Thread: Check if input is an integer using only loops

Hi there I have a homework that needs to verify if the input of the user is an integer using only loops no if statements Here's the problem:

A program is required that prompts the user for a number. The program will then print a series of asterisks to represent the number. If the user enters a number less than 1, the program stops. For example:

Enter a number: 5
*****
Enter a number: 3
***
Enter a number: 9
*********
Enter a number: 0

All user input must be validated:
• Check for non-numeric input when reading numeric input
• Check that values entered are within the expected range for their purpose, or in range based on the requirements statement

Post your current code, if you want help you are going to have to show some effort.

Code:

#include <stdio.h>
main()
{
   int num;   // The number entered by the user
   int i;     // The number of times the for loop has to loop until matching the
   
      do
      {
         printf ("\nEnter a number: ");       // Requests a number input from the user
         scanf ("%d", &num);                  // Stores the input to the variable num


           for (i = 0; i < num; i++)          // For loop that prints the asterisk depending on the num entered
           {
              printf("*");
           }


      }while (num != 0);                      // The problem is probably here where I could include && or || to verify if the input is an integer
}

How big can the number be? You mentioned earlier that there was an "expected range". If the range has to be between 0-9 then your program gets a little easier, if you have to account for any integer value then you are going to have to think a little bit more.

I'm not sure about the "expected range" I will have to deal with that later. Right now i'm just focused on validating the input if it is not an integer.

Originally Posted by mababaan

I'm not sure about the "expected range" I will have to deal with that later. Right now i'm just focused on validating the input if it is not an integer.

Right now scanf is doing that for you -- if they type in an integer it will read it; if they type in not an integer it won't[footnote1]. You have to listen to what scanf tells you, though, by looking at what it returns.

Code:

status = scanf("%d", &x);
if (status == 1) {
    //good times
} else {
    //bad input -- clear it out and try again
}

[1]Note that if they type something that starts with an integer, like 12x or even 14.73, then it will report success that it read an integer (and left the other stuff still waiting to be processed). If that's not what you want, then scanf is not the tool for the job.

Originally Posted by tabstop

Right now scanf is doing that for you -- if they type in an integer it will read it; if they type in not an integer it won't[footnote1]. You have to listen to what scanf tells you, though, by looking at what it returns.

Code:

status = scanf("%d", &x);
if (status == 1) {
    //good times
} else {
    //bad input -- clear it out and try again
}

[1]Note that if they type something that starts with an integer, like 12x or even 14.73, then it will report success that it read an integer (and left the other stuff still waiting to be processed). If that's not what you want, then scanf is not the tool for the job.

The only other problem with that is that the OP cannot use if statements :/

Originally Posted by SirPrattlepod

The only other problem with that is that the OP cannot use if statements :/

Good point. I had made a mental note to try to turn it into a while ((status = scanf()) == 1) statement or similar, but then forgot. I should start writing this stuff down.

Originally Posted by tabstop

Good point. I had made a mental note to try to turn it into a while ((status = scanf()) == 1) statement or similar, but then forgot. I should start writing this stuff down.

You're ruining my fun. I was going to write a solution using buckets of water :-(

Thanks tabstop, Im still new to c programming and im still in the loops chapter. I'm not sure if pointers or functions can be included here but it specifically says that:
"These programs are to be written using loops ONLY. Do not use any “if” statements in your code. For this assignment, a properly designed solution does not require “if” statements"

The prof showed the working program which can be accessed but the code cannot be shown and this are the sample inputs that i've tried:

Code:

Enter a number: 9
*********
Enter a number: 2.2
**
Enter a number: 2..2
**
Enter a number: 2.c
**
Enter a number: 2xx
**
Enter a number: xxx
Invalid number entered, please re-enter: 3?xc;
***
Enter a number:x
Invalid number entered, please re-enter: x3x
Invalid number entered, please re-enter: 3x3
***

Isn't it a little silly to restrict the use of if statements but still allow loops?

You can throw a loop into a function then have the loop return and it's practically the same as an if:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <errno.h>

long int checkNum(const char * s)
{
    long int n;
    char * e;

    /* Basically the same as if (!s) return -1 */
    while (!s) {
        printf("Enter a number\n");
        return -1;
    }

    n = strtol(s, &e, 10);

    while (errno == ERANGE) {
        printf("Number is out of possible range\n");
        return 0;
    }
    while (n == 0 && e == s) {
        printf("Not a number\n");
        return n;
    }

    return n;
}

int main(int argc, char ** argv)
{
    long int i, n = checkNum(argv[1]);
    for (i = 0; i < n; ++i)
        putchar('*');
    putchar('\n');
    return 0;
}