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Why is this an issue?
Certain bitwise operations
are not needed and should not be performed because their results are predictable.
Specifically, using & -1 with any value always results in the original value.
That is because the binary representation of -1 on a integral numeric type supporting
negative numbers, such as int or long, is based on two’s
complement and made of all 1s: 0b111…111.
Performing & between a value and 0b111…111 means applying the & operator to each bit of the value
and the bit 1, resulting in a value equal to the provided one, bit by bit.
anyValue & -1 // Noncompliant
anyValue // Compliant
Similarly, anyValue | 0 always results in anyValue, because the binary representation of 0 is always
0b000…000 and the | operator returns its first input when the second is 0.
anyValue | 0 // Noncompliant
anyValue // Compliant
The same applies to anyValue ^ 0: the ^ operator returns 1 when its two input bits are different
(1 and 0 or 0 and 1) and returns 0 when its two input bits are the same (both
0 or both 1). When ^ is applied with 0, the result would be 1 if the other input is
1, because the two input bits are different, and 0 if the other input bit is 0, because the two input are the
same. That results in returning anyValue.
anyValue ^ 0 // Noncompliant
anyValue // Compliant
Resources
Documentation
- Bitwise operations (C#
reference)
- And Operator (Visual Basic)
- Or Operator (Visual Basic)
- Xor Operator (Visual Basic)
- Integral numeric types (C#
reference)
- Numeric Data
Types (Visual Basic)
- Two’s complement
Articles & blog posts