Convert ascii char[] to hexadecimal char[] in C

16
votes

I am trying to convert a char[] in ASCII to char[] in hexadecimal.

Something like this:

hello --> 68656C6C6F

I want to read by keyboard the string. It has to be 16 characters long.

This is my code now. I don't know how to do that operation. I read about strol but I think it just convert str number to int hex...

#include <stdio.h>
main()
{
    int i = 0;
    char word[17];

    printf("Intro word:");

    fgets(word, 16, stdin);
    word[16] = '\0';
    for(i = 0; i<16; i++){
        printf("%c",word[i]);
    }
 }

I am using fgets because I read is better than fgets but I can change it if necessary.

Related to this, I am trying to convert the string read in a uint8_t array, joining each 2 bytes in one to get the hex number.

I have this function which I am using a lot in arduino so I think it should work in a normal C program without problems.

uint8_t* hex_decode(char *in, size_t len, uint8_t *out)
{
    unsigned int i, t, hn, ln;

    for (t = 0,i = 0; i < len; i+=2,++t) {

            hn = in[i] > '9' ? (in[i]|32) - 'a' + 10 : in[i] - '0';
            ln = in[i+1] > '9' ? (in[i+1]|32) - 'a' + 10 : in[i+1] - '0';

            out[t] = (hn << 4 ) | ln;
            printf("%s",out[t]);
    }
    return out;

}

But, whenever I call that function in my code, I get a segmentation fault.

Adding this code to the code of the first answer:

    uint8_t* out;
    hex_decode(key_DM, sizeof(out_key), out);

I tried to pass all necessary parameters and get in out array what I need but it fails...

4
chexascii
uint8_t* out; --> uint8_t* out = calloc(sizeof(out_key), sizeof(*out)); or strlen(key_DM)+1 instead of sizeof(out_key). – BLUEPIXY

4 Answers

14
votes
#include <stdio.h>
#include <string.h>

int main(void){
    char word[17], outword[33];//17:16+1, 33:16*2+1
    int i, len;

    printf("Intro word:");
    fgets(word, sizeof(word), stdin);
    len = strlen(word);
    if(word[len-1]=='\n')
        word[--len] = '\0';

    for(i = 0; i<len; i++){
        sprintf(outword+i*2, "%02X", word[i]);
    }
    printf("%s\n", outword);
    return 0;
}
5
votes

replace this

printf("%c",word[i]);

by

printf("%02X",word[i]);
4
votes

Use the %02X format parameter:

printf("%02X",word[i]);

More info can be found here: http://www.cplusplus.com/reference/cstdio/printf/

0
votes
void atoh(char *ascii_ptr, char *hex_ptr,int len)
{
    int i;

    for(i = 0; i < (len / 2); i++)
    {

        *(hex_ptr+i)   = (*(ascii_ptr+(2*i)) <= '9') ? ((*(ascii_ptr+(2*i)) - '0') * 16 ) :  (((*(ascii_ptr+(2*i)) - 'A') + 10) << 4);
        *(hex_ptr+i)  |= (*(ascii_ptr+(2*i)+1) <= '9') ? (*(ascii_ptr+(2*i)+1) - '0') :  (*(ascii_ptr+(2*i)+1) - 'A' + 10);

    }


}