C treats char as a letter, not a one-byte integer - c

Question

So let's say I need to take a number from user input, only from 0 to 9. Everything else is out of bounds.
I figure this number ought to fit within a one-byte integer, a char (0-255). Typically I see people use int here, but that's wasteful, isn't it? 32 bits (or whatever the implementation is) just to store the numbers 0-9?
What I find is that the code below (which uses shorts) works, but when I change it to use only a one-byte char, not a two-byte short, it begins to treat the user input as a string, so if a user enters 235, convert() no longer catches the error condition. It just returns the first digit of their number.
I'm aware that C implements strings as arrays of characters. Is there a way to use char to represent a single one-byte integer without treating it like a string? Should I just abandon the idea of using one-byte integers as numbers?
(I shortened this to test for 0-1, not 0-9, for the sake of brevity)
#include <stdio.h>
char *convert(short digit)
{
char *english;
switch (digit) {
case 0:
english = "zero";
break;
case 1:
english = "one";
break;
default:
english = "";
break;
}
return english;
}
void main(void)
{
short digit;
char *english;
printf("Enter a digit between 0 and 9: ");
scanf("%hu", &digit);
english = convert(digit);
if (english[0] == '\0') {
printf("\nYour number was invalid\n");
} else {
printf("\nYou entered the number %s\n", english);
}
}
Update:
Here's the attempted version that accepts a char, not a short:
#include <stdio.h>
char *convert(char digit)
{
char *english;
switch (digit) {
case 0:
english = "zero";
break;
case 1:
english = "one";
break;
default:
english = "";
break;
}
return english;
}
void main(void)
{
char digit;
char *english;
printf("Enter a digit between 0 and 9: ");
scanf("%c", &digit);
english = convert(digit);
if (english[0] == '\0') {
printf("\nYour number was invalid\n");
} else {
printf("\nYou entered the number %s\n", english);
}
}
The observed behavior of the above is that any number entered is flagged as invalid.

Answer 1

Change scanf("%c", &digit); to scanf("%hhd", &digit);
%c is the specifier to read a character and store that character. %d is the specifier to read a number written as base-10 digits and convert that to an integer (with hh meaning to write the answer into a char).
It is nothing to do with how "C treats a char", it is to do with how you instructed the scanf function to treat its input. A char is a one-byte integer.
Also, your code short digit; scanf("%hu", &digit); is incorrect; %hu is the specifier for unsigned short. Either use %hd or unsigned short digit;.

Answer 2

There is no way to use scanf to input a one byte integer. If you mean your format string is "%c" this is an (ascii) character, not a one byte integer. If you wish to treat it as a one byte integer you need to convert it yourself. c = c - '0'; will work find if a value 0-9 was provided.
If instead your intention is to scan in values within the full 8 bit range, you must scan in at least a short and you can take the lower 8 bits if you wish.

Answer 3

Typically I see people use int here, but that's wasteful, isn't it? 32 bits (or whatever the implementation is) just to store the numbers 0-9?
Using char instead of short won't save space and will be most likely counter productive in most cases.
For function parameters, the compiler will generate code that passes chars as ints (otherwise the stack will become unaligned which generates a fault on most platforms and severly slows down the machine on most others.
For most arithmetic operations on chars, the compiler will be forced to do integer promotion (extend char to int, do the operation and truncate the result).