How do I save a decimal number as hexadecimal on a single byte?

Question

#include <stdio.h>
int main()
{
    int x = 255;
    char hex[4] = {0};
    hex[0] = 0x02;
    hex[1] = 0x20;
    hex[2] = 0xef;
    hex[3] = x;
    for(int i = 0; i < 4; i++)
        printf("%X ", hex[i]);
}

This outputs " 2 20 FFFFFFEF FFFFFFFF". What should be changed so that it would output "02 20 EF FF"?

Claim Yang · Accepted Answer · 2017-01-01T04:32:50.323

0

The question is the data overflowed. char is -128 ~ 127, when 0xef = 239. So you can use type int or short or just unsigned char , like that:

#include <stdio.h>
int main()
{
    int x = 255;
    unsigned char hex[4] = {0};
    hex[0] = 0x02;
    hex[1] = 0x20;
    hex[2] = 0xef;
    hex[3] = x;
    for(int i = 0; i < 4; i++)
        printf("%02X ", hex[i]);
}

edited Jan 01 '17 at 04:32

answered Dec 31 '16 at 10:30

Claim Yang

309
1
9

It worked, thanks. Changing the char to unsigned char worked as well. – Matiz 12 Dec 31 '16 at 10:32
The signed-ness of `char` is implementation defined. And there is only a minimum guaranteed range for `((un)signed) char`. There is no guarantee the range is `-128..127`. – too honest for this site Dec 31 '16 at 14:18
Use `unsigned char` instead of int for x and hex if numbers are always below or equal to 255. – Sander Dec 31 '16 at 16:35

How do I save a decimal number as hexadecimal on a single byte?

1 Answers1