i dont understand why this code behavies the way it does

[acorn]

#include <iostream>
#include <limits>

using namespace std;

int main(int argc, char *arv[])
{

	cout << "enter 1 or 2: ";
	int response;
	cin >> response;

	if ((response != 1) || (response !=2))
		cout << "you must enter 1 or 2";

	cin.sync();
	cout << endl;
	cout << "Press ENTER to continue...";
	cin.ignore(numeric_limits<streamsize>::max(), '\n' );
}

by reading it i would think the if statement would fire if the response was not equal to one or not equal to 2. yet the if statement fires when i respond with 1. why is that.

[Turbine]

Use an and. &&

If response is not 1 or is not 2, invalid number. Will always be true.

[acorn]

ok || doesnt work like i think it did. i thought it meant 1 or the other. as in 1 = true or 2 = true results in true only if one or the other is true.

[Turbine]

It does mean one or the other, you're including every number except one, with two numbers. One value cannot be two at once.

int value = 4;

if (value == 4 && value != 4)
   //Similar circumstance, will never happen, impossible 

[acorn]

i will have to look at it over because it still doesnt make sense to me. if it says true condition if not 1 or not 2 then why would it also be true if i type 1.

[Zhuge]

Consider it this way. Let's say you input 1:
- Is 1 not equal to 1? No
- Is 1 not equal to 2? Yes
- Is one OR the other true? Yes, we run the if statement.

Let's say you input 2:
- Is 2 not equal to 1? Yes
- Is 2 not equal to 2? No
- Is one OR the other true? Yes, we run the if statement.

Any other number:
- Is X not equal to 1? Yes
- Is X not equal to 2? Yes
- Is one OR the other true? Yes, we run the if statement.

As you can see, the statement is always true.

[acorn]

thanks zhuge i feel like an idiot now.