Why is my c != 'o' || c != 'x' condition always true? [duplicate]

I have this loop statement, which I'll express using C-like syntax (C, C++, Java, JavaScript, PHP, etc. all use similar syntax):

while (c != 'o' || c != 'x') {
    c = getANewValue();
}

I want it to run until I get a 'o' or 'x', but it never exits, even when c is 'o' or 'x'. Why not?

I've also tried using if:

if (c != 'o' || c != 'x') {
    // Show an error saying it must be either 'o' or 'x'
}

but that also always shows the error message, even when c is 'o' or 'x'. Why?

Solution 1:

The condition (c != 'o' || c != 'x') can never be false. If c is 'o', then c != 'x' will be true and the OR condition is satisfied. If c is 'x', then c != 'o' will be true and the OR condition is satisfied.

You want && (AND), not || (OR):

while (c != 'o' && c != 'x') {
    // ...
}

"While c is NOT 'o' AND c is NOT `'x'..." (e.g., it's neither of them).

Or use De Morgan's law, covered here:

while (!(c == 'o' || c == 'x')) {
    // ...
}

"While it's NOT true that (c is 'o' or c is 'x')..."

Solution 2:

It must be if(c!='o' && c!='x') instead of if(c!='o' || c!='x'). If you use the or operator the boolean expression will be always true.

Solution 3:

Why is my c != 'o' || c != 'x' condition always true?

The expression combines two sub-expressions using the logical OR operator (||). The expression is true if at least one of the sub-expressions is true. In order to be false, both sub-expressions it connects must be false.

The sub-expressions are c != 'o' and c != 'x'.

The first sub-expression c != 'o' is false when c == 'o'. The same for the second one; it is false when c == 'x'.

Please note that they cannot be false on the same time because c cannot be 'o' and 'x' on the same time.