Why does || operator work in my situation

Question

Why is this an infinite loop using ||logical operator

char c;
std::cin >> c;
while( c != 'y' || c != 'Y' || c != 'N' || c != 'n') 

but this is not

while( c != 'y' && c != 'Y' && c != 'N' && c != 'n')

I don't understand why && operator work here because logically thinking || operator is better fit.

Answer 1

Lets just look at the very smallest part:

c != 'Y' || c != 'N'

If c is 'Y' then it is not 'N', if it is 'N' then it is not 'Y'. Aka:

c  |  c != 'Y' || c != 'N'
Y  |     1     ||   0      =  1
N  |     0     ||   1      =  1
?  |     1     ||   1      =  1

If your logic always returns 1 no matter what, the loop will run forever. I assume you are looking to wait until you get c as one of these values, so write it logically.

I want to wait until c is one of ['y', 'Y', ...]

And you might be able to write some nicer code:

std::array<char> options = {{'y', 'Y', 'n', 'N'}};

while (std::none_of(std::begin(options), std::end(options), 
         [&c](char check) { return check == c; };)) {
    std::cout << "Hey write the correct character!\n";
}

Untested!

Answer 2

You are checking the condition which is always true by using ||operator, ( c != 'y' || c != 'Y' || c != 'N' || c != 'n')

This is always true,because it would always satisfy at-least 3 conditions in your conditional statement.

if your char is n then it would satisfy remaining conditions c!='y' , c!='Y' , c!='N' and similarly for any other character