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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.

- Function:
`lsh`

*integer1 count* -
`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)`

- Function:
`ash`

*integer1 count* -
`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

- Function:
`logand`

*&rest ints-or-markers* This function returns the bitwise AND of the arguments: the

`n`th bit is 1 in the result if, and only if, the`n`th 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) ⇒ 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`

- Function:
`logior`

*&rest ints-or-markers* This function returns the bitwise inclusive OR of its arguments: the

`n`th bit is 1 in the result if, and only if, the`n`th 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

- Function:
`logxor`

*&rest ints-or-markers* This function returns the bitwise exclusive OR of its arguments: the

`n`th bit is 1 in the result if, and only if, the`n`th 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

- Function:
`lognot`

*integer* This function returns the bitwise complement of its argument: the

`n`th bit is one in the result if, and only if, the`n`th 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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