shift(@_) is a current $_

No. Here is the skeleton of the opening of the import subroutine in Win32::TieRegistry:

sub import { my $pkg = shift(@_); ... local( $_ ); while( @_ ) { $_= shift(@_); if( ... ) { ... } elsif( ... ) { ... ... } elsif( exists $_opt_subs{$_} ) { ... $registry->$_( shift(@_) ); } elsif( ... ) { ...

Each call to shift removes another element from the head of the @_ array, i.e., another argument from the list of those passed into sub import. So in the line:

$registry->$_( shift(@_) );

the variable $_ holds the last argument to have been removed from @_; but the expression shift(@_) removes the next argument from @_. $_ and shift(@_) will (most likely) have different values.

shift(@_) does not reset the value of $_:

17:43 >perl -wE "f(1, 2, 3); sub f { local($_) = shift(@_); my $c = sh +ift(@_); say '$_ = ', $_; say '$c = ', $c; say '@_ = ', @_; }" $_ = 1 $c = 2 @_ = 3 18:07 >

Hope that helps,

Athanasius <°(((><contra mundum Iustus alius egestas vitae, eros Piratica,


In reply to Re: Can't understand one line by Athanasius
in thread Can't understand one line by anaconda_wly

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