c - Convert two ASCII Hexadecimal Characters (Two ASCII bytes) in one byte

Question

I want to convert two ASCII bytes to one hexadecimal byte. eg.

0x30 0x43 => 0x0C , 0x34 0x46 => 0x4F ...

The ASCII bytes are a number between 0 and 9 or a letter between A and F (upper case only), so between 0x30 ... 0x39 and 0x41 ... 0x46

I know how "to construct" 0x4F with the numbers 0x34 and 0x46 : 0x4F = 0x34 * 0x10 + 0x46

So, in fact, i would to convert one ASCII byte in hexadecimal value.

For that, i can build and array to assign the hexadecimal value to the ASCII char :

0x30 => 0x00
0x31 => 0x01
...
0x46 => 0x0F

But, maybe it have a most ??proper ??solution.

The program will be run on an AVR μC and is compiled with avr-gcc, so scanf() / printf() solutions aren't suitable.

Have you got an idea ? Thanks

Answer 1

i can't make sense of your examples, but if you want to convert a string containing hexadecimal ascii characters to its byte value (e.g. so the string "56" becomes the byte 0x56, you can use this (which assumes your system is using ASCII)

uint8_t*
hex_decode(const 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] - 'A' + 10 : in[i] - '0';
                ln = in[i+1] > '9' ? in[i+1] - 'A' + 10 : in[i+1] - '0';

                out[t] = (hn << 4 ) | ln;
        }

        return out;
}

You'd use it like e.g.

char x[]="1234";
uint8_t res[2];
hex_decode(x,strlen(x),res);

And res (which must be at least half the length of the in parameter) now contains the 2 bytes 0x12,0x34

Note also that this code needs the hexadecimal letters A-F to be capital, a-f won't do (and it doesn't do any error checking - so you'll have to pass it valid stuff).