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Integers now at least 28 bits.
This commit is contained in:
Richard M. Stallman
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32ce36ad51
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@ -39,23 +39,22 @@ where Emacs does not support them.
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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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range is @minus{}8388608 to 8388607 (24 bits; i.e.,
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range is @minus{}8388608 to 8388607 (28 bits; i.e.,
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@ifinfo
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-2**23
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-2**27
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@end ifinfo
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@tex
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$-2^{23}$
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$-2^{27}$
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@end tex
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to
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@ifinfo
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2**23 - 1)
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2**27 - 1)
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@end ifinfo
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@tex
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$2^{23}-1$)
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$2^{27}-1$)
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@end tex
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on most machines, but on others it is @minus{}16777216 to 16777215 (25
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bits), or @minus{}33554432 to 33554431 (26 bits). Many examples in this
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chapter assume an integer has 24 bits.
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on most machines, but some machines may have a wider range. Many
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examples in this chapter assume an integer has 28 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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@ -66,7 +65,7 @@ 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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16777217 ; @r{Also the integer 1, due to overflow.}
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268435457 ; @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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@ -75,10 +74,10 @@ initial sign and optional final period.
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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 24-bit binary, the decimal integer 5 looks like this:
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In 28-bit binary, the decimal integer 5 looks like this:
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@example
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0000 0000 0000 0000 0000 0101
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0000 0000 0000 0000 0000 0000 0101
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@end example
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@noindent
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@ -88,12 +87,12 @@ 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
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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 24 ones. (This is called @dfn{two's
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@minus{}1 is represented as 28 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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@ -101,24 +100,24 @@ 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 1011
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1111 1111 1111 1111 1111 1111 1011
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@end example
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In this implementation, the largest 24-bit binary integer is the
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decimal integer 8,388,607. In binary, it looks like this:
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decimal integer 134,217,727. In binary, it looks like this:
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@example
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0111 1111 1111 1111 1111 1111
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0111 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 8,388,607, the value is the
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negative integer @minus{}8,388,608:
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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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@example
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(+ 1 8388607)
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@result{} -8388608
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@result{} 1000 0000 0000 0000 0000 0000
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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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@end example
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Many of the following functions accept markers for arguments as well
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@ -651,12 +650,12 @@ produces @minus{}2 on a 24-bit machine:
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@result{} -2
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@end example
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In binary, in the 24-bit implementation, the argument looks like this:
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In binary, in the 28-bit implementation, the argument looks like this:
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@example
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@group
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;; @r{Decimal 8,388,607}
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0111 1111 1111 1111 1111 1111
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;; @r{Decimal 134.217,727}
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0111 1111 1111 1111 1111 1111 1111
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@end group
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@end example
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@ -666,7 +665,7 @@ 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 1110
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1111 1111 1111 1111 1111 1111 1110
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@end group
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@end example
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@ -724,9 +723,9 @@ 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 1010
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1111 1111 1111 1111 1111 1111 1010
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@result{}
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1111 1111 1111 1111 1111 1101
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1111 1111 1111 1111 1111 1111 1101
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@end group
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@end example
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@ -735,11 +734,11 @@ 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{} 8388605
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;; @r{Decimal @minus{}6 becomes decimal 8,388,605.}
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1111 1111 1111 1111 1111 1010
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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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@result{}
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0111 1111 1111 1111 1111 1101
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0111 1111 1111 1111 1111 1111 1101
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@end group
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@end example
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@ -749,34 +748,34 @@ 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{ 24-bit binary values}
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; @r{ 28-bit binary values}
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(lsh 5 2) ; 5 = @r{0000 0000 0000 0000 0000 0101}
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@result{} 20 ; = @r{0000 0000 0000 0000 0001 0100}
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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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@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 1011}
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@result{} -20 ; = @r{1111 1111 1111 1111 1110 1100}
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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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(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 0101}
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@result{} 1 ; = @r{0000 0000 0000 0000 0000 0001}
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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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@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 1011}
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@result{} 4194302 ; = @r{0011 1111 1111 1111 1111 1110}
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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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@end group
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@group
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(ash -5 -2) ; -5 = @r{1111 1111 1111 1111 1111 1011}
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@result{} -2 ; = @r{1111 1111 1111 1111 1111 1110}
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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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@end group
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@end smallexample
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@end defun
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@ -813,23 +812,23 @@ because its binary representation consists entirely of ones. If
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@smallexample
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@group
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; @r{ 24-bit binary values}
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; @r{ 28-bit binary values}
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(logand 14 13) ; 14 = @r{0000 0000 0000 0000 0000 1110}
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; 13 = @r{0000 0000 0000 0000 0000 1101}
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@result{} 12 ; 12 = @r{0000 0000 0000 0000 0000 1100}
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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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@end group
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@group
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(logand 14 13 4) ; 14 = @r{0000 0000 0000 0000 0000 1110}
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; 13 = @r{0000 0000 0000 0000 0000 1101}
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; 4 = @r{0000 0000 0000 0000 0000 0100}
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@result{} 4 ; 4 = @r{0000 0000 0000 0000 0000 0100}
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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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@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}
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@result{} -1 ; -1 = @r{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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@ -845,18 +844,18 @@ passed just one argument, it returns that argument.
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@smallexample
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@group
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; @r{ 24-bit binary values}
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; @r{ 28-bit binary values}
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(logior 12 5) ; 12 = @r{0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0101}
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@result{} 13 ; 13 = @r{0000 0000 0000 0000 0000 1101}
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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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@end group
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@group
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(logior 12 5 7) ; 12 = @r{0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0101}
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; 7 = @r{0000 0000 0000 0000 0000 0111}
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@result{} 15 ; 15 = @r{0000 0000 0000 0000 0000 1111}
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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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@end group
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@end smallexample
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@end defun
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@ -872,18 +871,18 @@ 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{ 24-bit binary values}
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; @r{ 28-bit binary values}
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(logxor 12 5) ; 12 = @r{0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0101}
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@result{} 9 ; 9 = @r{0000 0000 0000 0000 0000 1001}
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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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@end group
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@group
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(logxor 12 5 7) ; 12 = @r{0000 0000 0000 0000 0000 1100}
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; 5 = @r{0000 0000 0000 0000 0000 0101}
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; 7 = @r{0000 0000 0000 0000 0000 0111}
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@result{} 14 ; 14 = @r{0000 0000 0000 0000 0000 1110}
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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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@end group
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@end smallexample
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@end defun
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@ -898,9 +897,9 @@ 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 0101}
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;; 5 = @r{0000 0000 0000 0000 0000 0000 0101}
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;; @r{becomes}
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;; -6 = @r{1111 1111 1111 1111 1111 1010}
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;; -6 = @r{1111 1111 1111 1111 1111 1111 1010}
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@end example
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@end defun
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@ -970,7 +969,10 @@ This function returns the logarithm of @var{arg}, with base 10. If
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@end defun
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@defun expt x y
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This function returns @var{x} raised to power @var{y}.
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This function returns @var{x} raised to power @var{y}. If both
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arguments are integers and @var{y} is positive, the result is an
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integer; in this case, it is truncated to fit the range of possible
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integer values.
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@end defun
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@defun sqrt arg
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