How make my program idiot-proof?

[nebula]

Hello guys. I wanted to know how to make my program safe against lazy people. For example, let's say I want the user to enter an int. Cause he's dumb, his input will be dfandoanfvoasdmpsnossjkdqwjdajdncasdfvdavdvasasdfd or 68469165465161616165161616165165165164164654 or 84+5+f44+6sdc4+ad65fbsdf+6bvsdf65fgs4df+g8sf+f+9g54dsf+dsf, and my program will explode. How can I prevent this to happen?

In advance, thank you for your help.

[docarchy]

I use cin.fail() to check if they entered something other than what my variable should hold. (like a letter into an integer). Then i clear the stream and the buffer and then ask them to enter a correct input again.

Something like this:

cin >> num;
if (cin.fail())
{
   cout << "Enter a number, not a letter dum dum!" << endl;
   cin.clear(); //clear stream
   cin.ignore(80, '\n'); //clear 80 characters until a \n is reached.
}

Something like that. Hope that helps.

[Cuddlebuddie928]

I use strings to get my integers, and a stringstream to convert them... looping until I get what I want.

#include <iostream>
#include <sstream>

int get_int(int min, int max, std::string prompt)
{
	int i;
	std::string str;

	while(true) {

		std::cout << prompt << std::endl;
		getline(std::cin, str, '\n'); //get a string terminated by end-of-line.

		std::stringstream ss(str); //copy the string to an input stream.

		//ss >> i converts the stream to an integer. This will fail something like "g45" but not fail if "45g" is entered.
		//!(ss >> str) checks to see if more characters are left; if so, it fails. The "45g" example would fail here.
		if(ss >> i && !(ss >> str) && i>=min && i<=max) return i;

		//if you get here, you didn't get your integer.
		std::cin.clear(); //clears any cin errors, like eof(). Protects against 'ctrl-z' etc.
		std::cerr << "You must enter an integer between " << min << " and " << max << ".\n";
	}
}

int main()
{
	int i=get_int(1, 10, "Enter a number between 1 and 10: ");
	std::cout << i;
	std::cin.get();

	return 0;
}