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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 with ash, 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 ⇒ 00001110
As 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 ⇒ 00000010
As 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 as lsh except when
integer1 and count are both negative. In that case,
ash puts ones in the empty bit positions on the left, while
lsh 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 “logical and” of the arguments: the nth bit is set in the result if, and only if, the nth bit is set in all the arguments. (“Set” means that the value of the bit is 1 rather than 0.)
For example, using 4-bit binary numbers, the “logical and” of 13 and 12 is 12: 1101 combined with 1100 produces 1100. In both the binary numbers, the leftmost two bits are set (i.e., they are 1’s), so the leftmost two bits of the returned value are set. However, for the rightmost two bits, each is zero in at least one of the arguments, so the rightmost two bits of the returned value are 0’s.
Therefore,
(logand 13 12)
⇒ 12
If logand is not passed any argument, it returns a value of
-1. This number is an identity element for logand
because its binary representation consists entirely of ones. If
logand 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 “inclusive or” of its arguments: the nth bit
is set in the result if, and only if, the nth bit is set in at least
one of the arguments. If there are no arguments, the result is zero,
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 “exclusive or” of its arguments: the
nth bit is set in the result if, and only if, the nth bit is
set 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
This function returns the logical complement of its argument: the nth bit is one in the result if, and only if, the nth bit is zero in integer, and vice-versa.
(lognot 5)
⇒ -6
;; 5 = 0000...000101 (30 bits total)
;; becomes
;; -6 = 1111...111010 (30 bits total)
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