passing array ref without variable

vit has asked for the wisdom of the Perl Monks concerning the following question:

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Re: passing array ref without variable by kennethk (Abbot) on May 03, 2010 at 14:34 UTC
Your syntax would probably work, depending on your expectations. Fortune favors the bold - make a mock-up script and see what happens! The syntax will pass an anonymous array whose contents are the keys of your hash. By using keys, however, you are breaking the link with the actual element itself (lvalue). This means that the function cannot change the values themselves, i.e. if this function normally removes all 'x's from the elements of the array, the hash keys will remain unchanged.	[reply]
Re: passing array ref without variable by AnomalousMonk (Archbishop) on May 03, 2010 at 15:00 UTC
Yes, just try it – and see what happens with Data::Dumper or any of its ilk. (Also nice to use strictures and warnings!) `>perl -wMstrict -le "use Data::Dumper; my %hash = qw(a 1 b 2 c 3 d 4); my_function([keys %hash]); sub my_function { my $arr_ref = shift; print Dumper $arr_ref; } " $VAR1 = [ 'c', 'a', 'b', 'd' ];` [download]	[reply] [d/l]
Re^2: passing array ref without variable by vit (Friar) on May 03, 2010 at 15:23 UTC
Can I pass somehow an array ref of values without separately forming an array? Maybe using foreach...	[reply]
Re^3: passing array ref without variable by kennethk (Abbot) on May 03, 2010 at 16:05 UTC
Why would you want to? Inline anonymous array construction is a trivial number of characters and is robust. This sounds like an XY Problem.	[reply]
Re^3: passing array ref without variable by LanX (Saint) on May 03, 2010 at 17:03 UTC
I suppose what you really want/need is to pass the hash as reference, like this you don't need to pass the values and the keys separately. `call(\%hash)` If you explicitly want to pass the reference of a copy do this `call( { %hash } )` the `{ ... }` returns a ref of an anonymous hash of the list expanded from %hash, i.e. the ref of a hashcopy! Cheers Rolf	[reply] [d/l] [select]
Re^4: passing array ref without variable by vit (Friar) on May 03, 2010 at 21:07 UTC
Re^5: passing array ref without variable by FunkyMonk (Bishop) on May 03, 2010 at 21:15 UTC
Re^5: passing array ref without variable by LanX (Saint) on May 03, 2010 at 21:26 UTC
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Re^3: passing array ref without variable by AnomalousMonk (Archbishop) on May 03, 2010 at 16:15 UTC
Why would you want to? And how could you? It sounds like you (vit) are asking "Can I take a reference to an array that does not exist?".	[reply]
Re: passing array ref without variable by Anonymous Monk on May 03, 2010 at 14:22 UTC
Yes, just try it	[reply]
Re: passing array ref without variable by ambrus (Abbot) on May 04, 2010 at 13:11 UTC
Yes, that does work. What does not work so easily is when you want to pass the values of a hash to the function in such a way that it can write them. Eg. this works, the function can change the value in the hash: `my %d = ("foo", "bar"); sub fs { $_[0] = "quux"; } fs(values %d); prin +t join(" ", %d), $/; # OUTPUT: foo quux` [download] But this does not work: `my %d = ("foo", "bar"); sub fi { ${$_[0]}[0] = "quux"; } fi([values %d +]); print join(" ", %d), $/; # OUTPUT: foo bar` [download] Instead you could do `my %d = ("foo", "bar"); sub fi { ${$_[0]}[0] = "quux"; } sub capture { + \@_ } fi(capture(values %d)); print join(" ", %d), $/; # OUTPUT: foo quux` [download] But that's probably all irrelevant to your question.	[reply] [d/l] [select]