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

What you (apparently!) want to do is **convert** a pair of bytes to a single byte, by **interpreting** the two bytes as ASCII symbols which in turn represent hex digits, and emitting the corresponding hexadecimal value. Solving questions of this sort requires a precise understanding of what you are really doing; and precise communication helps a lot too :) You must be sure you understand, fundamentally, what data is. Like David said, bytes are just bytes; there are no ASCII bytes or hex bytes. — Karl Knechtel, Aug 03 '11 at 23:36
ASCII byte = byte that represent a ASCII char and hexadecimal byte = byte represented in hexadecimal. I know that bytes are just bytes; I just don't know how to explain it. But others have understood what I meant. Isn't the most important? — Loïc G., Aug 04 '11 at 18:35

nos · Accepted Answer · 2011-08-03T21:02:57.330

11

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).

edited Aug 03 '11 at 21:02

answered Aug 03 '11 at 20:47

nos

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Use `(in[i]|32)-'a'+10` and now you can support uppercase and lowercase (with just one more instruction in the machine code, most likely). – R.. GitHub STOP HELPING ICE Aug 03 '11 at 21:19
Isn't `uint8_t` a byte? Shouldn't that be `uint16_t` if it is supposed to contain two bytes? – Rudy Velthuis Aug 03 '11 at 21:47
@Rudy Velthuis It's supposed to be a byte, since it contains 1 byte, not 2 – nos Aug 04 '11 at 19:34
@nos: yes, I see it is an array of uint8_t. Brain fart, sorry. – Rudy Velthuis Aug 04 '11 at 19:54
Make sure to include cytpe.h – ojblass Apr 22 '16 at 17:31

Carl Norum · Answer 2 · 2011-08-03T20:51:29.093

10

You can use strtol(), which is part of avr-libc, or you can write just your specific case pretty easily:

unsigned char charToHexDigit(char c)
{
  if (c >= 'A')
    return c - 'A' + 10;
  else
    return c - '0';
}

unsigned char stringToByte(char c[2])
{
  return charToHexDigit(c[0]) * 16 + charToHexDigit(c[1]);
}

edited Aug 03 '11 at 20:51

answered Aug 03 '11 at 20:45

Carl Norum

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Ok, it was so easy ... I thought to the 4 bits left rotation for the number between 0 to 9 but the solution was even simpler ! Thanks – Loïc G. Aug 03 '11 at 20:59
It was a joke ? : $ man ascii Aucune entrée de manuel pour ascii – Loïc G. Aug 03 '11 at 21:02
@Loic: No, you must be missing the manual page. Here's the English version: http://linux.die.net/man/7/ascii – Adam Rosenfield Aug 03 '11 at 21:10
Is the value being returned from `stringToByte` getting cast to `unsigned char` due to the function being declared as such? – Tom Russell Aug 20 '22 at 01:54
Yes. Integer promotions make all the arithmetic inside `stringToByte` happen on `int` types, and then the result is returned as an `unsigned char`. – Carl Norum Aug 21 '22 at 00:13

score 0 · Answer 3 · answered Jan 02 '13 at 18:48

The task:

Convert a string containing hexadecimal ascii characters to its byte values so ascii "FF" becomes 0xFF and ascii "10" (x31x30x00) becomes 0x10

char asciiString[]="aaAA12fF";// input ascii hex string 
char result[4];               // byte equivalent of the asciiString (the size should be at half of asciiString[])

// the final result should be:

result[0] = 0xAA;
result[1] = 0xAA;       
result[2] = 0x12;
result[3] = 0xFF;

//1. Firt step: convert asciiString so it contains upper cases only:

// convert string to upper cases:
stringToUpperCases(asciiString);

use:

void stringToUpperCases(char *p)
{   
    for(int i=0; *(p+i) !='\0'; i++)
    {
        *(p+i) = (unsigned char) toupper( *(p+i) );
    }
}

//2. Convert a string containing hexadecimal ascii characters to its byte values:

// convert string to bytes:

int nrOfBytes = stringToBytes(asciiString,result);

//use:  
unsigned char charToHexDigit(char c)
{
if (c >= 'A')
    return (c - 'A' + 10);
else
    return (c - '0');
}

unsigned char ascii2HexToByte(char *ptr)
{
    return charToHexDigit( *ptr )*16 + charToHexDigit( *(ptr+1) );
}

int stringToBytes(char *string, char *result)
{
    int k=0;
    int strLen = strlen(string);

    for(int i = 0; i < strLen; i = i + 2)
    {
        result[k] = ascii2HexToByte( &string[i] );
        k++;
    }

    return k; // number of bytes in the result array 
}

//3. print result:

printNrOfBytes(nrOfBytes, result);

// use:

void printNrOfBytes(int nr, char *p)
{
   for(int i= 0; i < nr; i++)
    {
        printf( "0x%02X ", (unsigned char)*(p+i) );
    }
    printf( "\n");
}

//4. The result should be:

0xAA 0xAA 0x12 0xFF

//5. This is the test program:

char asciiString[]="aaAA12fF"; // input ascii hex string 
char result[4];                // result  
// convert string to upper cases:
stringToUpperCases(asciiString);

// convert string to bytes
int nrOfBytes = stringToBytes(asciiString,result);

// print result:
printNrOfBytes(nrOfBytes, result);

// result:
//  0xAA 0xAA 0x12 0xFF

score 0 · Answer 4 · edited Apr 15 '15 at 16:10

0

It's works, but could be much optimized !

inline uint8_t  twoAsciiByteToByte(const std::string& s)
{
    uint8_t r = 0;

    if (s.length() == 4)
    {
        uint8_t a = asciiToByte(s[0]);
        uint8_t b = asciiToByte(s[1]);
        uint8_t c = asciiToByte(s[2]);
        uint8_t d = asciiToByte(s[3]);

        int h = (a * 10 + b);
        int l = (c * 10 + d);

        if (s[0] == '3')
            h -= 30;
        else if (s[0] == '4')
            h -= 31;

        if (s[2] == '3')
            l -= 30;
        else if (s[2] == '4')
            l -= 31;

        r = (h << 4) | l;
    }

    return r;
}

edited Apr 15 '15 at 16:10

Opal

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answered Apr 15 '15 at 15:52

Dmitry

1

inline uint8_t asciiToByte(char c) { if (c >= '0' && c <= '9') return c - '0'; if (c >= 'A' && c <= 'F') return c - 'A' + 10; return 0; } – Dmitry Apr 15 '15 at 15:53

score 0 · Answer 5 · edited Aug 24 '19 at 10:53

Here's a version that works with both upper and lower-case hex strings:

void hex_decode(const char *in, size_t len, uint8_t *out)
{
  unsigned int i, hn, ln;
  char hc, lc;

  memset(out, 0, len);

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

    hc = in[i];
    if ('a' <= hc && hc <= 'f') hc = toupper(hc);
    lc = in[i+1];
    if ('a' <= lc && lc <= 'f') lc = toupper(lc);

    hn = hc > '9' ? hc - 'A' + 10 : hc - '0';
    ln = lc > '9' ? lc - 'A' + 10 : lc - '0';

    out[i >> 1] = (hn << 4 ) | ln;
  }
}

score 0 · Answer 6 · answered Apr 14 '20 at 11:19

Converting 2 hex chars to a byte is done in two steps:

Convert char a and b to their number (e.g. 'F' -> 0xF), which is done in two big if else branches, that check if the char is in the range '0' to '9', 'A' to 'F' or 'a' to 'f'.
In the 2nd step the two numbers are joined by shifting a (largest value is 0xF (0b0000_FFFF)) 4 to the left (a << 4 -> 0b1111_0000) and then apply the bitwise or operation on a and b ((a << 4) | b):

a: 0000_1111
b: 1111_0000
-> 1111_1111

#include <stdio.h>
#include <stdint.h>

#define u8 uint8_t
#define u32 uint32_t

u8 to_hex_digit(char a, char b) {
    u8 result = 0;

    if (a >= 0x30 && a <= 0x39) {
        result = (a - 0x30) << 4;
    } else if (a >= 0x41 && a <= 0x46) {
        result = (a - 0x41 + 10) << 4;
    } else if (a >= 0x61 && a <= 0x7A) {
        result = (a - 0x61 + 10) << 4;
    } else {
        printf("invalid hex digit: '%c'\n", a);
    }

    if (b >= 0x30 && b <= 0x39) {
        result |= b - 0x30;
    } else if (b >= 0x41 && b <= 0x46) {
        result |= b - 0x41 + 10;
    } else if (b >= 0x61 && b <= 0x7A) {
        result |= b - 0x61 + 10;
    } else {
        printf("invalid hex digit: '%c'\n", b);
    }

    return result;
}

u32 main() {
    u8 result = to_hex_digit('F', 'F');
    printf("0x%X (%d)\n", result, result);

    return 0;
}

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

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