integer(3p) Perl Programmers Reference Guide integer(3p)NAMEinteger - Perl pragma to use integer arithmetic instead of
floating point
SYNOPSIS
use integer;
$x = 10/3;
# $x is now 3, not 3.33333333333333333
DESCRIPTION
This tells the compiler to use integer operations from here
to the end of the enclosing BLOCK. On many machines, this
doesn't matter a great deal for most computations, but on
those without floating point hardware, it can make a big
difference in performance.
Note that this only affects how most of the arithmetic and
relational operators handle their operands and results, and
not how all numbers everywhere are treated. Specifically,
"use integer;" has the effect that before computing the
results of the arithmetic operators (+, -, *, /, %, +=, -=,
*=, /=, %=, and unary minus), the comparison operators (<,
<=, >, >=, ==, !=, <=>), and the bitwise operators (|, &, ^,
<<, >>, |=, &=, ^=, <<=, >>=), the operands have their frac-
tional portions truncated (or floored), and the result will
have its fractional portion truncated as well. In addition,
the range of operands and results is restricted to that of
familiar two's complement integers, i.e., -(2**31) ..
(2**31-1) on 32-bit architectures, and -(2**63) .. (2**63-1)
on 64-bit architectures. For example, this code
use integer;
$x = 5.8;
$y = 2.5;
$z = 2.7;
$a = 2**31 - 1; # Largest positive integer on 32-bit machines
$, = ", ";
print $x, -$x, $x + $y, $x - $y, $x / $y, $x * $y, $y == $z, $a, $a + 1;
will print: 5.8, -5, 7, 3, 2, 10, 1, 2147483647,
-2147483648
Note that $x is still printed as having its true non-integer
value of 5.8 since it wasn't operated on. And note too the
wrap-around from the largest positive integer to the largest
negative one. Also, arguments passed to functions and the
values returned by them are not affected by "use integer;".
E.g.,
srand(1.5);
$, = ", ";
print sin(.5), cos(.5), atan2(1,2), sqrt(2), rand(10);
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integer(3p) Perl Programmers Reference Guide integer(3p)
will give the same result with or without "use integer;"
The power operator "**" is also not affected, so that 2 **
.5 is always the square root of 2. Now, it so happens that
the pre- and post- increment and decrement operators, ++ and
--, are not affected by "use integer;" either. Some may
rightly consider this to be a bug -- but at least it's a
long-standing one.
Finally, "use integer;" also has an additional affect on the
bitwise operators. Normally, the operands and results are
treated as unsigned integers, but with "use integer;" the
operands and results are signed. This means, among other
things, that ~0 is -1, and -2 & -5 is -6.
Internally, native integer arithmetic (as provided by your C
compiler) is used. This means that Perl's own semantics for
arithmetic operations may not be preserved. One common
source of trouble is the modulus of negative numbers, which
Perl does one way, but your hardware may do another.
% perl -le 'print (4 % -3)'
-2
% perl -Minteger -le 'print (4 % -3)'
1
See "Pragmatic Modules" in perlmodlib, "Integer Arithmetic"
in perlop
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