# integer(3p) Perl Programmers Reference Guide integer(3p) #

integer(3p) Perl Programmers Reference Guide integer(3p)

## NNAAMMEE #

```
integer - Perl pragma to use integer arithmetic instead of floating point
```

## SSYYNNOOPPSSIISS #

```
use integer;
$x = 10/3;
# $x is now 3, not 3.33333333333333333
```

## DDEESSCCRRIIPPTTIIOONN #

```
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
ooppeerraattoorrss handle their operands and results, and nnoott 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 nnoott 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 uunnssiiggnneedd
integers, but with "use integer;" the operands and results are ssiiggnneedd.
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
```

perl v5.36.3 2014-11-17 integer(3p)