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Arguments are passed into a subroutine via @_. Accessing one element or more of that array can be done in two ways, non-destructive (the @_ array persists as passed in) or destructive (the @_ array is consumed during assignment):
# 1. non-destructive my $var = $_[0]; # first element of @_ my @list = @_; # complete @_ # 2. destructive my $var = shift; # first element gets removed + from @_ my ($foo, $bar) = map { shift } 1,2; # two elements get removed f +rom @_

If @_ needs to be intact for subsequent calls, as in

sub foo { my $foo = $_[0]; my $quux = baz(@_) return $foo ^= $quux; }

the non-destructive methods are used. Also, as the variables passed in via the vector @_ are references aliases to the thingies the caller provided, the non-destructive way is often used to do in-place transformations.

Otherwise, it just doesn't matter. Since after setting up the variables in the sub's scope @_ isn't looked at anymore, arguments may be shifted or not. In these cases saying $var = shift or $var = $_[0] does the same for the sub, although the impact on @_ is different.

So, saying $var = shift and @list = @_ is just caring about copying, but done that, not caring about @_ any more.

Using the arguments in a sub without prior assignment (i.e. without copying, as $_[0] .. $_[$#_]) modifies the thingies in the caller.


<update> changed reference to alias as per tye's post </update>

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In reply to Re: shift vs @_ by shmem
in thread shift vs @_ by Zadeh

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