Re^4: How to declare arrays and scalars together?

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Re^5: How to declare arrays and scalars together? by Anonymous Monk on Jun 19, 2012 at 01:59 UTC
That is my point. It is possible to pass multiple arrays, using array reference Well, you missed the point of Re: How to declare arrays and scalars together? which is, you're only passing lists, you cannot pass an array, an array and an arrayref aren't identical, you need to de-reference an arrayref, but not an array Given `my @one = (1,1); my @two = (2,2);` This `foo( @one, @two );` is basically the same as `foo( 1,1,2,2 );` With the appropriate prototype `foo( @one, @two );` could also be basically the same as `foo( \@one, \@two );` If you use `foo( \@one, \@two );` then in sub foo you would need to write `my( $oneref, $tworef ) = @_;` which is not the same as `my( @one, @two ) = @_;` You're always passing lists to subroutines	[reply] [d/l] [select]

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Re^5: How to declare arrays and scalars together?
by Anonymous Monk on Jun 19, 2012 at 01:59 UTC

That is my point. It is possible to pass multiple arrays, using array reference

Well, you missed the point of Re: How to declare arrays and scalars together? which is, you're only passing lists, you cannot pass an array, an array and an arrayref aren't identical, you need to de-reference an arrayref, but not an array

Given my @one = (1,1); my @two = (2,2);

This foo( @one, @two ); is basically the same as foo( 1,1,2,2 );

With the appropriate prototype foo( @one, @two ); could also be basically the same as foo( \@one, \@two );

If you use foo( \@one, \@two ); then in sub foo you would need to write my( $oneref, $tworef ) = @_; which is not the same as my( @one, @two ) = @_;

You're always passing lists to subroutines

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