如何在 C 中正确地将十六进制字符串转换为字节数组?

我需要将包含十六进制值作为字符的字符串转换为字节数组.虽然这已经被回答已经在这里作为第一个答案，我得到以下错误:

I need to convert a string, containing hex values as characters, into a byte array. Although this has been answered already here as the first answer, I get the following error:

warning: ISO C90 does not support the ‘hh’ gnu_scanf length modifier [-Wformat]

因为我不喜欢警告，省略hh只会产生另一个警告

Since I do not like warnings, and the omission of hh just creates another warning

warning: format ‘%x’ expects argument of type ‘unsigned int *’, but argument 3 has type ‘unsigned char *’ [-Wformat]

我的问题是:如何正确地做到这一点?为了完成，我再次在此处发布示例代码:

my question is: How to do this right? For completion, I post the example code here again:

#include <stdio.h>

int main(int argc, char **argv)
{
    const char hexstring[] = "deadbeef10203040b00b1e50", *pos = hexstring;
    unsigned char val[12];
    size_t count = 0;

     /* WARNING: no sanitization or error-checking whatsoever */
    for(count = 0; count < sizeof(val)/sizeof(val[0]); count++) {
        sscanf(pos, "%2hhx", &val[count]);
        pos += 2 * sizeof(char);
    }

    printf("0x");
    for(count = 0; count < sizeof(val)/sizeof(val[0]); count++)
        printf("%02x", val[count]);
    printf("\n");

    return(0);
}