Update for extra bit of integer range.
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1 changed files with 72 additions and 72 deletions
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@ -1,6 +1,6 @@
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@c -*-texinfo-*-
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@c This is part of the GNU Emacs Lisp Reference Manual.
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@c Copyright (C) 1990, 1991, 1992, 1993, 1994, 1995, 1998, 1999
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@c Copyright (C) 1990, 1991, 1992, 1993, 1994, 1995, 1998, 1999, 2003
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@c Free Software Foundation, Inc.
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@c See the file elisp.texi for copying conditions.
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@setfilename ../info/numbers
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@ -36,22 +36,22 @@ exact; they have a fixed, limited amount of precision.
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@section Integer Basics
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The range of values for an integer depends on the machine. The
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minimum range is @minus{}134217728 to 134217727 (28 bits; i.e.,
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minimum range is @minus{}268435456 to 268435455 (29 bits; i.e.,
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@ifnottex
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-2**27
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-2**28
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@end ifnottex
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@tex
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@math{-2^{27}}
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@math{-2^{28}}
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@end tex
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to
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@ifnottex
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2**27 - 1),
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2**28 - 1),
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@end ifnottex
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@tex
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@math{2^{27}-1}),
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@math{2^{28}-1}),
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@end tex
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but some machines may provide a wider range. Many examples in this
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chapter assume an integer has 28 bits.
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chapter assume an integer has 29 bits.
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@cindex overflow
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The Lisp reader reads an integer as a sequence of digits with optional
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@ -86,10 +86,10 @@ inclusively). Case is not significant for the letter after @samp{#}
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bitwise operators (@pxref{Bitwise Operations}), it is often helpful to
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view the numbers in their binary form.
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In 28-bit binary, the decimal integer 5 looks like this:
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In 29-bit binary, the decimal integer 5 looks like this:
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@example
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0000 0000 0000 0000 0000 0000 0101
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0 0000 0000 0000 0000 0000 0000 0101
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@end example
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@noindent
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@ -99,12 +99,12 @@ between groups of 8 bits, to make the binary integer easier to read.)
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The integer @minus{}1 looks like this:
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@example
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1111 1111 1111 1111 1111 1111 1111
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1 1111 1111 1111 1111 1111 1111 1111
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@end example
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@noindent
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@cindex two's complement
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@minus{}1 is represented as 28 ones. (This is called @dfn{two's
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@minus{}1 is represented as 29 ones. (This is called @dfn{two's
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complement} notation.)
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The negative integer, @minus{}5, is creating by subtracting 4 from
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@ -112,24 +112,24 @@ complement} notation.)
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@minus{}5 looks like this:
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@example
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1111 1111 1111 1111 1111 1111 1011
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1 1111 1111 1111 1111 1111 1111 1011
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@end example
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In this implementation, the largest 28-bit binary integer value is
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134,217,727 in decimal. In binary, it looks like this:
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In this implementation, the largest 29-bit binary integer value is
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268,435,455 in decimal. In binary, it looks like this:
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@example
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0111 1111 1111 1111 1111 1111 1111
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0 1111 1111 1111 1111 1111 1111 1111
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@end example
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Since the arithmetic functions do not check whether integers go
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outside their range, when you add 1 to 134,217,727, the value is the
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negative integer @minus{}134,217,728:
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outside their range, when you add 1 to 268,435,455, the value is the
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negative integer @minus{}268,435,456:
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@example
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(+ 1 134217727)
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@result{} -134217728
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@result{} 1000 0000 0000 0000 0000 0000 0000
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(+ 1 268435455)
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@result{} -268435456
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@result{} 1 0000 0000 0000 0000 0000 0000 0000
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@end example
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Many of the functions described in this chapter accept markers for
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@ -468,8 +468,8 @@ commonly used.
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if any argument is floating.
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It is important to note that in Emacs Lisp, arithmetic functions
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do not check for overflow. Thus @code{(1+ 134217727)} may evaluate to
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@minus{}134217728, depending on your hardware.
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do not check for overflow. Thus @code{(1+ 268435455)} may evaluate to
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@minus{}268435456, depending on your hardware.
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@defun 1+ number-or-marker
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This function returns @var{number-or-marker} plus 1.
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@ -788,19 +788,19 @@ value of a positive integer by two, rounding downward.
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The function @code{lsh}, like all Emacs Lisp arithmetic functions, does
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not check for overflow, so shifting left can discard significant bits
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and change the sign of the number. For example, left shifting
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134,217,727 produces @minus{}2 on a 28-bit machine:
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268,435,455 produces @minus{}2 on a 29-bit machine:
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@example
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(lsh 134217727 1) ; @r{left shift}
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(lsh 268435455 1) ; @r{left shift}
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@result{} -2
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@end example
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In binary, in the 28-bit implementation, the argument looks like this:
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In binary, in the 29-bit implementation, the argument looks like this:
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@example
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@group
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;; @r{Decimal 134,217,727}
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0111 1111 1111 1111 1111 1111 1111
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;; @r{Decimal 268,435,455}
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0 1111 1111 1111 1111 1111 1111 1111
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@end group
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@end example
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@ -810,7 +810,7 @@ which becomes the following when left shifted:
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@example
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@group
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;; @r{Decimal @minus{}2}
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1111 1111 1111 1111 1111 1111 1110
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1 1111 1111 1111 1111 1111 1111 1110
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@end group
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@end example
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@end defun
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@ -833,9 +833,9 @@ looks like this:
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@group
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(ash -6 -1) @result{} -3
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;; @r{Decimal @minus{}6 becomes decimal @minus{}3.}
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1111 1111 1111 1111 1111 1111 1010
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1 1111 1111 1111 1111 1111 1111 1010
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@result{}
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1111 1111 1111 1111 1111 1111 1101
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1 1111 1111 1111 1111 1111 1111 1101
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@end group
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@end example
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@ -844,11 +844,11 @@ In contrast, shifting the pattern of bits one place to the right with
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@example
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@group
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(lsh -6 -1) @result{} 134217725
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;; @r{Decimal @minus{}6 becomes decimal 134,217,725.}
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1111 1111 1111 1111 1111 1111 1010
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(lsh -6 -1) @result{} 268435453
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;; @r{Decimal @minus{}6 becomes decimal 268,435,453.}
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1 1111 1111 1111 1111 1111 1111 1010
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@result{}
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0111 1111 1111 1111 1111 1111 1101
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0 1111 1111 1111 1111 1111 1111 1101
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@end group
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@end example
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@ -858,34 +858,34 @@ Here are other examples:
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@c with smallbook but not with regular book! --rjc 16mar92
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@smallexample
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@group
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; @r{ 28-bit binary values}
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; @r{ 29-bit binary values}
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(lsh 5 2) ; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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@result{} 20 ; = @r{0000 0000 0000 0000 0000 0001 0100}
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(lsh 5 2) ; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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@result{} 20 ; = @r{0 0000 0000 0000 0000 0000 0001 0100}
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@end group
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@group
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(ash 5 2)
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@result{} 20
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(lsh -5 2) ; -5 = @r{1111 1111 1111 1111 1111 1111 1011}
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@result{} -20 ; = @r{1111 1111 1111 1111 1111 1110 1100}
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(lsh -5 2) ; -5 = @r{1 1111 1111 1111 1111 1111 1111 1011}
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@result{} -20 ; = @r{1 1111 1111 1111 1111 1111 1110 1100}
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(ash -5 2)
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@result{} -20
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@end group
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@group
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(lsh 5 -2) ; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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@result{} 1 ; = @r{0000 0000 0000 0000 0000 0000 0001}
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(lsh 5 -2) ; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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@result{} 1 ; = @r{0 0000 0000 0000 0000 0000 0000 0001}
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@end group
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@group
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(ash 5 -2)
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@result{} 1
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@end group
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@group
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(lsh -5 -2) ; -5 = @r{1111 1111 1111 1111 1111 1111 1011}
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@result{} 4194302 ; = @r{0011 1111 1111 1111 1111 1111 1110}
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(lsh -5 -2) ; -5 = @r{1 1111 1111 1111 1111 1111 1111 1011}
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@result{} 134217726 ; = @r{0 0111 1111 1111 1111 1111 1111 1110}
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@end group
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@group
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(ash -5 -2) ; -5 = @r{1111 1111 1111 1111 1111 1111 1011}
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@result{} -2 ; = @r{1111 1111 1111 1111 1111 1111 1110}
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(ash -5 -2) ; -5 = @r{1 1111 1111 1111 1111 1111 1111 1011}
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@result{} -2 ; = @r{1 1111 1111 1111 1111 1111 1111 1110}
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@end group
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@end smallexample
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@end defun
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@ -922,23 +922,23 @@ because its binary representation consists entirely of ones. If
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@smallexample
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@group
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; @r{ 28-bit binary values}
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; @r{ 29-bit binary values}
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(logand 14 13) ; 14 = @r{0000 0000 0000 0000 0000 0000 1110}
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; 13 = @r{0000 0000 0000 0000 0000 0000 1101}
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@result{} 12 ; 12 = @r{0000 0000 0000 0000 0000 0000 1100}
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(logand 14 13) ; 14 = @r{0 0000 0000 0000 0000 0000 0000 1110}
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; 13 = @r{0 0000 0000 0000 0000 0000 0000 1101}
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@result{} 12 ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
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@end group
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@group
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(logand 14 13 4) ; 14 = @r{0000 0000 0000 0000 0000 0000 1110}
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; 13 = @r{0000 0000 0000 0000 0000 0000 1101}
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; 4 = @r{0000 0000 0000 0000 0000 0000 0100}
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@result{} 4 ; 4 = @r{0000 0000 0000 0000 0000 0000 0100}
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(logand 14 13 4) ; 14 = @r{0 0000 0000 0000 0000 0000 0000 1110}
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; 13 = @r{0 0000 0000 0000 0000 0000 0000 1101}
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; 4 = @r{0 0000 0000 0000 0000 0000 0000 0100}
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@result{} 4 ; 4 = @r{0 0000 0000 0000 0000 0000 0000 0100}
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@end group
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@group
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(logand)
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@result{} -1 ; -1 = @r{1111 1111 1111 1111 1111 1111 1111}
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@result{} -1 ; -1 = @r{1 1111 1111 1111 1111 1111 1111 1111}
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@end group
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@end smallexample
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@end defun
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@ -954,18 +954,18 @@ passed just one argument, it returns that argument.
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@smallexample
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@group
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; @r{ 28-bit binary values}
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; @r{ 29-bit binary values}
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(logior 12 5) ; 12 = @r{0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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@result{} 13 ; 13 = @r{0000 0000 0000 0000 0000 0000 1101}
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(logior 12 5) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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@result{} 13 ; 13 = @r{0 0000 0000 0000 0000 0000 0000 1101}
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@end group
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@group
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(logior 12 5 7) ; 12 = @r{0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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; 7 = @r{0000 0000 0000 0000 0000 0000 0111}
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@result{} 15 ; 15 = @r{0000 0000 0000 0000 0000 0000 1111}
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(logior 12 5 7) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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; 7 = @r{0 0000 0000 0000 0000 0000 0000 0111}
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@result{} 15 ; 15 = @r{0 0000 0000 0000 0000 0000 0000 1111}
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@end group
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@end smallexample
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@end defun
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@ -981,18 +981,18 @@ result is 0, which is an identity element for this operation. If
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@smallexample
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@group
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; @r{ 28-bit binary values}
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; @r{ 29-bit binary values}
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(logxor 12 5) ; 12 = @r{0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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@result{} 9 ; 9 = @r{0000 0000 0000 0000 0000 0000 1001}
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(logxor 12 5) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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@result{} 9 ; 9 = @r{0 0000 0000 0000 0000 0000 0000 1001}
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@end group
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@group
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(logxor 12 5 7) ; 12 = @r{0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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; 7 = @r{0000 0000 0000 0000 0000 0000 0111}
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@result{} 14 ; 14 = @r{0000 0000 0000 0000 0000 0000 1110}
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(logxor 12 5 7) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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; 7 = @r{0 0000 0000 0000 0000 0000 0000 0111}
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@result{} 14 ; 14 = @r{0 0000 0000 0000 0000 0000 0000 1110}
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@end group
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@end smallexample
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@end defun
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@ -1007,9 +1007,9 @@ bit is one in the result if, and only if, the @var{n}th bit is zero in
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@example
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(lognot 5)
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@result{} -6
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;; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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;; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
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;; @r{becomes}
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;; -6 = @r{1111 1111 1111 1111 1111 1111 1010}
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;; -6 = @r{1 1111 1111 1111 1111 1111 1111 1010}
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@end example
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@end defun
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