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in reply to Little Perl Mysteries: what's your answer?

And my 2 cents worth..
A function prototyped to accept no arguments is treated as a candidate for inlining by the Perl compiler. In this case sub foo(){2} will be inlined because it returns a constant. The snippet will thus print 2, because print foo(); is turned into print 2; by the compiler, no matter what you do to the definition of sub foo later on at runtime.
Nice one, Ovid!

CU
Robartes-

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Re: Re: Little Perl Mysteries: what's your answer?
by dada (Chaplain) on Oct 31, 2002 at 08:54 UTC
    and the proof is: 
    c:\>perl -MO=Deparse -e "sub foo(){2};*::foo=sub(){3};print foo();"
sub foo () {
    2;
}
*foo = sub () {
    3;
}
;
print 2;
-e syntax OK

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    cheers,
    Aldo

    King of Laziness, Wizard of Impatience, Lord of Hubris

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[d/l]

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