LanX> when scalar is applied to a list,

Here "list" is to be read as comma separated (literal) list. (see scalar comma separator)

Sorry for nitpicking but there is too much confusion. .

What confusion where (where is the confusion)? And how is this distinction important? In this case?

I think its unimportant , esp in this case, a distinction without a difference ... list literal or literal list is not even in perlglossary

A different demo of the "null list"

## a literal list in list context
$ perl -le " print( qw/ a b c d / )"
abcd

## a literal list in scalar context
$ perl -le " print( scalar qw/ a b c d / )"
d

## a literal list in lvalue list context 
## a literal list in  list context  ( the left hand side list also hap
+pens to be an lvalue)
## a literal list in  list assignment (list context  )
$ perl -le " print( ( $q ) = qw/ a b c d / )"
a

## null list in scalar context counts, counting a list literal
$ perl -le " print( $q = () = qw/ a b c d / )"
4

## null list in scalar context counts, counting a "list"
$ perl -le " @f = qw/ a b c d /; print( $q = () = @f )"
4
## null list in scalar context counts
$ perl -le " print( scalar( () = qw/ a b c d / ) )"
4

## null list in list context discards (using literal list)
$ perl -le " print( () = qw/ a b c d / )"

## null list in list context discards (the array kind of list )
$ perl -le " @f = qw/ a b c d /; print( () = @f )"
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> What confusion where (where is the confusion)?

in the following you (could) see a flattened "list", but scalar doesn't return the last element, which should be 5.

  DB<118> @a=(4,5)
 => (4, 5)

  DB<119> scalar (1,2,3,@a)
 => 2
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in reality there is only the scalar comma operator returning the scalar of the last expression, which is here the size of @a.

Cheers Rolf

PS: Je suis Charlie!

LanX I think you just found a perl bug

$ perl -e"@f=4..5;print scalar( 1,2,3,@f) "
2
$ perl -e"@f=4..5;print scalar( 1,2,3,) "
3
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