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

Question

We get into unnecessary coding arguments at my work all-the-time. Today I asked if conditional AND (&&) or OR (||) had higher precedence. One of my coworkers insisted that they had the same precedence, I had doubts, so I looked it up.

According to MSDN AND (&&) has higher precedence than OR (||). But, can you prove it to a skeptical coworker?

http://msdn.microsoft.com/en-us/library/aa691323(VS.71).aspx

bool result = false || true && false; // --> false
// is the same result as
bool result = (false || true) && false; // --> false
// even though I know that the first statement is evaluated as 
bool result = false || (true && false); // --> false

So my question is how do you prove with code that AND (&&) has a higher precedence that OR (||)? If your answer is it doesn't matter, then why is it built that way in the language?

Answer 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

Answer 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 :)

Answer 3

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

bool result = true || false && false;

Answer 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?

Answer 5

false || true && true

Yields: true

false && true || true

Yields: true

Answer 6

You cannot just show the end result when your boolean expressions are being short-circuited. Here's a snippet that settles your case.

It relies on implementing & and | operators used by && and ||, as stated in MSDN 7.11 Conditional logical operators

public static void Test()
{
    B t = new B(true);
    B f = new B(false);

    B result = f || t && f;

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

public class B {
    bool val;
    public B(bool val) { this.val = val; }
    public static bool operator true(B b) { return b.val; }
    public static bool operator false(B b) { return !b.val; }
    public static B operator &(B lhs, B rhs) { 
        Console.WriteLine(lhs.ToString() + " & " + rhs.ToString());
        return new B(lhs.val & rhs.val); 
    }
    public static B operator |(B lhs, B rhs) { 
        Console.WriteLine(lhs.ToString() + " | " + rhs.ToString());
        return new B(lhs.val | rhs.val); 
    }
    public override string ToString() { 
        return val.ToString(); 
    }
}

The output should show that && is evaluated first before ||.

True & False
False | False
-----
False

For extra fun, try it with result = t || t && f and see what happens with short-circuiting.