Question

I have a program that has this code :

#include<stdio.h>
main(){
  int input;
  char g;

  do{
    printf("Choose a numeric value");
    printf(">");
    scanf("\n%c",&input);
    g=input-'0';
   }while((g>=-16 && g<=-1)||(g>=10 && g<=42)||(g>=43 && g<=79));

}

It basically uses ASCII manipulation to allow the program to accept numbers only .. ‘0’ is given the value 48 by default…the ASCII value – 48 gives a ranges of numbers above (in the while statement)

Anyway, whenever a user inputs numbers AND alphabets, such as :

abr39293afakvmienb23

The program ignores : a,b,r .. But takes ‘3’ as the first input.
For a b and r, the code under the do loop repeats. So for the above example, I get :

Choose a numeric value
>Choose a numeric value>
Choose a numeric value
>3

Is there a way I can stop this ??? I tried using \n%c to scan the character and account for whitespace, but that didnt work 🙁

Please help

thank you very much !

Answer 1

What I understand from your question is that when user enters abr3, the output you get is

Choose a numeric value
Choose a numeric value>
Choose a numeric value
3

Now you do not what these “excessive” Choose a numeric value to appear. If user has input abr3, its just take 3 from it and exit. Something like this.

$ Choose a numeric value> abr3

$

Now to do that you have to change format specifier in your scanf() statement. Try your program with

input = 0 ;
scanf( "%[0-9]" , &input );

It tells the scanf to accept all the contents within braces i.e. 0-9. Any other character outside this range is ignored. If user inputs abr3, a is not in the range, therefore input will remain equal to zero. Moreover, Choose a numeric value for a,b and r will not appear.

If user input is a number in the range of 0-9, it will be accepted and put into input.

You will have to add a check for input==0.

Hope it helps.