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Re: list returned from function

by japhy (Canon)
on Jan 17, 2002 at 21:12 UTC ( #139549=note: print w/replies, xml ) Need Help??


in reply to list returned from function

You need an extra layer of parens. $elem = (myFunc())[5];

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

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Re: Re: list returned from function
by count0 (Friar) on Jan 17, 2002 at 21:28 UTC
    In addition to japhy's excellent answer, I'd like to just point out what this is doing for anyone who might not quite understand why this works as it does.

    (my_func()) forces the output of this function into list context as well as groups it for the [5] we're about to append. Now that [5] takes a slice of the list.

    Here's a simple example
    sub foo { return qw(a b c d) } ( foo() )[1,3]; # Is the same as: ('a', 'b', 'c', 'd')[1,3]; # And effectively the same as @a = foo(); @a[1,3];
    You can read more about slicing in the book Effective Perl Programming (a *must* read =)

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[Eily]: you could tie a variable into not having the same value each time, if you like to make people who try to debug your code facepalm
[Corion]: perl -wle 'package o; use overload q("") => sub {warn "str"; ""}, bool => sub{warn "bool"; 1}; package main; my $o={}; bless $o => o; print "Yay" if ($o && !length($o))'
[Corion]: But people writing such code should document the objects they construct and why it makes sense for an object to be invisible as string while being true in a boolean context
[hippo]: That's equal parts clever and horrendous.
[Eily]: the overload version wouldn't return true with "$x" && !length $x though, I guess
[hippo]: The more I look at this code, the more $x is a plain old scalar and the more this condition will never be true. I'm calling it a bug at this point.
[hippo]: Thanks for your input which has soothed my sanity (a little)
[Corion]: Eily: Sure - if you force both things into stringy things, then you break that magic. But that would also mean that you changed the expression, as now $x = 0.00 will be true instead of false as it were before
[Corion]: Ah no, at least in my feeble experiments that doesn't change the meaning
[Corion]: We sell sanity in small packages ;)

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