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?
What is a Bitwise Operator? The Bitwise Operator in C is a type of operator that operates on bit arrays, bit strings, and tweaking binary values with individual bits at the bit level. For handling electronics and IoT-related operations, programmers use bitwise operators. It can operate faster at a bit level.
The bitwise AND & operator returns 1 if and only if both the operands are 1. Otherwise, it returns 0. The following table demonstrates the working of the bitwise AND operator. Let a and b be two operands that can only take binary values i.e. 1 and 0.
A promotion to int
occurs on byte
because binary complement is not defined for them.
See Unary numeric promotions and Bitwise complement operator.
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.
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
.
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