what's the scope of $_ array? 1

[1DMF]

Hi,

Can someone confirm what scope $_ is.

i.e.

if I call a sub with

mySub1(a,b,c);

the sub will have access to $_ array with elements indexed at 0,1,2 equaling a,b,c respectively.

but if in mySub1 I call another sub

mySub2(1,2,3);

Have I just trashed the $_ array for mySub1 as now indexes 0,1,2 equal 1,2,3 respectively?

thanks

1DMF

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[rharsh]

I'm not sure what kind of confirmation you are looking for, maybe this will help?

my $pattern = "%20s\t%s\n";
&mySub1 qw/a b c/;
&mySub2 qw/1 2 3/;


sub mySub1 {
	printf $pattern, "mySub1 before:", "@_";
	print "w/args -\n";
	&mySub2 qw/d e f/;
	print "w/o args -\n";
	&mySub2;
	printf $pattern, "mySub1 after:", "@_";
}

sub mySub2 {
	printf $pattern, "mySub2:", "@_";
}

 

[1DMF]

is $_ a global so when ever you call another sub, that array is overwritten with the new set of arguments being passed to the subroutine?

or is $_ local to each subroutine block?


"In complete darkness we are all the same, only our knowledge and wisdom separates us, don't let your eyes deceive you."

"If a shortcut was meant to be easy, it wouldn't be a shortcut, it would be the way!"

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[1DMF]

opps just realise my mistake I mean @_ !!!

"In complete darkness we are all the same, only our knowledge and wisdom separates us, don't let your eyes deceive you."

"If a shortcut was meant to be easy, it wouldn't be a shortcut, it would be the way!"

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[Kirsle]

@_ is local to every subroutine block.

The only exception is if you call a subroutine without parenthesis from within another sub, that sub gets the same @_ as the calling sub had. But the inner sub can't modify @_ for the calling sub even in this case.

$ perl
outer("hello", "world");

sub outer {
   # we get @_ = hello, world
   print "sub outer got: @_\n";

   inner(); # send an empty @_
   inner("goodbye", "mars");
   inner; # send it the same @_ we have

   print "at the end, sub outer has: @_\n";
}

sub inner {
   print "sub inner got: @_\n";

   # we might not be able to modify @_ in the case
   # of the second sub call where it is read-only
   eval {
      $_[0] = "bonjour";
   };
   if ($@) {
      warn $@;
   }

   print "end of sub inner, has: @_\n";
}
__END__
sub outer got: hello world
sub inner got: 
end of sub inner, has: bonjour
sub inner got: goodbye mars
Modification of a read-only value attempted at - line 19.
end of sub inner, has: goodbye mars
at the end, sub outer has: hello world

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perl -e '$|=$i=1;print" oo\n<|>\n_|_";x:sleep$|;print"\b",$i++%2?"/":"_";goto x;'

 

[Kirsle]

Actually, the last code isn't quite right.

-- inner; # send it the same @_ we have
++ &inner; # send it the same @_ we have

Perl was confused about the bareword "inner;" and didn't call it as a sub. Using the & disambiguates it and Perl calls the sub correctly.

New output:

sub outer got: hello world
sub inner got: 
end of sub inner, has: bonjour
sub inner got: goodbye mars
Modification of a read-only value attempted at - line 20.
end of sub inner, has: goodbye mars
sub inner got: hello world
Modification of a read-only value attempted at - line 20.
end of sub inner, has: hello world
at the end, sub outer has: hello world

Usually when you see a sub called with no parenthesis it's on a method of a module, like

print $obj->data;

so Perl isn't confused by it.

Kirsle.net | My personal homepage

perl -e '$|=$i=1;print" oo\n<|>\n_|_";x:sleep$|;print"\b",$i++%2?"/":"_";goto x;'

 

[1DMF]

Thanks Kirsle,

I'm glad that there is a local @_ for each subroutine block!

I have a sub which receives 3 arguments, I then call another sub passing it $_[0] as an argument, I was concerend when the sub returns back, that the @_ would have been trashed.

As @_ is local within the sub block, it isn't going to be a problem!



"In complete darkness we are all the same, only our knowledge and wisdom separates us, don't let your eyes deceive you."

"If a shortcut was meant to be easy, it wouldn't be a shortcut, it would be the way!"

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