Simple... TOO simple

[RaleTheBlade]

I have built a very simple Parse class to detect whether a users input can be input into an integer variable or not. Here is the code:

#include <iostream>

using namespace std;

class inputParser
{
public:
	bool IntParse(istream& _stream1, int& _int1)
	{
			_stream1 >> _int1;
			if (_stream1.fail() || _stream1.bad())
				return false;
			else
				return true;
	}
};

int main()
{
	int someInt;
	inputParser Input;

	cout << "Input an integer: ";
	if (Input.IntParse(cin, someInt))
		cout << "Is an integer";
	else
		cout << "Is NOT an integer";
	system("pause");
	return 0;
}

This code seems WAY too simple though, but it seems to work just fine... Am I missing something here or is this just how simple it should be?

[pontusen]

If you just want to check if it's a number, why not use std::isdigit() from cctype?

[RaleTheBlade]

Well I wanted it to put the number into the variable as well. This class function does that. Its kinda like C#'s Int16.TryParse() boolean function.

[Philip Nicoletti]

Unfortunately, your code will fail for input like this:

12ttt

[Paul McKenzie]

Well I wanted it to put the number into the variable as well. This class function does that. Its kinda like C#'s Int16.TryParse() boolean function.

Instead of using C# as a model, why not just do the obvious that any C or C++ programmer would do.

1) Get the input.
2) Call one of the multiple functions to test if the number is really an integer.

For 1), you read the data as a string.

For 2), you have strtol, strtoul, etc. that check if the number is an integer, converts it to an int, and if the number isn't an int, will report where the parse fails.

The problem of saying "this function exists in language X, so I'll write the same thing in C++" will in many cases cause you to write unnecessary code and reinvent the wheel.

Regards,

Paul McKenzie

[code_carnage]

I think this code will give correct result.

and it will not fail even for 12ttt...

In this case it will simply return false..

[davide++]

Hi all.

A part from it works or not, at the moment what you have done is a "function that pretends to be a class"; in your class there isn't data members, only methods, so it doesn't implement a real object. It isn't a class, but a simple collection of procedures grouped into a class.
In my opinion this isn't a good development principle. It's easier, and more correct, to write a function that checks if a value is an integer; better, to use a function that exists yet.

[Philip Nicoletti]

I think this code will give correct result.

and it will not fail even for 12ttt...

In this case it will simply return false..

What compiler are you using ? if you enter 12tt for in int variable
using operator >>, the result should be the following:

a) the int variable is given a value of 12
b) the stream position points at the first 't'

[sunnypalsingh]

One way to validate input for integers

#include <iostream>
#include <limits>
 
using namespace std;
 
int main() {
  int number = 0;
  cout << "Enter an integer: ";
  cin >> number;
  cin.ignore(numeric_limits<int>::max(), '\n');
 
  if (!cin || cin.gcount() != 1)
    cout << "Not a numeric value.";
  else
    cout << "Your entered number: " << number;
  return 0;
}