Perl long data type

Question

I am writing an app in Perl that requires long data type instead of integers. How can I achieve this. For example;

my $num = sprintf('%ld', 27823221234);

print $num;

The output is not a long, but an integer.

Answer 1

Here is some code that illustrates some of how Perl behaves - derived from your example:

use strict;
use warnings;

my $num = sprintf("%ld", 27823221234);
print "$num\n";

my $val = 27823221234;
my $str = sprintf("%ld", $val);
printf "%d = %ld = %f = %s\n", $val, $val, $val, $val;
printf "%d = %ld = %f = %s\n", $str, $str, $str, $str;

With a 64-bit Perl, this yields:

27823221234
27823221234 = 27823221234 = 27823221234.000000 = 27823221234
27823221234 = 27823221234 = 27823221234.000000 = 27823221234

If you really need big number (hundreds of digits), then look into the modules that support them. For example:

• Math::BigInt

Answer 2

Your options are:

• use a perl compiled for 64 bits
• use Math::Int64
• use Math::BigInt

update: ah, you can also use floats instead of integers:

printf("%.0f", 2**50)

IIRC, on most current architectures, floats can represent integers up to 2**54-1 precisely.

Answer 3

You are probably confused. Perl natively supports "long"-sized integer math, but I don't think its internal representation is where your problem is. What are you expecting your output to look like?

Answer 4

in your case 27823221234 is really represented as double, so when you try to feed to to sprintf you receive -1

my $x = 27823221234;

my $num = sprintf('%lf', $x);

print $num, "\n";

yields to

27823221234.000000

if you want to do math operations with large integers, consider using Math::Bigint module.