将 ASCII 字符数组转换为十六进制字符数组

Question

I am trying to convert an ASCII char array to hex char array to do something like this:

Input: char string[5]="abcd1234567"

Output: char buf[4096] = {0x63, 0x61, 0x62, 0x63, 0x64, 0x31, 0x32, 0x33 0x34, 0x35, 0x36, 0x37};

but my code didn't work and i couldn't figure it out.

code:

char string[20]="abcd";
   char buff[4000];
   

    sprintf(buff[0],"%02x",string[0]);

    printf("string[0]: %c",string[0]); 

it gives this error:

warning: passing argument 1 of 'sprintf' makes pointer from integer without a cast [-Wint-conversion]
   16 |     sprintf(buff[0],"%02x",string[0]);

Answer 1

There's no difference between 'c' and 0x63 on an ASCII-based machine, so all you need is the following:

char buf[4096];
strncpy(buf, string, sizeof(buf));

That technically does

char buf[4096] =
   {0x63, 0x61, 0x62, 0x63, 0x64, 0x31, 0x32, 0x33 0x34, 0x35, 0x36, 0x37, 0};
                                                                         ^^^
                                                                         |||

But that's equivalent giving the extra room in buf . Use memcpy if you want to avoid that extra NUL.

Answer 2

Implementation of converting each pair of hex string characters to their byte values. Explanations are in the comments:

void convertStringToByteArray(const char* string, unsigned char* output)
{
    // i+=2 will ignore a trailing single character if you don't have an even number of them,
    // such as the case in string above
    for (size_t i=0; i<strlen(string); i+=2)
    {
        // create a substring
        char temp[3];
        temp[0] = string[i];
        temp[1] = string[i+1];
        temp[2] = '\0';

        // there's quite a bit of error checking here that's possible, but since we're dealing
        // with bytes, we only really need to check if temp contains valid hex data. There
        // should be no danger of under or overflow
        char* endptr;
        // convert temp to a long. If there's a problem, *endptr will point to the invalid
        // char. Use 16 to specify this is a hex string
        long convert = strtol(temp, &endptr, 16);
        if (*endptr == '\0')
        {
            // no problems
            output[i/2] = convert;
        }
        else
        {
            // invalid hex, just fill in with 0? You can do what you want here
            output[i/2] = 0;
        }
    }
}

int main(void)
{
    // no need to specify a size for string here, let the compiler do it
    char string[]="abcd1234567";
    // every 2 chars in string are one byte.
    unsigned char output[strlen(string) / 2];
    convertStringToByteArray(string, output);

    return 0;
}

Demo