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

sub func {
    my ($delimiter, $text) = @_;
    return 1 and 0;
}

use Data::Dumper;
print Dumper(func());

[download]

guess what?

the result is $VAR1 = 1;

anyone knows the reason?

seems that Perl gives return the same priority as an ordinary sub,is this really a good design?

Replies are listed 'Best First'.
Re: Can you explain the result?
by choroba (Cardinal) on Dec 06, 2011 at 13:26 UTC
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[d/l]
[select]
      I know and has lower priority than &&,but it doesn't explain this issue.
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[d/l]
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        It does. Try with print instead of return to see.
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Re: Can you explain the result?
by RMGir (Prior) on Dec 06, 2011 at 13:20 UTC
    Sure. The result of your function is the value 1, and that's what you're passing to Dumper...

    Now, if you called 

    print Dumper(\&func);

    [download]
    on the function reference, you'd get a slightly more interesting result: 
    $VAR1 = sub { "DUMMY" };

    [download]
    I guess my version of Dumper isn't up to decompiling subs - I'm guessing that's expected.

    Mike
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      shouldn't  1 and 0 be false/0?
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[d/l]
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        As choroba said, return 1&&0 is different from return 1 and 0. 

        $ perl -MB::Deparse -e'
   sub f{return 1 and 0;}; 
   sub g{return 1 && 0}; 

   $deparse=B::Deparse->new(); 

   print "f(): ",$deparse->coderef2text(\&f),"\n"; 
   print "g(): ",$deparse->coderef2text(\&g),"\n";' 

f(): {
    0 if return 1;
}
g(): {
    return 0;
}

        [download]

        Mike
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[d/l]
        A reply falls below the community's threshold of quality. You may see it by logging in.
Re: Can you explain the result?
by TJPride (Pilgrim) on Dec 06, 2011 at 16:33 UTC
    Rather than worrying about why Perl does this, it may be simpler to just do return (1 and 0);
[reply]
[d/l]

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