map and smart match

toapole has asked for the wisdom of the Perl Monks concerning the following question:

Hello
Why this code works:

my @test = ( {id => 1}, {id => 2} );
my @arr = map $_->{id}, @test;
say 1 ~~ @arr;

and this doesn't (namely context in matching is scalar):

my @test = ( {id => 1}, {id => 2} );
say 1 ~~ map $_->{id}, @test;

?
or i'm just in state of sin? :)

Comment on map and smart match

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Re: map and smart match by ikegami (Patriarch) on Jul 09, 2009 at 22:41 UTC
Because `map $_->{id}, @test` is not an array. `say 1 ~~ @{[ map $_->{id}, @test ]};` should do the trick. I wonder if `say 1 ~~ [ map $_->{id}, @test ];` also works.	[reply] [d/l] [select]
Re^2: map and smart match by ikegami (Patriarch) on Jul 09, 2009 at 22:55 UTC
The passage I bolded from what is to be 5.10.1's perlsyn answers my own question: The behaviour of a smart match depends on what type of thing its arguments are. The behaviour is determined by the following table: the first row that applies determines the match behaviour (which is thus mostly determined by the type of the right operand). Note that the smart match implicitly dereferences any non-blessed hash or array ref, so the "Hash" and "Array" entries apply in those cases. (For blessed references, the "Object" entries apply.)	[reply]
Re: map and smart match by JavaFan (Canon) on Jul 10, 2009 at 10:57 UTC
Because in `say 1 ~~ map $_->{id}, @test;` [download] the right hand side is first evaluated (in scalar context), yielding 2. And `1 ~~ 2` is false.	[reply] [d/l] [select]