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Update Lispref for 30-bit integers.
* numbers.texi (Integer Basics, Bitwise Operations): * objects.texi (Integer Type): Update for integers now being 30-bit.
This commit is contained in:
Glenn Morris
parent
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@ -1,3 +1,8 @@
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2010-03-03 Glenn Morris <rgm@gnu.org>
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* numbers.texi (Integer Basics, Bitwise Operations):
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* objects.texi (Integer Type): Update for integers now being 30-bit.
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2010-02-27 Chong Yidong <cyd@stupidchicken.com>
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* display.texi (Low-Level Font): Document :otf font-spec property.
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@ -1,7 +1,8 @@
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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, 2001,
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@c 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010 Free Software Foundation, Inc.
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@c 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010
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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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@node Numbers, Strings and Characters, Lisp Data Types, Top
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@ -36,22 +37,22 @@ 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{}268435456 to 268435455 (29 bits; i.e.,
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minimum range is @minus{}536870912 to 536870911 (30 bits; i.e.,
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@ifnottex
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-2**28
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-2**29
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@end ifnottex
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@tex
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@math{-2^{28}}
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@math{-2^{29}}
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@end tex
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to
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@ifnottex
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2**28 - 1),
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2**29 - 1),
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@end ifnottex
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@tex
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@math{2^{28}-1}),
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@math{2^{29}-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 29 bits.
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chapter assume an integer has 30 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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@ -62,7 +63,7 @@ initial sign and optional final period.
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1. ; @r{The integer 1.}
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+1 ; @r{Also the integer 1.}
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-1 ; @r{The integer @minus{}1.}
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536870913 ; @r{Also the integer 1, due to overflow.}
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1073741825 ; @r{Also the integer 1, due to overflow.}
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0 ; @r{The integer 0.}
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-0 ; @r{The integer 0.}
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@end example
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@ -93,10 +94,10 @@ from 2 to 36. For example:
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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 29-bit binary, the decimal integer 5 looks like this:
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In 30-bit binary, the decimal integer 5 looks like this:
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@example
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0 0000 0000 0000 0000 0000 0000 0101
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00 0000 0000 0000 0000 0000 0000 0101
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@end example
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@noindent
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@ -106,12 +107,12 @@ 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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1 1111 1111 1111 1111 1111 1111 1111
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11 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 29 ones. (This is called @dfn{two's
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@minus{}1 is represented as 30 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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@ -119,24 +120,24 @@ complement} notation.)
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@minus{}5 looks like this:
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@example
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1 1111 1111 1111 1111 1111 1111 1011
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11 1111 1111 1111 1111 1111 1111 1011
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@end example
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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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In this implementation, the largest 30-bit binary integer value is
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536,870,911 in decimal. In binary, it looks like this:
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@example
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0 1111 1111 1111 1111 1111 1111 1111
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01 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 268,435,455, the value is the
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negative integer @minus{}268,435,456:
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outside their range, when you add 1 to 536,870,911, the value is the
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negative integer @minus{}536,870,912:
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@example
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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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(+ 1 536870911)
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@result{} -536870912
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@result{} 10 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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@ -820,19 +821,19 @@ 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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268,435,455 produces @minus{}2 on a 29-bit machine:
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536,870,911 produces @minus{}2 on a 30-bit machine:
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@example
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(lsh 268435455 1) ; @r{left shift}
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(lsh 536870911 1) ; @r{left shift}
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@result{} -2
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@end example
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In binary, in the 29-bit implementation, the argument looks like this:
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In binary, in the 30-bit implementation, the argument looks like this:
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@example
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@group
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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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;; @r{Decimal 536,870,911}
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01 1111 1111 1111 1111 1111 1111 1111
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@end group
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@end example
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@ -842,7 +843,7 @@ 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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1 1111 1111 1111 1111 1111 1111 1110
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11 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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@ -865,9 +866,9 @@ 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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1 1111 1111 1111 1111 1111 1111 1010
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11 1111 1111 1111 1111 1111 1111 1010
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@result{}
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1 1111 1111 1111 1111 1111 1111 1101
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11 1111 1111 1111 1111 1111 1111 1101
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@end group
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@end example
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@ -876,11 +877,11 @@ 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{} 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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(lsh -6 -1) @result{} 536870909
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;; @r{Decimal @minus{}6 becomes decimal 536,870,909.}
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11 1111 1111 1111 1111 1111 1111 1010
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@result{}
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0 1111 1111 1111 1111 1111 1111 1101
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01 1111 1111 1111 1111 1111 1111 1101
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@end group
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@end example
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@ -890,34 +891,34 @@ 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{ 29-bit binary values}
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; @r{ 30-bit binary values}
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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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(lsh 5 2) ; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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@result{} 20 ; = @r{00 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{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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(lsh -5 2) ; -5 = @r{11 1111 1111 1111 1111 1111 1111 1011}
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@result{} -20 ; = @r{11 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{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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(lsh 5 -2) ; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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@result{} 1 ; = @r{00 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{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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(lsh -5 -2) ; -5 = @r{11 1111 1111 1111 1111 1111 1111 1011}
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@result{} 268435454 ; = @r{00 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{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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(ash -5 -2) ; -5 = @r{11 1111 1111 1111 1111 1111 1111 1011}
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@result{} -2 ; = @r{11 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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@ -952,23 +953,23 @@ because its binary representation consists entirely of ones. If
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@smallexample
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@group
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; @r{ 29-bit binary values}
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; @r{ 30-bit binary values}
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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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(logand 14 13) ; 14 = @r{00 0000 0000 0000 0000 0000 0000 1110}
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; 13 = @r{00 0000 0000 0000 0000 0000 0000 1101}
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@result{} 12 ; 12 = @r{00 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{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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(logand 14 13 4) ; 14 = @r{00 0000 0000 0000 0000 0000 0000 1110}
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; 13 = @r{00 0000 0000 0000 0000 0000 0000 1101}
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; 4 = @r{00 0000 0000 0000 0000 0000 0000 0100}
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@result{} 4 ; 4 = @r{00 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{1 1111 1111 1111 1111 1111 1111 1111}
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@result{} -1 ; -1 = @r{11 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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@ -982,18 +983,18 @@ passed just one argument, it returns that argument.
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@smallexample
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@group
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; @r{ 29-bit binary values}
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; @r{ 30-bit binary values}
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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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(logior 12 5) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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@result{} 13 ; 13 = @r{00 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{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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(logior 12 5 7) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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; 7 = @r{00 0000 0000 0000 0000 0000 0000 0111}
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@result{} 15 ; 15 = @r{00 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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@ -1007,18 +1008,18 @@ 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{ 29-bit binary values}
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; @r{ 30-bit binary values}
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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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(logxor 12 5) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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@result{} 9 ; 9 = @r{00 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{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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(logxor 12 5 7) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
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; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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; 7 = @r{00 0000 0000 0000 0000 0000 0000 0111}
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@result{} 14 ; 14 = @r{00 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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@ -1031,9 +1032,9 @@ 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{0 0000 0000 0000 0000 0000 0000 0101}
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;; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
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;; @r{becomes}
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;; -6 = @r{1 1111 1111 1111 1111 1111 1111 1010}
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;; -6 = @r{11 1111 1111 1111 1111 1111 1111 1010}
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@end example
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@end defun
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@ -165,24 +165,24 @@ latter are unique to Emacs Lisp.
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@node Integer Type
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@subsection Integer Type
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The range of values for integers in Emacs Lisp is @minus{}268435456 to
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268435455 (29 bits; i.e.,
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The range of values for integers in Emacs Lisp is @minus{}536870912 to
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536870911 (30 bits; i.e.,
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@ifnottex
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-2**28
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-2**29
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@end ifnottex
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@tex
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@math{-2^{28}}
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@math{-2^{29}}
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@end tex
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to
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@ifnottex
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2**28 - 1)
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2**29 - 1)
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@end ifnottex
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@tex
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@math{2^{28}-1})
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@math{2^{29}-1})
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@end tex
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on most machines. (Some machines may provide a wider range.) It is
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important to note that the Emacs Lisp arithmetic functions do not check
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for overflow. Thus @code{(1+ 268435455)} is @minus{}268435456 on most
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for overflow. Thus @code{(1+ 536870911)} is @minus{}536870912 on most
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machines.
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The read syntax for integers is a sequence of (base ten) digits with an
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@ -196,7 +196,7 @@ leading @samp{+} or a final @samp{.}.
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1 ; @r{The integer 1.}
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1. ; @r{Also the integer 1.}
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+1 ; @r{Also the integer 1.}
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536870913 ; @r{Also the integer 1 on a 29-bit implementation.}
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1073741825 ; @r{Also the integer 1 on a 30-bit implementation.}
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@end group
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@end example
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