Rejecting a floating point

[ForgotMySemiColon]

Hello, I am very new to C++ and in the process of trying to write a programme. Part of the programme will be to work out a factorial amount which I have working, however I want the user to have to enter a whole number (not 0, negative or floating point) For simplicity I have written a basic example but I can not figure out how to write a "If entry = float, then error message. Am i missing something? unfortunately I have been trying to learn via google and youtube. Any help would be appreciated, even if its a "You need to google X,Y,Z" Thanks in advance

 
#include <iostream>
#include <string>

using namespace std;

int main()
{
 int n;
 
 cout <<"Please enter a number:  " << endl;
 cin >> n ; 
 
 if(n == 0)
 
 {
 cout << "You entered 0, please enter a whole number greater than 0" << endl; 
 }
 
 if(n < 0) {
     
     cout <<"You have entered a negative number" << endl;
     
 }
 
 else {
     
     cout <<"you entered: " << n << endl;
     
 }
 return 0;
 
}

[helios]

If you >> into an int then you can't detect specifically if the user entered a fractional value. You can only detect if they entered a non-number.

if (!(std::cin >> n))
    //something other than an integer was entered 

The comparison will be true if the user enters anything that can't be an integer, including "0.1" and "foo".

[ForgotMySemiColon]

Ah I see, I might have to go back to the drawing board then. Thank you for responding.

[seeplus]

To determine the type of input entered when the type is an int, try:

#include <iostream>

using namespace std;

int main()
{
	int n {};

	cout << "Please enter a number:  ";

	if (!(cin >> n) || (cin.peek() != '\n' && cin.peek() != '.'))
		cout << "You have not entered a number\n";
	else
		if (cin.peek() == '.')
			cout << "You have entered a float number\n";
		else
			if (n == 0)
				cout << "You entered 0\n";
			else
				if (n < 0)
					cout << "You have entered a negative number\n";
				else
					cout << "You entered: " << n << '\n';

}

However, if you only want to determine if the entered number is valid or not, then consider:

#include <iostream>

using namespace std;

int main()
{
	int n {};

	cout << "Please enter a number:  ";

	if (!(cin >> n) || cin.peek() != '\n' || n <= 0)
		cout << "You have not entered a valid number\n";
	else
		cout << "You entered: " << n << '\n';
}

[Manga]

#include <iostream>
#include <string>
#include <limits>

using namespace std;

int main()
{
 int n;
 
 do {
     cout << "Enter a number greater than 0. ";
     cin >> n;
     
     if(!cin) {
         cout << "Not valid." <<endl;
         cin.clear();
         cin.ignore(numeric_limits<streamsize>::max(), '\n');
         n = 0;
         }
     
     
 } while(n <= 0);
 
 
 return 0;
 
}

It should also be noted that you need not worry about anyone putting in a float value. It will be converted to an int automatically. If someone puts 5.8 for example it will take only the value of 5.