C# conditional AND (&&) OR (||) precedence

Solution 1:

Change the first false by true. I know it seems stupid to have (true || true) but it proves your point.

bool result = true || true && false;   // --> true 
     result = (true || true) && false; // --> false
     result = true || (true && false); // --> true

Solution 2:

If you really want to freak him out try:

bool result = True() | False() && False();

Console.WriteLine("-----");
Console.WriteLine(result);

static bool True()
{
    Console.WriteLine(true);
    return true;
}

static bool False()
{
    Console.WriteLine(false);
    return false;
}

This will print:

True
False
False
-----
False

Edit:

In response to the comment:

In C#, | is a logical operator that performs the same boolean logic as ||, but does not short-circuit. Also in C#, the | operator has a higher precedence than both || and &&.

By printing out the values, you can see that if I used the typical || operator, only the first True would be printed - followed by the result of the expression which would have been True also.

But because of the higher precedence of |, the true | false is evaluated first (resulting in true) and then that result is &&ed with false to yield false.

I wasn't trying to show the order of evaluation, just the fact that the right half of the | was evaluated period when it normally wouldn't be :)

Solution 3:

Wouldn't this get you what you're after? Or maybe I'm missing something...

bool result = true || false && false;

Solution 4:

You don't prove it with code but with logic. AND is boolean multiplication whereas OR is boolean addition. Now which one has higher precedence?