in reply to Re: Operator overloading with returning lists?
in thread Operator overloading with returning lists?

x behaves different in list context as well

Not a very good example. It's more the LHS operand than the context which determines what is returned. 

>perl -le"$x = 'x' x 5; print $x"
xxxxx

>perl -le"$x = ( 'x' x 5 )[0]; print $x"
xxxxx

>perl -le"$x = ( ('x') x 5 )[0]; print $x"
x

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The only time context matters, x behaves normally. 

>perl -le"$x = ('x') x 5; print $x"
xxxxx

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Re^3: Operator overloading with returning lists?
by JavaFan (Canon) on Dec 01, 2008 at 10:54 UTC
    x behaves different in list context as well
    Not a very good example. It's more the LHS operand than the context which determines what is returned.
    That's not all I said. I immediately followed it (not even stopping to start a new sentence) with
    but only if its LHS has parens, making it an odd one
    Both context and the LHS matter, just as I said: 
    my $w =  'x'  x 5; say "$w";   # Scalar context, no parens.
my $x = ('x') x 5; say "$x";   # Scalar context, parens.
my @y =  'x'  x 5; say "@y";   # List context, no parens.
my @z = ('x') x 5; say "@z";   # List context, parens.
__END__
xxxxx
xxxxx
xxxxx
x x x x x

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[reply]
[d/l]
      since when does scalar context of a list result in a join??? 
      
  DB<88> p Dumper (('x') x 5)
$VAR1 = 'x';
$VAR2 = 'x';
$VAR3 = 'x';
$VAR4 = 'x';
$VAR5 = 'x';

  DB<89> p Dumper scalar (('x') x 5)
$VAR1 = 'xxxxx';

  DB<90> p Dumper scalar (@x=('x') x 5)
$VAR1 = 5;

  DB<91> p Dumper (scalar qw/x x x x x/)
$VAR1 = 'x';

      [download]

      mysterious perl ... 8(

      Cheers Rolf

      UPDATE: 

        DB<97> p Dumper scalar ( (1,2) x 5 )
$VAR1 = '22222';


  DB<98> p Dumper ( (1,2) x 5 )
$VAR1 = 1;
$VAR2 = 2;
$VAR3 = 1;
$VAR4 = 2;
$VAR5 = 1;
$VAR6 = 2;
$VAR7 = 1;
$VAR8 = 2;
$VAR9 = 1;
$VAR10 = 2;

      [download]

      The scalar is applied to the list before the x-op can act. The parens are ignored...

      Thats a bug or a feature?

[reply]
[d/l]
[select]

        Perl doesn't see things as left-to-right as you do.

        "scalar" in "scalar ( (1,2) x 5 )" acts on "x", not "(1,2)".
        The return value of "()x" in scalar context is (join '', map LHS, 1..RHS).

        Similarly, "scalar" in "scalar ( ($a) = 5 )" acts on "=", not "($a)".
        The return value of "()=" in scalar context is the number of items returned by its RHS.

        Not a bug. In both case, the code works as designed and as intended.

[reply]
[d/l]
[select]
        Neither. 
        C:\>perl  -MData::Dumper -le"print Dumper scalar ( 1, 2 ) "
$VAR1 = 2;


C:\>perl  -MData::Dumper -le"print Dumper scalar ( 1, 2 ) x 5"
$VAR1 = '22222';


C:\>perl  -MData::Dumper -le"print Dumper 2 x 5"
$VAR1 = '22222';

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        "List" Is a Four-Letter Word
[reply]
[d/l]
        since when does scalar context of a list result in a join???
        It does? Which of your examples do you think supports this notion?
        The scalar is applied to the list before the x-op can act. The parens are ignored...
        I do not know what you mean by this.
        Thats a bug or a feature?
        From 'man perlop': 
               Binary "x" is the repetition operator.  In scalar context or if the
       left operand is not enclosed in parentheses, it returns a string
       consisting of the left operand repeated the number of times specified
       by the right operand.  In list context, if the left operand is enclosed
       in parentheses or is a list formed by "qw/STRING/", it repeats the
       list.  If the right operand is zero or negative, it returns an empty
       string or an empty list, depending on the context.
[reply]

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