compare number to letter - comparison

Question

Is it possible to compare a number to a letter and make a difference out of it?
What I'm trying to do is:
Ask the user for a number between 0-10 (choose from a menu)
Check what that number was, if the number isn't between 0-10 it will
ask the user for another number
This all works very well until the user inputs a letter (for example 'A').
I'm using scanf() to store the value that the user inputs in an integer variable. So if the user inputs 'A', the value 65 gets stored in the variable.
This is causing me a lot of headache, because I want to make a difference between letters and numbers..
Here's my code for checking the input:
int checkNumber(int input,int low,int high){
int noPass=0,check=input;
if(check<low){
noPass=1;
}
if(check>high){
noPass=1;
}
if(noPass==1){
while(noPass==1){
printf("Input number between %d - %d \n",low,high);
scanf("%d",&check);
if((check>low)&&(check<high)){
noPass=0;
}
}
}
return check;
}
What happens if the user inputs a letter inside the while loop in this function; it starts looping endlessly asking for an input between low and high.
I want to somehow filter out letters, without actually filtering out the letter's values (65 and above).
-Is this possible?

Answer 1

So I continued to wrestle this problem and I came up with this solution:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
//pre: stdlib.h and ctype.h needs to be included, input cannot be initialized to a value within low and high, low cannot be greater than high
//post: returns an integer value that ranges between low and high
int checkNumber(int input,int low,int high){
int noPass=0,check=input;
if(low>high){
printf("Low is greater than high, abort! \n");
exit(EXIT_FAILURE);
}
if(isdigit(check)){
noPass=1;
}
if((check<low)||(check>high)){
noPass=1;
}
if(noPass==1){
while(noPass==1){
printf("Input a number between %d - %d \n",low,high);
scanf("%d",&check);
getchar();
if((check>=low)&&(check<=high)){
noPass=0;
}
}
}
return check;
}
int main(int argc, char *argv[]){
int i=2147483647;
printf("Choose an alternative: \n");
printf("1. Happy Fun time! \n");
printf("2. Sad, sad time! \n");
printf("3. Indifference.. \n");
printf("4. Running out of ideas. \n");
printf("5. Placeholder \n");
printf("6. Hellow World? \n");
printf("0. -Quit- \n");
scanf("%d",&i);
getchar();
i=checkNumber(i,0,6);
if(i==0){
printf("You chose 0! \n");
}
if(i==1){
printf("You chose 1! \n");
}
if(i==2){
printf("You chose 2! \n");
}
if(i==3){
printf("You chose 3! \n");
}
if(i==4){
printf("You chose 4! \n");
}
if(i==5){
printf("You chose 5! \n");
}
if(i==6){
printf("You chose 6! \n");
}
return 0;
}
It works the way I want it to, but it's not perfect. The biggest flaw is that the variable for the value in input (int i, in main()) cannot be initialized to a value between low and high.
For example: if int i=3; low=0 and high=6, and the user writes a letter: the value of i remains at 3. 3 is sent to checkNumber, and is immediately passed as 3.
I chose to initialize i as 2147483647, which is an unlikely number - but it is still possible.
In conclusion: it works, but it is flawed.

Answer 2

The char automatically become casted into the ASCII code (http://www.c-howto.de/tutorial-anhang-ascii-tabelle.html) . Like u can see, the numbers for the chars are all over the 10 u accept, so the easiest way is to only check if the number is between 0-10 like u said