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Re: list assignment to list in scalar context

by TGI (Parson)
on Nov 09, 2012 at 03:33 UTC ( #1003053=note: print w/replies, xml ) Need Help??

in reply to list assignment to list in scalar context

I can't really answer "why" Larry and crew chose the things they did. But, I might be able to shed a little light.

Let's look at your examples

scalar((1,3,5)) #evaluates to the last element of the list --- 5 if( (5,3,0) ) #evaluates to the last element of the list --- 0, +thus evaluates to false $val=(5,6,7); #evaluates to the last element of the list --- 7

These use the comma operator ',' in scalar context, returning whatever is on the right of the comma. In the case of multiple commas, you get the value of the rightmost expression.

There is no list here at all. You just think there is because you have been trained to think of the comma as meaning 'list'.

$ret = (($i, $j, $k)=(5,6,8,9)); #evaluates to the number of el +ements on the right side list --- 4 if( ($k, $v) = ( 1, 0 ) ) #evaluates to the number of el +ements on the right side list --- 2, thus evaluates to TRUE while ( ($key, $value) = each %map ) #still evaluates to the number + of elements on the right side hash

Here we actually have some lists, and thus actual list assignment.

List assignment is scalar context gives us the size of the list that was passed into the assignment. So here $foo = ($bar, $baz) = (1..50); $foo is equal to 50.

In list context you get the contents of the resulting list. So  @foo = () = (1..50) leaves @foo empty.

Why is it this way? I don't know. But I can tell you that getting the size of the assigned from list can be useful.

my $count_foo = () = /foo/g;

Written a bit more compactly, that construct gets called "the goatse operator": $foo =()= /foo/g; If you don't get the joke, don't look it up.

I hope this helps a bit.

TGI says moo

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