fivemack (fivemack) wrote,

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hair-tearing perl question

What I want: a subroutine footle such that, if you call footle(a,b) twice with the same a,b, it does nothing the second time

What I did:
use strict;
sub footle
  my ($a,$b,%done) = @_;
  my $concat = $a.$b;
  if ($done{$concat} == 0)
    print "footling $a $b";
    $done{$concat} = 1;

my %isdone = ();


But this doesn't work because parameters are passed by value.

But if I call as footle("bootle","bumtrinket",\%isdone), which passes isdone by reference, it still does the footling twice.

Even if I put $_[2]=%done before the end of the subroutine, it still does the footling twice.

And if I put print join "*",(keys %done); at the start of the subroutine, it says HASH(0x8188110)footling bootle bumtrinket

So how do I really pass the parameter by reference, as if I'd said void footle(int a, int b, set<string>& done) in C++?
Tags: geek, perl
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