How do you use atoi to assign individual elements of a char array?

Question

So as we all probably know, the atoi converts a char to a number. But, what do you do if you only want one of the array elements instead of the whole array?

Answer 1

Besides just using its integral value as shown by James, you could put it in a seperate buffer:

char buf[2] = { temp[h], '\0' };
num[h] = atoi(buf);

Answer 2

If you only want to convert a single character you don't need to use atoi():

if (temp[h] >= '0' && temp[h] <= '9')
{
    num[h] = temp[h] - '0';
}
else
{
    // handle error:  character was not a digit
}

In C, the value of each digit is one greater than the value of the previous digit, so this is guaranteed to work.

The reason that atoi() does not work is because it takes a const char* as its argument, not a char. That pointer has to point to a null terminated string.