C++, Enum and OR condition - c++

Question

I have an enum with 3 values:
enum InputState { Pressed, Released, Held };
And I'm using it in this code:
//GetState returns an InputState
if(myInput.GetState(keyCode) == InputState::Pressed)
{
//This means "keyCode" has the state "Pressed"
}
Why doesn't this work?
if(myInput.GetState(keyCode) == (InputState::Pressed || InputState::Held))
{
//This is always false
}
if((myInput.GetState(keyCode) == InputState::Pressed) || (myInput.GetState(keyCode) == InputState::Held))
{
//This works as intended, triggers when "keyCode" is either Pressed OR Held
}
As a test, I did:
//Using the same values from the enum, but as int now
if(1 == (1 || 2))
{
//This works as intended
}
Am I missing something?

Answer 1

|| is a binary operation that expects two boolean values. In your example, the boolean values are the result of testing for equality with ==.
To see why your simplified example works, let's evaluate the expression
1 == (1 || 2)
We must start inside the parentheses first, so we are going to first evaluate (1 || 2). In C++, any non-zero value is equivalent to true when it is used in a boolean expression, so (1 || 2) is equivalent to (true || true) which evaluates to true. So now our expression is
1 == true
Again, 1 is equivalent to true in this context, so this comparison is the same as true == true, which of course evaluates to true.

Answer 2

Yes, you are missing something. This worked by total accident:
(1 == (1 || 2))
It is NOT set comparison. It simply calculated (1 || 2) as true, then converted true to its integral value (1).
The same thing would have happened with
(1 == (1 || 4))
(1 == (0 || 1))
(1 == (0 || 4))
and they all are true.
But
(2 == (1 || 2))
is false.