c char array to uint8_t array - c

Question

I have a basic question regarding a c problem I'm having. My input char array would be something like:
'DABC95C1'
and I want to make an uint8_t array out of it
0xDA 0xBC 0x95 0xC1
I have easily access to each char but I dont know how I can form 0xDA. Is there function in c or can i just cast it?

Answer 1

Use the strtoull function to convert a string to a number in a given base. Then just shift out the desired bytes. Such as:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
unsigned long long res = strtoull("DABC95C1", NULL, 16);
printf("%hhx, %hhx, %hhx, %hhx",
(unsigned char)res,
(unsigned char)((res >> 8) & 0xFF),
(unsigned char)((res >> 16) & 0xFF),
(unsigned char)((res >> 24) & 0xFF)
);
return 0;
}
result:
c1, 95, bc, da
Demo
Notes:
As your requirement is to get an array of bytes, you might be tempted to do something like
uint8_t *arr = (uint8_t*)&res;
But here are two caveats in this:
1) I is a strict aliasing rule violation (you can somehow to work around it by replacing uint8_t with char)
2) The order of the returned bytes will be implementation specific (endianness dependent) and thus not portable. Also note that the result is unsigned long long, so you might get extra padding zeros as either the beginning of the array or in the end of it.

Answer 2

Any size string in the chosen order. Portable digit conversion and it optimizes very well on the ASCII systems. https://godbolt.org/g/Ycah1e
#include <stdio.h>
#include <stdint.h>
#include <string.h>
int CharToDigit(const char c);
void *StringToTable(const char *str, const void *buff, const int order)
{
uint8_t *ptr = (uint8_t *)buff;
size_t len;
int incr = order ? 1 : -1;
if(buff && str)
{
len = strlen(str);
if(len &1) return NULL;
ptr += order ? 0 : len / 2 - 1;
while(*str)
{
int d1 = CharToDigit(*str++);
int d2 = CharToDigit(*str++);
if(d1 == -1 || d2 == -1) return NULL;
*ptr = d1 * 16 + d2;
ptr += incr;
}
}
return buff;
}
int main(void) {
int index = 0;
char *str = "78deAc8912fF0f3B";
uint8_t buff[strlen(str) / 2];
StringToTable(str, buff, 0);
printf("String: %s\nResult: ", str);
for(index = 0; index < strlen(str) / 2; index++ )
{
printf("[0x%02hhx]", buff[index] );
}
printf("\n");
StringToTable(str, buff, 1);
printf("String: %s\nResult: ", str);
for(index = 0; index < strlen(str) / 2; index++ )
{
printf("[0x%02hhx]", buff[index] );
}
printf("\n");
return 0;
}
int CharToDigit(const char c)
{
switch(c)
{
case 'a':
case 'A':
return 10;
case 'b':
case 'B':
return 11;
case 'c':
case 'C':
return 12;
case 'd':
case 'D':
return 13;
case 'e':
case 'E':
return 14;
case 'f':
case 'F':
return 15;
case '0':
return 0;
case '1':
return 1;
case '2':
return 2;
case '3':
return 3;
case '4':
return 4;
case '5':
return 5;
case '6':
return 6;
case '7':
return 7;
case '8':
return 8;
case '9':
return 9;
default:
return -1;
}
}

Answer 3

You can conver a character to an int like so
static inline int char2int(char Ch)
{
return(Ch>='0'&&Ch<='9')?(Ch-'0'):(Ch-'A'+10);
//assuming correct input with no lowercase letters
}
Two characters then with
static inline
int chars2int(unsigned char const Chars[2])
{
return (char2int(Chars[0])<<4)|(char2int(Chars[1]));
}
And several characters by converting each pair:
static inline int char2int(char Ch)
{
return(Ch>='0'&&Ch<='9')?(Ch-'0'):(Ch-'A'+10);
}
static inline
int chars2int(unsigned char const Chars[2])
{
return (char2int(Chars[0])<<4)|(char2int(Chars[1]));
}
#include <stdio.h>
#include <string.h>
#include <assert.h>
int main()
{
char const inp[] = "DABC95C1";
assert((sizeof(inp)-1)%2==0);
unsigned i;
unsigned char out[(sizeof(inp)-1)/2];
for(i=0;i<sizeof(inp);i+=2){
out[i/2]=chars2int((unsigned char*)inp+i);
}
for(i=0;i<sizeof(out);i++)
printf("%2x\n", out[i]);
}