How does Bitwise AND interact with boolean values?

Question

I have come across a piece of code that I can't wrap my head around:

public void Connect()
{
    if (!(!string.IsNullOrEmpty(_vpnConnectionName) & !string.IsNullOrEmpty(_ipToPing)))
    {
        return;
    }

    GetConnected();
}

I've researched the single ampersand on SO and elsewhere, and found it is a Bitwise AND operand. However the explanations all revolve around integers.

I can understand the applicability of bitwise operations on integers, but how are they applicable to booleans?

Try to avoid asking two questions in the same question. Typically only one gets answered — Eric Lippert, Apr 10 '14 at 14:13
Look up demorgans law and you'll find that your transformation is correct. — Eric Lippert, Apr 10 '14 at 14:20
There is no distinction between *int* and *bool* after the C# compiler is done with it. The CLR treats them as ints with the expected outcome when you logically-and them. Also the reason that VB.NET's True is -1, a language that historically did not distinguish between logical and bit-wise AND. So the answer to Eric's comment "are bools bits" is Yes. — Hans Passant, Apr 12 '14 at 22:41

score 8 · Accepted Answer · answered Apr 10 '14 at 14:10

8

It is a logical operator on bools that evaluates both sides regardless of the value of the left side.

answered Apr 10 '14 at 14:10

Eric Lippert

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@Eric I believe it is a bitwise operator, certainly on the IL level. This comes to play with booleans that are not 0 or 1 which are legal on the CLR. Is that true? See http://pastebin.com/U7vqpn6G This would be reproducible using verifiable, safe IL. – usr Apr 10 '14 at 14:14
@usr people who store things other than zero or one in a boolean are looking for trouble and should not be surprised when they find it. – Eric Lippert Apr 10 '14 at 14:19

score 3 · Answer 2 · answered Apr 10 '14 at 14:09

3

For bool, & isn't a bitwise AND operator - it's the non-short-circuiting logical AND operator - it always evaluates both arguments - as opposed to the "usual" && operator that only evaluates its second argument if its first argument is true.

(Realised that it may not be obvious, that the first & above is linked to the actual documentation)

answered Apr 10 '14 at 14:09

Damien_The_Unbeliever

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@PaulyGlott - no, `&&` doesn't evaluate its second argument if the first argument is `false`. – Damien_The_Unbeliever Apr 10 '14 at 14:11
Also though `&` *behaves* like a **logical** `AND` for booleans, it certainly remains to *be* a **bitwise** `AND` operator. The bools simply are defined as such, that the bitwise operation on their bits behaves like a logical one. – MrPaulch Apr 10 '14 at 14:31

How does Bitwise AND interact with boolean values?

2 Answers2