Merge: Document wide integers better.
This commit is contained in:
Paul Eggert
commit
001903b549
8 changed files with 99 additions and 74 deletions
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@ -1,3 +1,10 @@
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2011-06-02 Paul Eggert <eggert@cs.ucla.edu>
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Document wide integers better.
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* buffers.texi (Buffers):
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* files.texi (Visiting): Document maxima for 64-bit machines,
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and mention virtual memory limits.
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2011-05-28 Chong Yidong <cyd@stupidchicken.com>
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* custom.texi (Hooks): Reorganize. Mention Prog mode.
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@ -43,8 +43,11 @@ can be different from the value in other buffers. @xref{Locals}.
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A buffer's size cannot be larger than some maximum, which is defined
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by the largest buffer position representable by the @dfn{Emacs
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integer} data type. This is because Emacs tracks buffer positions
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using that data type. For 32-bit machines, the largest buffer size is
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512 megabytes.
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using that data type. For typical 64-bit machines, the maximum buffer size
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enforced by the data types is @math{2^61 - 2} bytes, or about 2 EiB.
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For typical 32-bit machines, the maximum is @math{2^29 - 2} bytes, or
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about 512 MiB. Buffer sizes are also limited by the size of Emacs's
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virtual memory.
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@menu
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* Select Buffer:: Creating a new buffer or reselecting an old one.
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@ -209,7 +209,8 @@ to reread it.
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about 10 megabytes), Emacs asks you for confirmation first. You can
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answer @kbd{y} to proceed with visiting the file. Note, however, that
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Emacs cannot visit files that are larger than the maximum Emacs buffer
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size, which is around 512 megabytes on 32-bit machines
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size, which is limited by the amount of memory Emacs can allocate
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and by the integers that Emacs can represent
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(@pxref{Buffers}). If you try, Emacs will display an error message
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saying that the maximum buffer size has been exceeded.
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@ -1,3 +1,15 @@
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2011-06-03 Paul Eggert <eggert@cs.ucla.edu>
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Document wide integers better.
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* files.texi (File Attributes): Document ino_t values better.
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ino_t values no longer map to anything larger than a single cons.
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* numbers.texi (Integer Basics, Integer Basics, Arithmetic Operations):
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(Bitwise Operations):
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* objects.texi (Integer Type): Use a binary notation that is a bit easier
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to read, and that will port better if 62-bits becomes the default.
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Fix or remove incorrect examples.
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* os.texi (Time Conversion): Document time_t values better.
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2011-05-31 Lars Magne Ingebrigtsen <larsi@gnus.org>
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* processes.texi (Process Information): Document
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@ -1237,11 +1237,12 @@ deleted and recreated; @code{nil} otherwise.
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@item
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The file's inode number. If possible, this is an integer. If the
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inode number is too large to be represented as an integer in Emacs
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Lisp, but still fits into a 32-bit integer, then the value has the
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Lisp but dividing it by @math{2^16} yields a representable integer,
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then the value has the
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form @code{(@var{high} . @var{low})}, where @var{low} holds the low 16
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bits. If the inode is wider than 32 bits, the value is of the form
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bits. If the inode number is too wide for even that, the value is of the form
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@code{(@var{high} @var{middle} . @var{low})}, where @code{high} holds
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the high 24 bits, @var{middle} the next 24 bits, and @var{low} the low
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the high bits, @var{middle} the middle 24 bits, and @var{low} the low
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16 bits.
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@item
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@ -50,8 +50,9 @@ to
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@tex
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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 30 bits.
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but some machines provide a wider range. Many examples in this
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chapter assume that an integer has 30 bits and that floating point
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numbers are IEEE double precision.
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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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@ -97,17 +98,18 @@ view the numbers in their binary form.
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In 30-bit binary, the decimal integer 5 looks like this:
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@example
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00 0000 0000 0000 0000 0000 0000 0101
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0000...000101 (30 bits total)
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@end example
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@noindent
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(We have inserted spaces between groups of 4 bits, and two spaces
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between groups of 8 bits, to make the binary integer easier to read.)
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(The @samp{...} stands for enough bits to fill out a 30-bit word; in
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this case, @samp{...} stands for twenty 0 bits. Later examples also
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use the @samp{...} notation to make binary integers easier to read.)
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The integer @minus{}1 looks like this:
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@example
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11 1111 1111 1111 1111 1111 1111 1111
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1111...111111 (30 bits total)
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@end example
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@noindent
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@ -120,14 +122,14 @@ complement} notation.)
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@minus{}5 looks like this:
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@example
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11 1111 1111 1111 1111 1111 1111 1011
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1111...111011 (30 bits total)
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@end example
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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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01 1111 1111 1111 1111 1111 1111 1111
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0111...111111 (30 bits total)
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@end example
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Since the arithmetic functions do not check whether integers go
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@ -137,7 +139,7 @@ negative integer @minus{}536,870,912:
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@example
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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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@result{} 1000...000000 (30 bits total)
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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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@ -508,8 +510,8 @@ 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+ 268435455)} may evaluate to
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@minus{}268435456, depending on your hardware.
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do not check for overflow. Thus @code{(1+ 536870911)} may evaluate to
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@minus{}536870912, 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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@ -829,19 +831,19 @@ 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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536,870,911 produces @minus{}2 on a 30-bit machine:
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536,870,911 produces @minus{}2 in the 30-bit implementation:
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@example
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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 30-bit implementation, the argument looks like this:
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In binary, the argument looks like this:
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@example
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@group
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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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0111...111111 (30 bits total)
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@end group
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@end example
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@ -851,7 +853,7 @@ 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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11 1111 1111 1111 1111 1111 1111 1110
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1111...111110 (30 bits total)
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@end group
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@end example
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@end defun
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@ -874,9 +876,9 @@ 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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11 1111 1111 1111 1111 1111 1111 1010
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1111...111010 (30 bits total)
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@result{}
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11 1111 1111 1111 1111 1111 1111 1101
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1111...111101 (30 bits total)
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@end group
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@end example
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@ -887,9 +889,9 @@ In contrast, shifting the pattern of bits one place to the right with
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@group
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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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1111...111010 (30 bits total)
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@result{}
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01 1111 1111 1111 1111 1111 1111 1101
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0111...111101 (30 bits total)
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@end group
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@end example
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@ -899,34 +901,35 @@ 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{ 30-bit binary values}
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; @r{ 30-bit binary values}
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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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(lsh 5 2) ; 5 = @r{0000...000101}
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@result{} 20 ; = @r{0000...010100}
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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{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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(lsh -5 2) ; -5 = @r{1111...111011}
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@result{} -20 ; = @r{1111...101100}
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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{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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(lsh 5 -2) ; 5 = @r{0000...000101}
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@result{} 1 ; = @r{0000...000001}
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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{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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(lsh -5 -2) ; -5 = @r{1111...111011}
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@result{} 268435454
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; = @r{0011...111110}
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@end group
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@group
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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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(ash -5 -2) ; -5 = @r{1111...111011}
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@result{} -2 ; = @r{1111...111110}
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@end group
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@end smallexample
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@end defun
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@ -961,23 +964,23 @@ because its binary representation consists entirely of ones. If
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@smallexample
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@group
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; @r{ 30-bit binary values}
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; @r{ 30-bit binary values}
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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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(logand 14 13) ; 14 = @r{0000...001110}
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; 13 = @r{0000...001101}
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@result{} 12 ; 12 = @r{0000...001100}
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@end group
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@group
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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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(logand 14 13 4) ; 14 = @r{0000...001110}
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; 13 = @r{0000...001101}
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; 4 = @r{0000...000100}
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@result{} 4 ; 4 = @r{0000...000100}
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@end group
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@group
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(logand)
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@result{} -1 ; -1 = @r{11 1111 1111 1111 1111 1111 1111 1111}
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@result{} -1 ; -1 = @r{1111...111111}
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@end group
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@end smallexample
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@end defun
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@ -991,18 +994,18 @@ passed just one argument, it returns that argument.
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@smallexample
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@group
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; @r{ 30-bit binary values}
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; @r{ 30-bit binary values}
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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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(logior 12 5) ; 12 = @r{0000...001100}
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; 5 = @r{0000...000101}
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@result{} 13 ; 13 = @r{0000...001101}
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@end group
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@group
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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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(logior 12 5 7) ; 12 = @r{0000...001100}
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; 5 = @r{0000...000101}
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; 7 = @r{0000...000111}
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@result{} 15 ; 15 = @r{0000...001111}
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@end group
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@end smallexample
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@end defun
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@ -1016,18 +1019,18 @@ 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{ 30-bit binary values}
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; @r{ 30-bit binary values}
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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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(logxor 12 5) ; 12 = @r{0000...001100}
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; 5 = @r{0000...000101}
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@result{} 9 ; 9 = @r{0000...001001}
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@end group
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@group
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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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(logxor 12 5 7) ; 12 = @r{0000...001100}
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; 5 = @r{0000...000101}
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; 7 = @r{0000...000111}
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@result{} 14 ; 14 = @r{0000...001110}
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@end group
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@end smallexample
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@end defun
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@ -1040,9 +1043,9 @@ 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{00 0000 0000 0000 0000 0000 0000 0101}
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;; 5 = @r{0000...000101} (30 bits total)
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;; @r{becomes}
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;; -6 = @r{11 1111 1111 1111 1111 1111 1111 1010}
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;; -6 = @r{1111...111010} (30 bits total)
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@end example
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@end defun
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@ -179,10 +179,9 @@ to
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@tex
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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+ 536870911)} is @minus{}536870912 on most
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machines.
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on typical 32-bit machines. (Some machines provide a wider range.)
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Emacs Lisp arithmetic functions do not check for overflow. Thus
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@code{(1+ 536870911)} is @minus{}536870912 if Emacs integers are 30 bits.
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The read syntax for integers is a sequence of (base ten) digits with an
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optional sign at the beginning and an optional period at the end. The
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@ -195,7 +194,6 @@ 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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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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@ -203,8 +201,8 @@ leading @samp{+} or a final @samp{.}.
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As a special exception, if a sequence of digits specifies an integer
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too large or too small to be a valid integer object, the Lisp reader
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reads it as a floating-point number (@pxref{Floating Point Type}).
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For instance, on most machines @code{536870912} is read as the
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floating-point number @code{536870912.0}.
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For instance, if Emacs integers are 30 bits, @code{536870912} is read
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as the floating-point number @code{536870912.0}.
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@xref{Numbers}, for more information.
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@ -1193,11 +1193,11 @@ to calendrical information and vice versa. You can get time values
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from the functions @code{current-time} (@pxref{Time of Day}) and
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@code{file-attributes} (@pxref{Definition of file-attributes}).
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Many operating systems are limited to time values that contain 32 bits
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Many 32-bit operating systems are limited to time values that contain 32 bits
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of information; these systems typically handle only the times from
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1901-12-13 20:45:52 UTC through 2038-01-19 03:14:07 UTC. However, some
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operating systems have larger time values, and can represent times far
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in the past or future.
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1901-12-13 20:45:52 UTC through 2038-01-19 03:14:07 UTC. However, 64-bit
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and some 32-bit operating systems have larger time values, and can
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represent times far in the past or future.
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Time conversion functions always use the Gregorian calendar, even
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for dates before the Gregorian calendar was introduced. Year numbers
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Add table
Reference in a new issue