How to manipulate char - C

Question

In my program, I need the user to enter coordinates, starting with a letter and finishing by a number. The format is "e6", for example. I want to write a function that converts the number into an int. With this example, i want the "6" to be converted to int 6. My problem comes from the fact that the number can go up to 11.

I tried this function:

int function(char c[]) {

    if (c[2]=='0') {
        return (int)c[1] * 10;
    }

    else if (c[2]=='1') {
        return (int)c[1]*10 + 1;
    }

    else {
        return (int)(c[1]-'0');
    }
}

The problem is that it returns numbers unrelated to the input, like "490" if I enter "f10".

I hope my problem is clear enough so you can help me !

Answer: I understand now my error, and what -'0' means. The new working function is:

int function(char c[]) {

    if (c[2]=='0') {
        return (int)(c[1]-'0') * 10;
    }

    else if (c[2]=='1') {
        return (int)(c[1]-'0')*10 + 1;
    }

    else {
        return (int)(c[1]-'0');
    }
}

Thanks to all of you !

Answer 1

Here you convert a digit character to its value by subtracting '0' :

return (int)(c[1]-'0');

That's correct (although the cast is completely unnecessary).

But here you just use the digit character as though it were a value:

return (int)c[1] * 10;

Since '1' is 49, the output is not surprising.