Re: Declaring a code ref with arguments

have you ever tested your code? this is broken!!
however you should look at perldoc perlref on how to build code refs..
this is a working version:

#!/usr/bin/perl -w

use strict;

sub print_me {
  my $arg = shift || "Hello World!\n";
  print $arg;
}

my %hash = (no_arg => sub { &print_me()},
        with_arg => sub {&print_me("Hello again!\n")}
        );

$hash{no_arg}->();
$hash{with_arg}->();
[download]

Comment on Re: Declaring a code ref with arguments Download Code

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Re: Re: Declaring a code ref with arguments by !unlike (Beadle) on Aug 05, 2003 at 09:35 UTC
The example I provided did not `use strict`. My apologies for that oversight. With that being the case the code worked fine (even with `-w`) Cheers !unlike I write my Perl code like how I like my sex: fast and dirty. ;)	[reply] [d/l] [select]
Re: Re: Declaring a code ref with arguments by cfreak (Chaplain) on Aug 05, 2003 at 13:35 UTC
His argument passing is wrong but it sounds like you are saying that this: `my %hash = ( foo =>\&do_foo, bar =>\&do_bar, );` [download] is wrong, (correct me if I'm misinterpreting your post) which I'm not sure where you find that in perlref as that works, even while using strict. To pass the arguments you can do this: `$hash{foo}->('foo','bar','baz');` [download] Lobster Aliens Are attacking the world!	[reply] [d/l] [select]
Re: Re: Re: Declaring a code ref with arguments by cianoz (Friar) on Aug 05, 2003 at 14:55 UTC
no, at all, if you look carefully at what he posted you'll find this: `%hash = {no_arg => \&print_me, with_arg => \&print_me("Hello World, again!\n" };` [download] as you can see he uses curly brackets to build a hash and lacks right parenthesis just after "Hello World, again!\n" what i say is that this code will never work (maybe is just a wrong trascription) and that you can't do `bar =>\&do_bar("argument")` [download] so you have to do `bar => sub { do_bar("argument") }` [download] that's all.. update of course (foo => \&foo) with no arguments is perfectly legal..	[reply] [d/l] [select]


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