000000000000000000000000
New 32-bit integer: 327680

[lolgasm]

Thanks for suggestions!

This line should alternate little/big endians as you described.

for ( j = 4; j > 0; j--)
        {
            for ( i = 8; i > 0; i--)
            {
                big[(8 * j) - i] = little[(8 * (5 - j)) - i];
            }
        }