in reply to Re^2: Match operator giving unexpected output
in thread Match operator giving unexpected output

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 the error message is what is confusing 

$ perl -MO=Deparse -wE "say scalar('a', 'e', 'i', 'o', 'u');"
Useless use of a constant ("a") in void context at -e line 1.
Useless use of a constant ("e") in void context at -e line 1.
Useless use of a constant ("i") in void context at -e line 1.
Useless use of a constant ("o") in void context at -e line 1.
BEGIN { $^W = 1; }
use feature 'current_sub', 'evalbytes', 'fc', 'say', 'state', 'switch'
+, 'unicode_strings', 'unicode_eval';
say scalar('???', '???', '???', '???', 'u');
-e syntax OK

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Void context? scalar(...) is clearly not void :D

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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Re^4: Match operator giving unexpected output
by LanX (Saint) on Jan 13, 2015 at 12:02 UTC
    > 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!

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[d/l]
      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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[reply]
[d/l]
        Nevermind the warnings warn and not using numbers is clear 
        $ perl -we"@f=( qw/ r p n /);print scalar( q/a/, q/b/, q/c/ ,@f) "
Useless use of a constant ("a") in void context at -e line 1.
Useless use of a constant ("b") in void context at -e line 1.
Useless use of a constant ("c") in void context at -e line 1.
3

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[d/l]

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