Any arguments passed in show up in the array @_. Therefore, if you called a function with two arguments, those would be stored in $_[0] and $_[1]. The array @_ is a local array, but its elements are aliases for the actual scalar parameters. In particular, if an element $_[0] is updated, the corresponding argument is updated (or an error occurs if it is not updatable). If an argument is an array or hash element which did not exist when the function was called, that element is created only when (and if) it is modified or a reference to it is taken. (Some earlier versions of Perl created the element whether or not the element was assigned to.) Assigning to the whole array @_ removes that aliasing, and does not update any arguments.

reading above paragraph from perlsub, and i don't really quite get this sentence..

If an argument is an array or hash element which did not exist when the function was called, that element is created only when (and if) it is modified or a reference to it is taken.

does it mean that the following should print 1 (but it doesn't)? perl -le 'x(@y);sub x{$_[0]=1;print @y}'

In reply to perlsub question.. by Anonymous Monk

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