Why two `{` is required here?

Question

$text =~ s/(cat|tomatoes)/ ${{ qw<tomatoes cat cat tomatoes> }}{$1} /ge;

And I can't replace ${{ qw<tomatoes cat cat tomatoes> }}{$1} with { qw<tomatoes cat cat tomatoes> }->{$1},why?

UPDATE

  5 @array = qw<a b c d>;
  6 $ref = \@array;
  7 @{$ref} = qw<1 2 3 4>;
  8 #@$ref = qw<1 2 3 4>;//also works
  9 print "@array";

So it indicates neither {} nor ${} is required to dereferencing,the {} is only required when ambiguity arises,and $ only in scalar context.

Answer 1

${{ qw<tomatoes cat cat tomatoes> }}{$1} 

is

my $ref = { qw<tomatoes cat cat tomatoes> };
${ $ref }{$key}

The inner brackets form an anonymous hash constructor. It creates a hash, assigns the contents of the brackets to it, then returns a reference to it.

The outer brackets are part of the hash dereference. They can be omitted (e.g. $$ref{$key} instead of ${$ref}{$key}) when unambiguous (e.g. when dereferencing a simple scalar), but this is not such a circumstance.

One can also dereference using the arrow notation, so one could also have used

{ qw<tomatoes cat cat tomatoes> }->{$1} 

The difference is that the version being used is simply a variable lookup, so it doesn't require /e, while the latter is Perl code, so it does require /e.

---

If you had just

${ qw<tomatoes cat cat tomatoes> }{$1} 

that would be the same as

${ "tomatoes" }{$1} 

since qw() in scalar context returns the last value. That, in turn, is the same as

$tomatoes{$1} 

(except that use strict; wouldn't allow it) and that's obviously not what you want.

Answer 2

The outer brackets dereference the anonymous hash created by the inner brackets.

Update for clarification: The second format you use would work if you give the compiler a clue by putting a + in front of it:

+{ qw<tomatoes cat cat tomatoes }->{$1}