How to allow & take only two digit numbers as Input, in C?

Question

I'm currently doing a project in which I've to take & consider only two digit numbers which an user is entering. I don't want to take an input & then check whether it's LESS THAN 100 or not.

Instead of doing this checking, is there any other way of filter out only 2-digit numbers & accept it as input, while programming in C-language? As, I don't even want to allow the user to enter 3 digit number. For example, if the user wants to enter 123, he won't be able to do it. Only if he is entering 12 or 23, then only the input is accepted.

Thank you.

Answer 1

We could truncate the input to 2 digits but that is NOT what the question asked. It said inputs of more than 2 digits are to be ignored.

Additionally, the problem with simply restricting the number of digits input, is that any truncated digits are taken by the next input. So for example with

scanf("%2d",&myNum);

if I enter 678 the 8 remains in the buffer and is read for the next input. We could clear the input buffer but that is not what the question asked: you don't want numbers truncated to 2 digits, you want them ignored.

Here is my answer, which reads the digits as a text string and rejects those which do not have 2 digits, and those that are not digits.

#include <stdio.h>
#include <string.h>
#include <ctype.h>

int main()
{
    int myNum;
    int c;
    char str[4];

    do {
        myNum = 0;
        printf("Enter number: ");
        scanf("%3s",str);
        while((c = getchar()) != '\n' && c != EOF);   // flush the input
        if (strlen(str) != 2 || !isdigit(str[0]) || !isdigit(str[1]))
            continue;
        if (sscanf(str, "%d", &myNum) != 1)
            continue;
        printf("You entered %d\n\n", myNum);
    } while (myNum != 10);
    return 0;
}

Sample session:

Enter number: 42
You entered 42

Enter number: 678
Enter number: 1
Enter number: 10
You entered 10

You can see thet 678 and 1 were not accepted.

Answer 2

You need to limit digits while having scanf. I hope below code snippet will work for you.

int main()
{
    int myNum;
    printf("Enter number");
    scanf("%2d",&myNum);      // here I am taking input of two digits only
    printf("%d", myNum);      // Verifying input by printing 
}

I hope for single digits you will input as 01 or 1, still it will work out same.

Answer 3

SOLUTION:

scanf("%2d",&variableName);

i think it'll work....

Answer 4

If you are looking for a program that disables the alphabet input and restrict numeric input only to two digits, I guess that needs some kind of interfacing with keyboard(with terminal I/O settings) which is slightly harder. Other than that I guess there are two simpler ways to do so in addition to scanf() :

1) Use a character array to let the user enter the number, but you flush out all the extraneous digits except first two and feed the array to atoi() to make it real integer value for further use in your program.

2) In order to avoid user to scribble the console, you can use ncurses facility which provides getche() like use. Refer What is Equivalent to getch() & getche() in Linux?

Finally after taking two characters you should be able to convert them as integer and use.

In both the approaches you can validate if user has actually entered numerics, and then convert them to integers.

Answer 5

#include<stdio.h>   
#define CHOICE 2   
int digit(int num)  
{  
    int i=0;  
    while(num>0)  
    {  
        num=num/10;  
        ++i;  
    }  
    return i;  
}  
int main()  
{  
    int num=0;  
    do  
    {  
       printf("\nInput two digit number: ");  
       scanf("%d",&num);  
       if(digit(num)==CHOICE)  
       {  
          break;  
       }  
       else  
       {  
           printf("\nInvalid Input");  
           continue;  
       }  
    }  
    while(1);  
    printf("You entered %d",num);  
}