在C中，如何将char *转换为十六进制

Question

I have extracted a MAC address into a char* array such that each section of the array is a pair of char values.

mac[0] = "a1"
mac[1] = "b2"
...
mac[5] = "f6"

Basically I need to take the char arrays and convert them to an unsigned char such that the hex representation is the same as the original char values.

a1 in ascii -> 0xa1

What is the best way to convert char* to hex in C?

Answer 1

My C isn't the best, but this works using strtol(start, [stop], base)

#include <stdlib.h>
#include <stdio.h>

int main(char ** argv, int argc) {
  char * input = "a1";
  printf("%d\n", (unsigned char)strtol(input, NULL, 16));
}

Answer 2

#include <stdlib.h>
#include <stdio.h>

---

unsigned char mac_uchar[6];

for(i = 0;i < 6; ++i) {
    mac_uchar = (unsigned char) strtol (&mac[n], (char **) NULL, 16);
}

I was slow in answering...

Answer 3

You can use sscanf for this purpose.

int mac_byte;
unsigned char byte;
sscanf(mac[0],"%x",&mac_byte);
byte = mac_byte & 0xFF;