Converting ascii hex string to byte array - c

Question

I have a char array say char value []={'0','2','0','c','0','3'};
I want to convert this into a byte array like unsigned char val[]={'02','0c','03'}
This is in an embedded application so i can't use string.h functions. How can i do this?

Answer 1

Sicne you talk about an embedded application I assume that you want to save the numbers as values and not as strings/characters. So if you just want to store your character data as numbers (for example in an integer), you can use sscanf.
This means you could do something like this:
char source_val[] = {'0','A','0','3','B','7'} // Represents the numbers 0x0A, 0x03 and 0xB7
uint8 dest_val[3]; // We want to save 3 numbers
for(int i = 0; i<3; i++)
{
sscanf(&source_val[i*2],"%x%x",&dest_val[i]); // Everytime we read two chars --> %x%x
}
// Now dest_val contains 0x0A, 0x03 and 0xB7
However if you want to store it as a string (like in your example), you can't use unsigned char
since this type is also just 8-Bit long, which means it can only store one character. Displaying 'B3' in a single (unsigned) char does not work.
edit: Ok according to comments, the goal is to save the passed data as a numerical value. Unfortunately the compiler from the opener does not support sscanf which would be the easiest way to do so. Anyhow, since this is (in my opinion) the simplest approach, I will leave this part of the answer at it is and try to add a more custom approach in this edit.
Regarding the data type, it actually doesn't matter if you have uint8. Even though I would advise to use some kind of integer data type, you can also store your data into an unsigned char. The problem here is, that the data you get passed, is a character/letter, that you want to interpret as a numerical value. However, the internal storage of your character differs. You can check the ASCII Table, where you can check the internal values for every character.
For example:
char letter = 'A'; // Internally 0x41
char number = 0x61; // Internally 0x64 - represents the letter 'a'
As you can see there is also a differnce between upper an lower case.
If you do something like this:
int myVal = letter; //
myVal won't represent the value 0xA (decimal 10), it will have the value 0x41.
The fact you can't use sscanf means you need a custom function. So first of all we need a way to conver one letter into an integer:
int charToInt(char letter)
{
int myNumerical;
// First we want to check if its 0-9, A-F, or a-f) --> See ASCII Table
if(letter > 47 && letter < 58)
{
// 0-9
myNumerical = letter-48;
// The Letter "0" is in the ASCII table at position 48 -> meaning if we subtract 48 we get 0 and so on...
}
else if(letter > 64 && letter < 71)
{
// A-F
myNumerical = letter-55
// The Letter "A" (dec 10) is at Pos 65 --> 65-55 = 10 and so on..
}
else if(letter > 96 && letter < 103)
{
// a-f
myNumerical = letter-87
// The Letter "a" (dec 10) is at Pos 97--> 97-87 = 10 and so on...
}
else
{
// Not supported letter...
myNumerical = -1;
}
return myNumerical;
}
Now we have a way to convert every single character into a number. The other problem, is to always append two characters together, but this is rather easy:
int appendNumbers(int higherNibble, int lowerNibble)
{
int myNumber = higherNibble << 4;
myNumber |= lowerNibbler;
return myNumber;
// Example: higherNibble = 0x0A, lowerNibble = 0x03; -> myNumber 0 0xA3
// Of course you have to ensure that the parameters are not bigger than 0x0F
}
Now everything together would be something like this:
char source_val[] = {'0','A','0','3','B','7'} // Represents the numbers 0x0A, 0x03 and 0xB7
int dest_val[3]; // We want to save 3 numbers
int temp_low, temp_high;
for(int i = 0; i<3; i++)
{
temp_high = charToInt(source_val[i*2]);
temp_low = charToInt(source_val[i*2+1]);
dest_val[i] = appendNumbers(temp_high , temp_low);
}
I hope that I understood your problem right, and this helps..

Answer 2

If you have a "proper" array, like value as declared in the question, then you loop over the size of it to get each character. If you're on a system which uses the ASCII alphabet (which is most likely) then you can convert a hexadecimal digit in character form to a decimal value by subtracting '0' for digits (see the linked ASCII table to understand why), and subtracting 'A' or 'a' for letters (make sure no letters are higher than 'F' of course) and add ten.
When you have the value from the first hexadeximal digit, then convert the second hexadecimal digit the same way. Multiply the first value by 16 and add the second value. You now have single byte value corresponding to two hexadecimal digits in character form.
Time for some code examples:
/* Function which converts a hexadecimal digit character to its integer value */
int hex_to_val(const char ch)
{
if (ch >= '0' && ch <= '9')
return ch - '0'; /* Simple ASCII arithmetic */
else if (ch >= 'a' && ch <= 'f')
return 10 + ch - 'a'; /* Because hex-digit a is ten */
else if (ch >= 'A' && ch <= 'F')
return 10 + ch - 'A'; /* Because hex-digit A is ten */
else
return -1; /* Not a valid hexadecimal digit */
}
...
/* Source character array */
char value []={'0','2','0','c','0','3'};
/* Destination "byte" array */
char val[3];
/* `i < sizeof(value)` works because `sizeof(char)` is always 1 */
/* `i += 2` because there is two digits per value */
/* NOTE: This loop can only handle an array of even number of entries */
for (size_t i = 0, j = 0; i < sizeof(value); i += 2, ++j)
{
int digit1 = hex_to_val(value[i]); /* Get value of first digit */
int digit2 = hex_to_val(value[i + 1]); /* Get value of second digit */
if (digit1 == -1 || digit2 == -1)
continue; /* Not a valid hexadecimal digit */
/* The first digit is multiplied with the base */
/* Cast to the destination type */
val[j] = (char) (digit1 * 16 + digit2);
}
for (size_t i = 0; i < 3; ++i)
printf("Hex value %lu = %02x\n", i + 1, val[i]);
The output from the code above is
Hex value 1 = 02
Hex value 2 = 0c
Hex value 3 = 03
A note about the ASCII arithmetic: The ASCII value for the character '0' is 48, and the ASCII value for the character '1' is 49. Therefore '1' - '0' will result in 1.

Answer 3

It's easy with strtol():
#include <stdlib.h>
#include <assert.h>
void parse_bytes(unsigned char *dest, const char *src, size_t n)
{
/** size 3 is important to make sure tmp is \0-terminated and
the initialization guarantees that the array is filled with zeros */
char tmp[3] = "";
while (n--) {
tmp[0] = *src++;
tmp[1] = *src++;
*dest++ = strtol(tmp, NULL, 16);
}
}
int main(void)
{
unsigned char d[3];
parse_bytes(d, "0a1bca", 3);
assert(d[0] == 0x0a);
assert(d[1] == 0x1b);
assert(d[2] == 0xca);
return EXIT_SUCCESS;
}
If that is not available (even though it is NOT from string.h), you could do something like:
int ctohex(char c)
{
if (c >= '0' && c <= '9') {
return c - '0';
}
switch (c) {
case 'a':
case 'A':
return 0xa;
case 'b':
case 'B':
return 0xb;
/**
* and so on
*/
}
return -1;
}
void parse_bytes(unsigned char *dest, const char *src, size_t n)
{
while (n--) {
*dest = ctohex(*src++) * 16;
*dest++ += ctohex(*src++);
}
}

Answer 4

Assuming 8-bit bytes (not actually guaranteed by the C standard, but ubiquitous), the range of `unsigned char` is 0..255, and the range of `signed char` is -128..127. ASCII was developed as a 7-bit code using values in the range 0-127, so the same value can be represented by both `char` types.
For the now discovered task of converting a counted hex-string from ascii to unsigned bytes, here's my take:
unsigned int atob(char a){
register int b;
b = a - '0'; // subtract '0' so '0' goes to 0 .. '9' goes to 9
if (b > 9) b = b - ('A' - '0') + 10; // too high! try 'A'..'F'
if (b > 15) b = b - ('a' - 'A); // too high! try 'a'..'f'
return b;
}
void myfunc(const char *in, int n){
int i;
unsigned char *ba;
ba=malloc(n/2);
for (i=0; i < n; i+=2){
ba[i/2] = (atob(in[i]) << 4) | atob(in[i+1]);
}
// ... do something with ba
}