How can I do 64-bit arithmetic in Perl?

Question

I am a perl newbie,

Can I simply use 64-bit arithmetic in Perl?

For example

$operand1 = 0xFFFFFFFFFFFF;   # 48 bit value
$operand2 = 0xFFFFFFFFFFFF;   # 48 bit value

$Result = $operand1 * $operand2;

I am basically looking for a replacement for the int64_t in perl.
Is there any way to mention, if the variable is signed or unsigned?

Do you really want 64-bit arithmetic or big numbers? Do you want it to overflow after 64 bits? — brian d foy, Dec 09 '09 at 00:54

score 14 · Accepted Answer · answered Dec 08 '09 at 02:40

14

Yes, however you need to have Perl compiled with 64-bit support.

answered Dec 08 '09 at 02:40

Alan Haggai Alavi

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"use bigint" will ensure that it always works. (If you want absolute speed, you picked the wrong programming language.) – jrockway Dec 08 '09 at 02:42
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Perl is often used in bioinformatics and other fields requiring high-performance. If your numbers are sure to fit within 64 bits, then this answer will create better performing code. Just make sure to use a recent version of Perl, and you may want to compile it yourself. "64-bit support is now considered to be mature" http://dev.perl.org/perl5/news/2002/07/18/580ann/ – Paul Dec 08 '09 at 02:57
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Even with 64 bit support, you'd need a `use integer;` to avoid getting a (rounded) floating point result and instead get a (high-bits truncated) integer result. – ysth Dec 08 '09 at 03:37

score 11 · Answer 2 · edited Sep 30 '16 at 16:16

11

See bigint:

Transparent BigInteger support for Perl...

All operators (including basic math operations) except the range operator .. are overloaded. Integer constants are created as proper BigInts.

Floating point constants are truncated to integer. All parts and results of expressions are also truncated.

Unlike integer, this pragma creates integer constants that are only limited in their size by the available memory and CPU time...

edited Sep 30 '16 at 16:16

gnat

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answered Dec 08 '09 at 02:42

Sinan Ünür

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+1 thank you, but I could not use functions like hex() etc ~~~ – Alphaneo Dec 08 '09 at 08:50
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Alphaneo, actually you can do `$a = Math::BigInt->new("0x".$your_int)` and `$a->as_hex()` – osgx Sep 22 '11 at 16:42

Greg Hewgill · Answer 3 · 2009-12-08T08:12:45.353

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Yes, Perl automatically handles large integer arithmetic for you. However, Perl does not offer a distinction between signed and unsigned types (there's no need, since there are not fixed bounds on large integer range).

The perlnumber manual page has more information about the different numeric formats supported by Perl.

edited Dec 08 '09 at 08:12

answered Dec 08 '09 at 02:38

Greg Hewgill

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It's not clear what you mean by "distinction between signed and unsigned values"? – ysth Dec 08 '09 at 06:39
Right, I should have said "types" instead of "values". Fixed. – Greg Hewgill Dec 08 '09 at 08:12

score 0 · Answer 4 · answered May 16 '13 at 11:57

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Be aware, 64bit arithmetic in Perl is one, but what it is displayed by sprintf %d %u and %s, is second. Current perl version supports 64bits without problems, but sprintf %d format not, %b likewise .

answered May 16 '13 at 11:57

Znik

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score 0 · Answer 5 · answered Feb 07 '18 at 17:08

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use bigint will make Perl handle arbitrary size integers correctly without integer overflow.

Eg.:

use bigint;
print 1 << 256;

will print:

115792089237316195423570985008687907853269984665640564039457584007913129639936

answered Feb 07 '18 at 17:08

wvdz

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How can I do 64-bit arithmetic in Perl?

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