comment on

As JavaFan says, contrary to most expectations \ actually provides LIST context ! So:

#!/usr/bin/perl -w
use strict ; use warnings ;

sub foo {
  my ($tag) = @_ ;
  print "$tag: ", wantarray ? "LIST" : "SCALAR", "\n" ;
  return wantarray ? ('a', 'b') : \78 ;
} ;

my $a = \foo('SCALAR \foo()') ;
my @a = \foo('LIST   \foo()') ;

my $b = ${foo('SCALAR ${foo()}')} ;
my @b = (${foo('LIST   ${foo()}')}) ;
[download]

gives:

SCALAR \foo(): LIST
LIST   \foo(): LIST
SCALAR ${foo()}: SCALAR
LIST   ${foo()}: SCALAR

which also shows that ${...} does provide SCALAR context as expected.

This behaviour is not mentioned in perlop and I cannot see it in perlref either :-(

It is documented that \($a, @b, %c) is the same as (\$a, \@b, \%c). Which requires the , to be operating in LIST context -- so perhaps we should not be surprised. (It's also worth remembering that LISTs are interpolated when they are evaluated, not when they are constructed...)

In reply to Re^3: Why does foo() evaluate in array context in "${\foo()}"? by gone2015
in thread Why does foo() evaluate in array context in "${\foo()}"? by ELISHEVA

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