在 C/C++ 中将用 ASCII 字符表示的十六进制数转换为十进制整数

Question

my problem is as follows.

I'm reading a piece of ascii data from a sensor, let's say it's "400". It's stored in an array of characters. In hex (ascii) that would be { 0x34, 0x30, 0x30 }.

What I'm trying to get from that set of characters is an integer in decimal representative of hex 0x400, which would be 1024. All the other numeric values in this array of ascii characters are represented in decimal, so I've been using this:

int num_from_ascii(char reading[], int start, int length){
    printf("++++++++num_from_ascii+++++++++\n");
    char radar_block[length];
    for(int i = 0; i < length; i++){
        radar_block[i] = reading[start + i];
        printf("%02x ", reading[start + i]);
    }
  printf("\n");
  return atoi(radar_block);
}

This obviously just gives me back 400, but I need a decimal integer from a hex value. Any advice?

Answer 1

As Eugene has suggested, all you need to do is replace atoi(radar_block) by strtol(radar_block, NULL, 16) . That takes a "base" argument, which can be 10 for decimal, 16 for hex (which is what you want), etc or 0 to auto-detect using the C++ rules (leading "0x" for hex, leading "0" for octal).

You should never use atoi anyway because it does not handle invalid inputs safely. strtol does everything that atoi does, has well defined errno for all edge cases, and also allows you to distinguish "0" from non-numeric input.

As user3121023 mentioned, don't forget to NUL-terminate the string you pass to strtol (this is a serious bug in your code calling atoi as well).