Re: NEWBIE Brain Teaser #2, by nysus

[merlyn]

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

-- Randal L. Schwartz, Perl hacker

[Eureka_sg]

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

[merlyn]

SPOILER ALERT REPLY

Not to that, but to something else the teaser is using.

-- Randal L. Schwartz, Perl hacker

[japhy]

What a qweer hint that was.

japhy -- Perl and Regex Hacker

[MrScrooge]

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.

[merlyn]

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

[artist]

$_ = 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.

[merlyn]

Try:

print qw(1 2 3)[0]

under 5.005 and 5.6.

I presume you mean "try" as in "note that it won't even compile under 5.5".

-- Randal L. Schwartz, Perl hacker