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

Thats a bug or a feature?