CODEWITHSUNDEEP
c#binarybit-manipulationoperator-keyword
S. Valmont
S. Valmont

Reputation: 1021

Bitwise "~" Operator in C#

Consider this unit test code:

    [TestMethod]
    public void RunNotTest()
    {

        // 10101100 = 128 + 32 + 8 + 4 = 172
        byte b = 172;

        // 01010011 = 64 + 16 + 2 + 1 = 83
        Assert.AreEqual(83, (byte)~b);
    }

This test passes. However without the byte cast it fails because the "~" operator returns a value of -173. Why is this?

Upvotes: 16

Views: 3724

Answers (2)

Tugrul Ates
Tugrul Ates

Reputation: 9687

A promotion to int occurs on byte because binary complement is not defined for them.

See the documentation for bitwise and shift operators and numeric promotions.

Intrinsically, when you call ~ on the unsigned 8 bit value 10101100, it is promoted to the 32-bit signed value 0...010101100. Its complement is the 32-bit value 1...101010011, which is equal to -173 for int. A cast of this result into byte is a demotion to the unsigned 8-bit value 01010011, losing the most significant 24 bits. The end result is interpreted as 83 in an unsigned representation.

Upvotes: 19

Yahia
Yahia

Reputation: 70369

Because ~ returns an int. See ~ Operator (C# Reference) (MSDN)

It is only predefined for int, uint, long, and ulong - so there is an implicit cast when using it on byte.

Upvotes: 4

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