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Do you know where your variables are?

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One thing I have always wanted (in every language I've used except assembler) is a function that returned both result and remainder of an integer division.

Both are usually calculated and left in seperate cpu registers after a DIV instruction, but I am forced by to use "/" and "%" to get at them, which means that the division is being done twice.

I did once create a function using in-line assembler in C to do this, but the single return value from a C function meant that the function was messy to use.

Perl's natural ability to return a list and assign lists mean that the idea resurrected itself in my brain when I saw this post and the answers to it.

My solution to this was:

use strict; sub div_mod { return int( $_[0]/$_[1]) , ($_[0] % $_[1]); } my $i = 1332443; my ( $s, $m, $h, $d ); ($m,$s) = div_mod($i, 60); ($h,$m) = div_mod($m, 60); ($d,$h) = div_mod($h, 24); print "$d days, $h hours, $m minutes, $s seconds.\n";

Which I think is rather elegant, but still possible less efficient than the other solutions, and less efficient than it could be as I am still performing the division twice to get at the information.

I took a look at the inline-c stuff, but that doesn't help for the same reasons.

I also took a quick look at the overloading capabilities but I am not familiar enough with Perl to understand that yet.

I am wondering if it is possible to define a new operator for Perl? Maybe "/%" that would do the same thing as my div_mod() sub in the code above but accessing the appropriate registers to save a division?

Possibly better (and more Perlish) would be to have both values returned only if the mod (%) operator was called in a LIST context, especially as mod (%) is an inherently integer operator?

Does this have any merit? If so, would this have to be done deep in the guts of the compiler/interpreter or could it be added as a module in some way?

In reply to A better mod (%) operator? by BrowserUk

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