Convert int to unsigned char array and backwards

Question

#include <stdio.h>
#include <stdlib.h>

int main(){
    int n = 56789000;
    unsigned char bytes[4];

    bytes[0] = (n >> 24) & 0xFF;
    bytes[1] = (n >> 16) & 0xFF;
    bytes[2] = (n >> 8) & 0xFF;
    bytes[3] = n & 0xFF;

    int test = (bytes[3] << 24) | (bytes[2] << 16) | (bytes[1] << 8) | (bytes[0]);

    printf("%d\n",n);
    printf("%d\n", test);
}

Output is:

56789000
143155715

Hello,

I'm trying to store an integer into an unsigned char array and want to convert it back to an integer later. I found some code-snippets which lead me to the code above but the output is not as expected. Can you please help to fix the code above. I'm not very familiar with C and don't know what is wrong with the code.

Thank in advance

I wouldn't since you're setting bytes[0] to the top 8 bits but then re-assembling with it as the lower 8 bits (and the same for the others). Try printing it in hex (%x) and all will become obvious. — The Archetypal Paul
– The Archetypal Paul, Commented Dec 18, 2014 at 11:02

Iharob Al Asimi · Accepted Answer · 2014-12-18 11:00:57Z

1

This line is in the opposite order

int test = (bytes[3] << 24) | (bytes[2] << 16) | (bytes[1] << 8) | (bytes[0]);

so

int test = (bytes[0] << 24) | (bytes[1] << 16) | (bytes[2] << 8) | (bytes[3]);

should work.

answered Dec 18, 2014 at 11:00

Iharob Al Asimi

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1 Comment

aQuip Over a year ago

Thanks a lot. Should have tried that. Have nice day.

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