Update Lispref for 30-bit integers.

* numbers.texi (Integer Basics, Bitwise Operations):
* objects.texi (Integer Type): Update for integers now being 30-bit.

doc/lispref/ChangeLog

+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.

doc/lispref/numbers.texi

 @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
 

doc/lispref/objects.texi

@@ -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