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### 6.8 Bitwise Operations

Bitwise operators operate on integers, treating each bit independently. They are not allowed for floating-point types.

The examples in this section use binary constants, starting with ‘0b’ (see Integer Constants). They stand for 32-bit integers of type int.

~a

Unary operator for bitwise negation; this changes each bit of a from 1 to 0 or from 0 to 1.

~0b10101000 ⇒ 0b11111111111111111111111101010111
~0 ⇒ 0b11111111111111111111111111111111
~0b11111111111111111111111111111111 ⇒ 0
~ (-1) ⇒ 0

It is useful to remember that ~x + 1 equals -x, for integers, and ~x equals -x - 1. The last example above shows this with -1 as x.

a & b

Binary operator for bitwise “and” or “conjunction.” Each bit in the result is 1 if that bit is 1 in both a and b.

0b10101010 & 0b11001100 ⇒ 0b10001000
a | b

Binary operator for bitwise “or” (“inclusive or” or “disjunction”). Each bit in the result is 1 if that bit is 1 in either a or b.

0b10101010 | 0b11001100 ⇒ 0b11101110
a ^ b

Binary operator for bitwise “xor” (“exclusive or”). Each bit in the result is 1 if that bit is 1 in exactly one of a and b.

0b10101010 ^ 0b11001100 ⇒ 0b01100110

To understand the effect of these operators on signed integers, keep in mind that all modern computers use two’s-complement representation (see Integer Representations) for negative integers. This means that the highest bit of the number indicates the sign; it is 1 for a negative number and 0 for a positive number. In a negative number, the value in the other bits increases as the number gets closer to zero, so that 0b111111 is -1 and 0b100000 is the most negative possible integer.

Warning: C defines a precedence ordering for the bitwise binary operators, but you should never rely on it. You should never rely on how bitwise binary operators relate in precedence to the arithmetic and shift binary operators. Other programmers don’t remember this precedence ordering, so always use parentheses to explicitly specify the nesting.

For example, suppose offset is an integer that specifies the offset within shared memory of a table, except that its bottom few bits (LOWBITS says how many) are special flags. Here’s how to get just that offset and add it to the base address.

shared_mem_base + (offset & (-1 << LOWBITS))

Thanks to the outer set of parentheses, we don’t need to know whether ‘&’ has higher precedence than ‘+’. Thanks to the inner set, we don’t need to know whether ‘&’ has higher precedence than ‘<<’. But we can rely on all unary operators to have higher precedence than any binary operator, so we don’t need parentheses around the left operand of ‘<<’.

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