integer - Perl pragma to use integer arithmetic instead of floating point
use integer;
$x = 10/3;
# $x is now 3, not 3.33333333333333333
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 fractional 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);
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