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In a computer, an integer is represented as a binary number, a sequence of bits (digits which are either zero or one). A bitwise operation acts on the individual bits of such a sequence. For example, shifting moves the whole sequence left or right one or more places, reproducing the same pattern moved over.
The bitwise operations in Emacs Lisp apply only to integers.
lsh
, which is an abbreviation for logical shift, shifts the bits in integer1 to the left count places, or to the right if count is negative, bringing zeros into the vacated bits. If count is negative,lsh
shifts zeros into the leftmost (most-significant) bit, producing a positive result even if integer1 is negative. Contrast this withash
, below.Here are two examples of
lsh
, shifting a pattern of bits one place to the left. We show only the low-order eight bits of the binary pattern; the rest are all zero.(lsh 5 1) ⇒ 10 ;; Decimal 5 becomes decimal 10. 00000101 ⇒ 00001010 (lsh 7 1) ⇒ 14 ;; Decimal 7 becomes decimal 14. 00000111 ⇒ 00001110As the examples illustrate, shifting the pattern of bits one place to the left produces a number that is twice the value of the previous number.
Shifting a pattern of bits two places to the left produces results like this (with 8-bit binary numbers):
(lsh 3 2) ⇒ 12 ;; Decimal 3 becomes decimal 12. 00000011 ⇒ 00001100
On the other hand, shifting one place to the right looks like this:
(lsh 6 -1) ⇒ 3 ;; Decimal 6 becomes decimal 3. 00000110 ⇒ 00000011 (lsh 5 -1) ⇒ 2 ;; Decimal 5 becomes decimal 2. 00000101 ⇒ 00000010As the example illustrates, shifting one place to the right divides the value of a positive integer by two, rounding downward.
The function
lsh
, like all Emacs Lisp arithmetic functions, does not check for overflow, so shifting left can discard significant bits and change the sign of the number. For example, left shifting 536,870,911 produces −2 in the 30-bit implementation:(lsh 536870911 1) ; left shift ⇒ -2
In binary, the argument looks like this:
;; Decimal 536,870,911 0111...111111 (30 bits total)
which becomes the following when left shifted:
;; Decimal −2 1111...111110 (30 bits total)
ash
(arithmetic shift) shifts the bits in integer1 to the left count places, or to the right if count is negative.
ash
gives the same results aslsh
except when integer1 and count are both negative. In that case,ash
puts ones in the empty bit positions on the left, whilelsh
puts zeros in those bit positions.Thus, with
ash
, shifting the pattern of bits one place to the right looks like this:(ash -6 -1) ⇒ -3 ;; Decimal −6 becomes decimal −3. 1111...111010 (30 bits total) ⇒ 1111...111101 (30 bits total)
In contrast, shifting the pattern of bits one place to the right with
lsh
looks like this:(lsh -6 -1) ⇒ 536870909 ;; Decimal −6 becomes decimal 536,870,909. 1111...111010 (30 bits total) ⇒ 0111...111101 (30 bits total)
Here are other examples:
; 30-bit binary values (lsh 5 2) ; 5 = 0000...000101 ⇒ 20 ; = 0000...010100 (ash 5 2) ⇒ 20 (lsh -5 2) ; -5 = 1111...111011 ⇒ -20 ; = 1111...101100 (ash -5 2) ⇒ -20 (lsh 5 -2) ; 5 = 0000...000101 ⇒ 1 ; = 0000...000001 (ash 5 -2) ⇒ 1 (lsh -5 -2) ; -5 = 1111...111011 ⇒ 268435454 ; = 0011...111110 (ash -5 -2) ; -5 = 1111...111011 ⇒ -2 ; = 1111...111110
This function returns the bitwise AND of the arguments: the nth bit is 1 in the result if, and only if, the nth bit is 1 in all the arguments.
For example, using 4-bit binary numbers, the bitwise AND of 13 and 12 is 12: 1101 combined with 1100 produces 1100. In both the binary numbers, the leftmost two bits are both 1 so the leftmost two bits of the returned value are both 1. However, for the rightmost two bits, each is 0 in at least one of the arguments, so the rightmost two bits of the returned value are both 0.
Therefore,
(logand 13 12) ⇒ 12If
logand
is not passed any argument, it returns a value of −1. This number is an identity element forlogand
because its binary representation consists entirely of ones. Iflogand
is passed just one argument, it returns that argument.; 30-bit binary values (logand 14 13) ; 14 = 0000...001110 ; 13 = 0000...001101 ⇒ 12 ; 12 = 0000...001100 (logand 14 13 4) ; 14 = 0000...001110 ; 13 = 0000...001101 ; 4 = 0000...000100 ⇒ 4 ; 4 = 0000...000100 (logand) ⇒ -1 ; -1 = 1111...111111
This function returns the bitwise inclusive OR of its arguments: the nth bit is 1 in the result if, and only if, the nth bit is 1 in at least one of the arguments. If there are no arguments, the result is 0, which is an identity element for this operation. If
logior
is passed just one argument, it returns that argument.; 30-bit binary values (logior 12 5) ; 12 = 0000...001100 ; 5 = 0000...000101 ⇒ 13 ; 13 = 0000...001101 (logior 12 5 7) ; 12 = 0000...001100 ; 5 = 0000...000101 ; 7 = 0000...000111 ⇒ 15 ; 15 = 0000...001111
This function returns the bitwise exclusive OR of its arguments: the nth bit is 1 in the result if, and only if, the nth bit is 1 in an odd number of the arguments. If there are no arguments, the result is 0, which is an identity element for this operation. If
logxor
is passed just one argument, it returns that argument.; 30-bit binary values (logxor 12 5) ; 12 = 0000...001100 ; 5 = 0000...000101 ⇒ 9 ; 9 = 0000...001001 (logxor 12 5 7) ; 12 = 0000...001100 ; 5 = 0000...000101 ; 7 = 0000...000111 ⇒ 14 ; 14 = 0000...001110