

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**311) on 32bit architectures, and
(2**63) .. (2**631) on 64bit architectures. For example, this code
use integer; $x = 5.8; $y = 2.5; $z = 2.7; $a = 2**31  1; # Largest positive integer on 32bit 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 noninteger value of
5.8 since it wasn't operated on. And note too the wraparound 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 longstanding 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 the perlmodlib manpage, Integer Arithmetic in the perlop manpage