将char数组中的char转换为ANSI C中自己的char数组

Question

I have a char array representing a double precision floating point number in hex form.

char *hex = ""402499999999999A"

I want to extract each char in hex as its own char array and read it into an unsigned int num . For example, I tried

sscanf((char *)&hex[3], "%X", &num);

But this doesn't give me the 4th char as an individual char array, it gives me the sub char array from the 4th position on, which I suppose is because arrays are given by the pointer of their first element.

Is there a better way to do this? I looked at strcpy and it seems that I can only copy the first n chars, so that's no good.

Answer 1

You can do this in many ways. One way is as follows (which is the correct way of how you were doing it):

char only_1_char[2] = {'\0', '\0'};
only_1_char[0] = hex[3];
sscanf(only_1_char, "%X", &num);

and a more efficient solution:

if (hex[3] <= '9')
    num = hex[3] - '0';
else
    num = hex[3] - 'A' + 10;

This is just a sample, though. In truth you need to take care of invalid input and lower cases if that is a possibility.

Answer 2

Try something like this:

for(i = 0; src[i] != 0; i++) {
    if(src[i]) <= '9') {
        dest[i] = src[i] - '0';
    } else {
        dest[i] = toupper(src[i]) - 'A' + 10;
    }
}

It can be improved with error handling (eg detect if "src[i]" contains a valid/sane character).