1package integer; 2 3our $VERSION = '1.00'; 4 5=head1 NAME 6 7integer - Perl pragma to use integer arithmetic instead of floating point 8 9=head1 SYNOPSIS 10 11 use integer; 12 $x = 10/3; 13 # $x is now 3, not 3.33333333333333333 14 15=head1 DESCRIPTION 16 17This tells the compiler to use integer operations from here to the end 18of the enclosing BLOCK. On many machines, this doesn't matter a great 19deal for most computations, but on those without floating point 20hardware, it can make a big difference in performance. 21 22Note that this only affects how most of the arithmetic and relational 23B<operators> handle their operands and results, and B<not> how all 24numbers everywhere are treated. Specifically, C<use integer;> has the 25effect that before computing the results of the arithmetic operators 26(+, -, *, /, %, +=, -=, *=, /=, %=, and unary minus), the comparison 27operators (<, <=, >, >=, ==, !=, <=>), and the bitwise operators (|, &, 28^, <<, >>, |=, &=, ^=, <<=, >>=), the operands have their fractional 29portions truncated (or floored), and the result will have its 30fractional portion truncated as well. In addition, the range of 31operands and results is restricted to that of familiar two's complement 32integers, i.e., -(2**31) .. (2**31-1) on 32-bit architectures, and 33-(2**63) .. (2**63-1) on 64-bit architectures. For example, this code 34 35 use integer; 36 $x = 5.8; 37 $y = 2.5; 38 $z = 2.7; 39 $a = 2**31 - 1; # Largest positive integer on 32-bit machines 40 $, = ", "; 41 print $x, -$x, $x + $y, $x - $y, $x / $y, $x * $y, $y == $z, $a, $a + 1; 42 43will print: 5.8, -5, 7, 3, 2, 10, 1, 2147483647, -2147483648 44 45Note that $x is still printed as having its true non-integer value of 465.8 since it wasn't operated on. And note too the wrap-around from the 47largest positive integer to the largest negative one. Also, arguments 48passed to functions and the values returned by them are B<not> affected 49by C<use integer;>. E.g., 50 51 srand(1.5); 52 $, = ", "; 53 print sin(.5), cos(.5), atan2(1,2), sqrt(2), rand(10); 54 55will give the same result with or without C<use integer;> The power 56operator C<**> is also not affected, so that 2 ** .5 is always the 57square root of 2. Now, it so happens that the pre- and post- increment 58and decrement operators, ++ and --, are not affected by C<use integer;> 59either. Some may rightly consider this to be a bug -- but at least it's 60a long-standing one. 61 62Finally, C<use integer;> also has an additional affect on the bitwise 63operators. Normally, the operands and results are treated as 64B<unsigned> integers, but with C<use integer;> the operands and results 65are B<signed>. This means, among other things, that ~0 is -1, and -2 & 66-5 is -6. 67 68Internally, native integer arithmetic (as provided by your C compiler) 69is used. This means that Perl's own semantics for arithmetic 70operations may not be preserved. One common source of trouble is the 71modulus of negative numbers, which Perl does one way, but your hardware 72may do another. 73 74 % perl -le 'print (4 % -3)' 75 -2 76 % perl -Minteger -le 'print (4 % -3)' 77 1 78 79See L<perlmodlib/"Pragmatic Modules">, L<perlop/"Integer Arithmetic"> 80 81=cut 82 83$integer::hint_bits = 0x1; 84 85sub import { 86 $^H |= $integer::hint_bits; 87} 88 89sub unimport { 90 $^H &= ~$integer::hint_bits; 91} 92 931; 94