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checked in by ghudson, 24 years ago
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1 | package integer; |
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2 | |
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3 | =head1 NAME |
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4 | |
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5 | integer - Perl pragma to compute arithmetic in integer instead of double |
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6 | |
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7 | =head1 SYNOPSIS |
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8 | |
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9 | use integer; |
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10 | $x = 10/3; |
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11 | # $x is now 3, not 3.33333333333333333 |
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12 | |
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13 | =head1 DESCRIPTION |
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14 | |
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15 | This tells the compiler to use integer operations |
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16 | from here to the end of the enclosing BLOCK. On many machines, |
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17 | this doesn't matter a great deal for most computations, but on those |
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18 | without floating point hardware, it can make a big difference. |
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19 | |
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20 | Note that this affects the operations, not the numbers. If you run this |
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21 | code |
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22 | |
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23 | use integer; |
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24 | $x = 1.5; |
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25 | $y = $x + 1; |
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26 | $z = -1.5; |
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27 | |
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28 | you'll be left with C<$x == 1.5>, C<$y == 2> and C<$z == -1>. The $z |
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29 | case happens because unary C<-> counts as an operation. |
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30 | |
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31 | Native integer arithmetic (as provided by your C compiler) is used. |
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32 | This means that Perl's own semantics for arithmetic operations may |
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33 | not be preserved. One common source of trouble is the modulus of |
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34 | negative numbers, which Perl does one way, but your hardware may do |
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35 | another. |
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36 | |
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37 | % perl -le 'print (4 % -3)' |
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38 | -2 |
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39 | % perl -Minteger -le 'print (4 % -3)' |
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40 | 1 |
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41 | |
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42 | See L<perlmod/Pragmatic Modules>. |
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43 | |
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44 | =cut |
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45 | |
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46 | $integer::hint_bits = 0x1; |
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47 | |
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48 | sub import { |
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49 | $^H |= $integer::hint_bits; |
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50 | } |
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51 | |
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52 | sub unimport { |
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53 | $^H &= ~$integer::hint_bits; |
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54 | } |
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55 | |
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56 | 1; |
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