1

Currently, I have a loop in my program that has the format:

  #include<stdio.h>

    int main()
    {
    int n;

         while(1)
         {
         printf("Enter a positive number, n");

         scanf("%d",&n);

         if(n>0)
             {
              break;
             }
          }
      }

My intention in using the if(n>0) statement was to exit the loop if and only if the user enters a positive integer. However, if the user types in the character "g", for example, the loop will still break, as the ASCII value will be interpreted. How do I form this loop so that it breaks just when the user enters a positive integer value for n?

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3
  • In my case it's not breaking up when you enter "g" or any char ... ? Commented Oct 26, 2012 at 20:37
  • @Omkant Because n is not initialised to anything. You are seeing undefined behaviour. See the answer and comments below. Commented Oct 26, 2012 at 20:40
  • stackoverflow.com/a/22236689/3388865 I have posted a solution to your question in the link above. Commented Mar 7, 2014 at 3:28

1 Answer 1

Reset to default
4

Check the return value of scanf which returns the number of successful inputs read:

int rc = scanf("%d",&n);

if (rc==1 && n>0) {
  ...
}

I would suggest you to use fgets to read the input into a string and convert into integer using strtol as scanf is a poor choice for the reason said by @paddy.

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1 Comment

scanf("%d",&n) won't remove non-integer bytes from the input stream. The OP wants to continue looping until an integer is entered. So you really need to read a single character if the integer input fails and then loop again.

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