Only accept whole numbers for input

[brenP]

I am completing an assignment for a class where the prompt states Fractional scores, such as 8.3 are not allowed another requirement was to not allow scores < 0 or scores >10

I think I was able to successfully make it so the program does not allow scores < 0 or scores >10 but I do not know how to not accept a fractional score

If the user enters a decimal I would like to display a message saying "only whole numbers are accepted" and prompt the user to enter the number again correctly

  int getJudgeData(string judge)
{
        int input = -1;
        while ( input < 0 || input > 10 )
        {
            cout << "Enter the score from " << judge <<endl;
            cin >> input;
        
        }
    
        return input;
}

[disturbedfuel15]

Since there are such a small amount of accepted inputs, have you considered just typing out the numbers 1-10?
I.E.

	while (input != 1 && input != 2 &&
		input != 3 && input != 4 &&
		input != 5 && input != 6 &&
		input != 7 && input != 8 &&
		input != 9 && input != 10)
	{
		cout << "Enter the score from 1-10" << endl;
		cin >> input;
	}

Here is another way, but make sure to #include <cmath>
the ceil() function rounds the number up.
The only way it won't round up is if it is whatever.0 (already rounded!)
So you make it round up all decimals.
If the user entered number does not equal the rounded up number, it must not be whole.
;)

	double input = -1.0;
	while (input < 0.0 || input > 10.0 || ceil(input) != input)
	{
		cout << "Enter the score from 1-10" << endl;
		cin >> (double)input;

	}

[brenP]

@disturbedfuel15 that is a good idea and it works. Thank you for the suggestion

[Esslercuffi]

in order to check if they entered a non-whole number, you need to have a data type which can receive it. if you make input a double, then you can check with :
while( input < 0 || input > 10 || (input - (int)input) > 0 ) to see if they entered a floating point value.

Seems kinda silly to do this though.

[brenP]

@Esslercuffi
That also works very well. Thank you

[Lowest0ne]

There are three things going on here, get a number, make sure it is an integer, and make sure it is in range. Therefore, you want to make three functions:

#include <iostream>

double getNumber(void);
int    getInteger(void);
int    getIntegerInRange(const int low, const int high);

int main(void)
{
	const int low  = 0;
	const int high = 10;

	std::cout << "Please enter an integer in range " << low << " and " << high << ": ";
	int number = getIntegerInRange(low, high);
	std::cout << "You entered: " << number << '\n';

	std::cin.get(); // because I'm using Visual Studio
	return 0;
}

double getNumber(void)
{
	double number;
	while (!(std::cin >> number))  // the user entered a letter
	{
		std::cin.clear();          // clear cin's error state
		std::cin.ignore(80, '\n'); // ignore whatever the user typed

		std::cout << "Only numbers are accepted, try again: ";
	}

	std::cin.ignore(80, '\n');     // remeber to clear the input buffer

	return number;
}

int getInteger(void)
{
	double number = getNumber();
	while (number != static_cast< int >(number)) // casting to int changes the number
	{
		std::cout << "Only integers are accepted, try again: ";
		number = getNumber();
	}

	return static_cast< int >(number);  // return number casted to int
}

int getIntegerInRange(const int low, const int high)
{
	int number = getInteger();
	while (number < low || number > high)
	{
		std::cout << "Number must be in range ( " << low << ", " << high << " ), try again: ";
		number = getInteger();
	}

	return number;
}