//I don't mean ASCII code here.

#include <stdio.h>
//blah, blah, blah...
char getter;
getter = getchar();

/*NOW: if getter is a number, make it into an int. Could we do this? Something using isalpha(), maybe?*/
/*...actually, maybe use the ASCII code and if it's within the number range (wherever that is--one can check) you subtract something from it...?*/
//Any ideas?

You evidentally already know the solution, since you just said it. :)

try the "itoa" and the "atoi" commands.

AFAIK, the C standard specifies that the characters representing the numbers from 0 to 10 shall be in a sequential block starting with 0. Therefore, 'n'-'0', where n can be any number, will always equal n.

If I understood your question, you want to read a number as a character and then convert it to an int that has the value of the number red. The solution I propose is:

-find the ASCII code for 1.

You can do this by writting a simple program that contains:
int i;
for(i=1;i<=255;i++)
printf("%d = %c",i,i);

-then you do something like this:

getter-=ascii_code_for_1; // a-=b is equivalent to a=a-b;
// (just in case there is
// somebody that doesn't know)

This will still be a char, but it contanis the value you need, so you can assign this value to an int variable.

I don't have time to test this, but I think I used this in one of my programs a long time ago, and it should work. If it doesn't please tell me. It's an interesting problem.

The ASCii values for 0-9 are 48-57 (or, in hexadecimal "translation", 30-39).

Hope that helps!

Happy coding! :)

It's not advisable to ever type ASCII codes into your programs. Instead, simply use '0' for the ASCII code for 0, etc.. This makes things much more portable.