Commit 1ddd6622 authored by Glenn Morris's avatar Glenn Morris

Update Lispref for 30-bit integers.

* numbers.texi (Integer Basics, Bitwise Operations):
* objects.texi (Integer Type): Update for integers now being 30-bit.
parent 855dc98c
2010-03-03 Glenn Morris <rgm@gnu.org>
* numbers.texi (Integer Basics, Bitwise Operations):
* objects.texi (Integer Type): Update for integers now being 30-bit.
2010-02-27 Chong Yidong <cyd@stupidchicken.com>
* display.texi (Low-Level Font): Document :otf font-spec property.
......
@c -*-texinfo-*-
@c This is part of the GNU Emacs Lisp Reference Manual.
@c Copyright (C) 1990, 1991, 1992, 1993, 1994, 1995, 1998, 1999, 2001,
@c 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010 Free Software Foundation, Inc.
@c 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010
@c Free Software Foundation, Inc.
@c See the file elisp.texi for copying conditions.
@setfilename ../../info/numbers
@node Numbers, Strings and Characters, Lisp Data Types, Top
......@@ -36,22 +37,22 @@ exact; they have a fixed, limited amount of precision.
@section Integer Basics
The range of values for an integer depends on the machine. The
minimum range is @minus{}268435456 to 268435455 (29 bits; i.e.,
minimum range is @minus{}536870912 to 536870911 (30 bits; i.e.,
@ifnottex
-2**28
-2**29
@end ifnottex
@tex
@math{-2^{28}}
@math{-2^{29}}
@end tex
to
@ifnottex
2**28 - 1),
2**29 - 1),
@end ifnottex
@tex
@math{2^{28}-1}),
@math{2^{29}-1}),
@end tex
but some machines may provide a wider range. Many examples in this
chapter assume an integer has 29 bits.
chapter assume an integer has 30 bits.
@cindex overflow
The Lisp reader reads an integer as a sequence of digits with optional
......@@ -62,7 +63,7 @@ initial sign and optional final period.
1. ; @r{The integer 1.}
+1 ; @r{Also the integer 1.}
-1 ; @r{The integer @minus{}1.}
536870913 ; @r{Also the integer 1, due to overflow.}
1073741825 ; @r{Also the integer 1, due to overflow.}
0 ; @r{The integer 0.}
-0 ; @r{The integer 0.}
@end example
......@@ -93,10 +94,10 @@ from 2 to 36. For example:
bitwise operators (@pxref{Bitwise Operations}), it is often helpful to
view the numbers in their binary form.
In 29-bit binary, the decimal integer 5 looks like this:
In 30-bit binary, the decimal integer 5 looks like this:
@example
0 0000 0000 0000 0000 0000 0000 0101
00 0000 0000 0000 0000 0000 0000 0101
@end example
@noindent
......@@ -106,12 +107,12 @@ between groups of 8 bits, to make the binary integer easier to read.)
The integer @minus{}1 looks like this:
@example
1 1111 1111 1111 1111 1111 1111 1111
11 1111 1111 1111 1111 1111 1111 1111
@end example
@noindent
@cindex two's complement
@minus{}1 is represented as 29 ones. (This is called @dfn{two's
@minus{}1 is represented as 30 ones. (This is called @dfn{two's
complement} notation.)
The negative integer, @minus{}5, is creating by subtracting 4 from
......@@ -119,24 +120,24 @@ complement} notation.)
@minus{}5 looks like this:
@example
1 1111 1111 1111 1111 1111 1111 1011
11 1111 1111 1111 1111 1111 1111 1011
@end example
In this implementation, the largest 29-bit binary integer value is
268,435,455 in decimal. In binary, it looks like this:
In this implementation, the largest 30-bit binary integer value is
536,870,911 in decimal. In binary, it looks like this:
@example
0 1111 1111 1111 1111 1111 1111 1111
01 1111 1111 1111 1111 1111 1111 1111
@end example
Since the arithmetic functions do not check whether integers go
outside their range, when you add 1 to 268,435,455, the value is the
negative integer @minus{}268,435,456:
outside their range, when you add 1 to 536,870,911, the value is the
negative integer @minus{}536,870,912:
@example
(+ 1 268435455)
@result{} -268435456
@result{} 1 0000 0000 0000 0000 0000 0000 0000
(+ 1 536870911)
@result{} -536870912
@result{} 10 0000 0000 0000 0000 0000 0000 0000
@end example
Many of the functions described in this chapter accept markers for
......@@ -820,19 +821,19 @@ value of a positive integer by two, rounding downward.
The function @code{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
268,435,455 produces @minus{}2 on a 29-bit machine:
536,870,911 produces @minus{}2 on a 30-bit machine:
@example
(lsh 268435455 1) ; @r{left shift}
(lsh 536870911 1) ; @r{left shift}
@result{} -2
@end example
In binary, in the 29-bit implementation, the argument looks like this:
In binary, in the 30-bit implementation, the argument looks like this:
@example
@group
;; @r{Decimal 268,435,455}
0 1111 1111 1111 1111 1111 1111 1111
;; @r{Decimal 536,870,911}
01 1111 1111 1111 1111 1111 1111 1111
@end group
@end example
......@@ -842,7 +843,7 @@ which becomes the following when left shifted:
@example
@group
;; @r{Decimal @minus{}2}
1 1111 1111 1111 1111 1111 1111 1110
11 1111 1111 1111 1111 1111 1111 1110
@end group
@end example
@end defun
......@@ -865,9 +866,9 @@ looks like this:
@group
(ash -6 -1) @result{} -3
;; @r{Decimal @minus{}6 becomes decimal @minus{}3.}
1 1111 1111 1111 1111 1111 1111 1010
11 1111 1111 1111 1111 1111 1111 1010
@result{}
1 1111 1111 1111 1111 1111 1111 1101
11 1111 1111 1111 1111 1111 1111 1101
@end group
@end example
......@@ -876,11 +877,11 @@ In contrast, shifting the pattern of bits one place to the right with
@example
@group
(lsh -6 -1) @result{} 268435453
;; @r{Decimal @minus{}6 becomes decimal 268,435,453.}
1 1111 1111 1111 1111 1111 1111 1010
(lsh -6 -1) @result{} 536870909
;; @r{Decimal @minus{}6 becomes decimal 536,870,909.}
11 1111 1111 1111 1111 1111 1111 1010
@result{}
0 1111 1111 1111 1111 1111 1111 1101
01 1111 1111 1111 1111 1111 1111 1101
@end group
@end example
......@@ -890,34 +891,34 @@ Here are other examples:
@c with smallbook but not with regular book! --rjc 16mar92
@smallexample
@group
; @r{ 29-bit binary values}
; @r{ 30-bit binary values}
(lsh 5 2) ; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
@result{} 20 ; = @r{0 0000 0000 0000 0000 0000 0001 0100}
(lsh 5 2) ; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
@result{} 20 ; = @r{00 0000 0000 0000 0000 0000 0001 0100}
@end group
@group
(ash 5 2)
@result{} 20
(lsh -5 2) ; -5 = @r{1 1111 1111 1111 1111 1111 1111 1011}
@result{} -20 ; = @r{1 1111 1111 1111 1111 1111 1110 1100}
(lsh -5 2) ; -5 = @r{11 1111 1111 1111 1111 1111 1111 1011}
@result{} -20 ; = @r{11 1111 1111 1111 1111 1111 1110 1100}
(ash -5 2)
@result{} -20
@end group
@group
(lsh 5 -2) ; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
@result{} 1 ; = @r{0 0000 0000 0000 0000 0000 0000 0001}
(lsh 5 -2) ; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
@result{} 1 ; = @r{00 0000 0000 0000 0000 0000 0000 0001}
@end group
@group
(ash 5 -2)
@result{} 1
@end group
@group
(lsh -5 -2) ; -5 = @r{1 1111 1111 1111 1111 1111 1111 1011}
@result{} 134217726 ; = @r{0 0111 1111 1111 1111 1111 1111 1110}
(lsh -5 -2) ; -5 = @r{11 1111 1111 1111 1111 1111 1111 1011}
@result{} 268435454 ; = @r{00 0111 1111 1111 1111 1111 1111 1110}
@end group
@group
(ash -5 -2) ; -5 = @r{1 1111 1111 1111 1111 1111 1111 1011}
@result{} -2 ; = @r{1 1111 1111 1111 1111 1111 1111 1110}
(ash -5 -2) ; -5 = @r{11 1111 1111 1111 1111 1111 1111 1011}
@result{} -2 ; = @r{11 1111 1111 1111 1111 1111 1111 1110}
@end group
@end smallexample
@end defun
......@@ -952,23 +953,23 @@ because its binary representation consists entirely of ones. If
@smallexample
@group
; @r{ 29-bit binary values}
; @r{ 30-bit binary values}
(logand 14 13) ; 14 = @r{0 0000 0000 0000 0000 0000 0000 1110}
; 13 = @r{0 0000 0000 0000 0000 0000 0000 1101}
@result{} 12 ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
(logand 14 13) ; 14 = @r{00 0000 0000 0000 0000 0000 0000 1110}
; 13 = @r{00 0000 0000 0000 0000 0000 0000 1101}
@result{} 12 ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
@end group
@group
(logand 14 13 4) ; 14 = @r{0 0000 0000 0000 0000 0000 0000 1110}
; 13 = @r{0 0000 0000 0000 0000 0000 0000 1101}
; 4 = @r{0 0000 0000 0000 0000 0000 0000 0100}
@result{} 4 ; 4 = @r{0 0000 0000 0000 0000 0000 0000 0100}
(logand 14 13 4) ; 14 = @r{00 0000 0000 0000 0000 0000 0000 1110}
; 13 = @r{00 0000 0000 0000 0000 0000 0000 1101}
; 4 = @r{00 0000 0000 0000 0000 0000 0000 0100}
@result{} 4 ; 4 = @r{00 0000 0000 0000 0000 0000 0000 0100}
@end group
@group
(logand)
@result{} -1 ; -1 = @r{1 1111 1111 1111 1111 1111 1111 1111}
@result{} -1 ; -1 = @r{11 1111 1111 1111 1111 1111 1111 1111}
@end group
@end smallexample
@end defun
......@@ -982,18 +983,18 @@ passed just one argument, it returns that argument.
@smallexample
@group
; @r{ 29-bit binary values}
; @r{ 30-bit binary values}
(logior 12 5) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
@result{} 13 ; 13 = @r{0 0000 0000 0000 0000 0000 0000 1101}
(logior 12 5) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
@result{} 13 ; 13 = @r{00 0000 0000 0000 0000 0000 0000 1101}
@end group
@group
(logior 12 5 7) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
; 7 = @r{0 0000 0000 0000 0000 0000 0000 0111}
@result{} 15 ; 15 = @r{0 0000 0000 0000 0000 0000 0000 1111}
(logior 12 5 7) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
; 7 = @r{00 0000 0000 0000 0000 0000 0000 0111}
@result{} 15 ; 15 = @r{00 0000 0000 0000 0000 0000 0000 1111}
@end group
@end smallexample
@end defun
......@@ -1007,18 +1008,18 @@ result is 0, which is an identity element for this operation. If
@smallexample
@group
; @r{ 29-bit binary values}
; @r{ 30-bit binary values}
(logxor 12 5) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
@result{} 9 ; 9 = @r{0 0000 0000 0000 0000 0000 0000 1001}
(logxor 12 5) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
@result{} 9 ; 9 = @r{00 0000 0000 0000 0000 0000 0000 1001}
@end group
@group
(logxor 12 5 7) ; 12 = @r{0 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
; 7 = @r{0 0000 0000 0000 0000 0000 0000 0111}
@result{} 14 ; 14 = @r{0 0000 0000 0000 0000 0000 0000 1110}
(logxor 12 5 7) ; 12 = @r{00 0000 0000 0000 0000 0000 0000 1100}
; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
; 7 = @r{00 0000 0000 0000 0000 0000 0000 0111}
@result{} 14 ; 14 = @r{00 0000 0000 0000 0000 0000 0000 1110}
@end group
@end smallexample
@end defun
......@@ -1031,9 +1032,9 @@ bit is one in the result if, and only if, the @var{n}th bit is zero in
@example
(lognot 5)
@result{} -6
;; 5 = @r{0 0000 0000 0000 0000 0000 0000 0101}
;; 5 = @r{00 0000 0000 0000 0000 0000 0000 0101}
;; @r{becomes}
;; -6 = @r{1 1111 1111 1111 1111 1111 1111 1010}
;; -6 = @r{11 1111 1111 1111 1111 1111 1111 1010}
@end example
@end defun
......
......@@ -165,24 +165,24 @@ latter are unique to Emacs Lisp.
@node Integer Type
@subsection Integer Type
The range of values for integers in Emacs Lisp is @minus{}268435456 to
268435455 (29 bits; i.e.,
The range of values for integers in Emacs Lisp is @minus{}536870912 to
536870911 (30 bits; i.e.,
@ifnottex
-2**28
-2**29
@end ifnottex
@tex
@math{-2^{28}}
@math{-2^{29}}
@end tex
to
@ifnottex
2**28 - 1)
2**29 - 1)
@end ifnottex
@tex
@math{2^{28}-1})
@math{2^{29}-1})
@end tex
on most machines. (Some machines may provide a wider range.) It is
important to note that the Emacs Lisp arithmetic functions do not check
for overflow. Thus @code{(1+ 268435455)} is @minus{}268435456 on most
for overflow. Thus @code{(1+ 536870911)} is @minus{}536870912 on most
machines.
The read syntax for integers is a sequence of (base ten) digits with an
......@@ -196,7 +196,7 @@ leading @samp{+} or a final @samp{.}.
1 ; @r{The integer 1.}
1. ; @r{Also the integer 1.}
+1 ; @r{Also the integer 1.}
536870913 ; @r{Also the integer 1 on a 29-bit implementation.}
1073741825 ; @r{Also the integer 1 on a 30-bit implementation.}
@end group
@end example
......
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