How to initialize char array with hexadecimal values in C

1 Answer

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#include <stdio.h>

int main(void) {
    unsigned char arr[32] = {0x00  ,0x01  ,0x02  ,0x03  ,0x04  ,0x05  ,0x06  ,0x07  ,0x08  ,0x09  ,0x0a  ,0x0b  ,0x0c  ,0x0d  ,0x0e  ,0x0f};
    
    for (int i = 0; i < 16; i ++)
        printf("%x - %d\n", arr[i], arr[i]);
}



/*
run:

0 - 0
1 - 1
2 - 2
3 - 3
4 - 4
5 - 5
6 - 6
7 - 7
8 - 8
9 - 9
a - 10
b - 11
c - 12
d - 13
e - 14
f - 15

*/

 



answered Jan 21, 2021 by avibootz

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