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Re: NEWBIE Brain Teaser #2, by nysus

by merlyn (Sage)
on Apr 15, 2001 at 21:01 UTC ( #72674=note: print w/ replies, xml ) Need Help??


in reply to NEWBIE Brain Teaser #2, by nysus

Hint: The answer to Part C will depend on whether you're on 5.5 or 5.6.

-- Randal L. Schwartz, Perl hacker


Comment on Re: NEWBIE Brain Teaser #2, by nysus
Re: Re: NEWBIE Brain Teaser #2, by nysus
by Eureka_sg (Monk) on Apr 15, 2001 at 21:07 UTC

    SPOILER ALERT!

    From my understanding, the existing @_ will be passed by default when you use &subroutine instead. Is there a change in 5.6?

    Edit: chipmunk 2001-04-16

      You are correct that @_ will be passed to a subroutine if you call it without using the parenthesis. This is true of both 5.6 and 5.005 at least.

      However the result of the attempted assignment of a list to $_ differ depending on Perl version, and this DOES affect the output you will see.

        However the result of the attempted assignment of a list to $_ differ depending on Perl version, and this DOES affect the output you will see.
        Perl never attempts to assign a list to $_. Ever.

        Understanding is demonstrated when correct phrasing is used. {grin}

        -- Randal L. Schwartz, Perl hacker

Re: Re: NEWBIE Brain Teaser #2, by nysus
by artist (Parson) on Apr 16, 2001 at 16:07 UTC
    $_ = qw(nothing nothing)
    Translates to:
    1. Under 5.005
    @_ = qw(nothing nothing)
    2. Under 5.6
    No effect on @_;
    Resulting
    call for &subroutine
    1. Under 5.005
    prints "..nothing..nothing"
    2. Under 5.6
    prints "..alpha ..omega"

    Try:

    print qw(1 2 3)[0]
    under 5.005 and 5.6.

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