How to Validate an Integer Input

Integer validation for input

When the reading fails, you set valid to false, so the condition in the while loop is false and the program returns input (which is not initialized, by the way).

You also have to empty the buffer before using it again, something like:

#include <iostream>
#include <limits>

using namespace std;

double read_input()
{
    double input = -1;
    bool valid= false;
    do
    {
        cout << "Enter a number: " << flush;
        cin >> input;
        if (cin.good())
        {
            //everything went well, we'll get out of the loop and return the value
            valid = true;
        }
        else
        {
            //something went wrong, we reset the buffer's state to good
            cin.clear();
            //and empty it
            cin.ignore(numeric_limits<streamsize>::max(),'\n');
            cout << "Invalid input; please re-enter." << endl;
        }
    } while (!valid);

    return (input);
}

How do i check the input is an integer in java

You use an while loop and break on sucess

int age;

while (true) {   // wil break on sucess 
   Scanner input = new Scanner(System.in);
   System.out.print("\nEnter Age       : "); 
   String AGE = input.nextLine();
    try{
      age = Integer.parseInt(AGE);
      break;  
    }catch (NumberFormatException ex){
        System.out.print("Invalid input " + AGE + " is not a number");
    }
}

Validation of integer with character input in C program

Some notes:
1) You aren't checking the scanf() return value, and this is very usable: the negative return means that the entered characters can't be converted to int (because of "%d" format), and the return value equal 0 means that the input is empty (no characters entered).

2) In case that the user entered wrong character(s) (not digit(s)), the input buffer will remain busy until you read it in other way. Good idea is to use additional scanf("%s") here to read any characters as string, so buffer will be empty after this call. Using rewind() is not enough here.

3) There is no need to additional checking of input in selectionChecking() for isdigit(), because "%d" format in scanf() doesn't allow to read anything else but number.

4) There is no need to pass pointer to selection value in selectionChecking() call - it will be enough to pass it as value.

So, try this below:

// declaration of 'selectionCheck()'
int selectionCheck(int input, int min, int max);

void main()
{
    int selection;
    while (1)
    {
        while (1)
        {
            printf("Enter Your Selection (0-4) > ");

            int ret = scanf("%d", &selection);
            if (ret < 0) // invalid characters on input
            {
                printf("Invalid characters\n");
                scanf("%s"); // empty buffer, reading it as string and putting readed characters to nowhere ;)
                continue; // go to top of loop
            }

            if (ret == 0) // empty input
            {
                printf("No (empty) input\n");
                continue; // go to top of loop
            }

            // here 'ret' is greather than 0, so valid number was entered

            if (selectionCheck(selection, 0, 4)) // is value between 0 and 4 ?
                break; // yes, success, break current loop!

            printf("Invalid value\n");
        }

        printf("Success\n");
    }

    system("pause");
}

int selectionCheck(int input, int min, int max)
{
    if (input < min || input > max)
        return 0;
    else 
        return 1;
}

Of course, you can write 'selectionCheck()' more condensed:

int selectionCheck(int input, int min, int max)
{
    return (input < min || input > max) ? 0 : 1;
}

or simply:

int selectionCheck(int input, int min, int max)
{
    return (input >= min && input <= max);
}

JavaScript validating input only contains integers on submit

To check if a number is an integer try this. It returns a boolean.

Number.isInteger(yourNumber) 

docs: https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest/Synchronous_and_Asynchronous_Requests

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