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Re: map sub to list?

by BrowserUk (Patriarch)
on Sep 09, 2013 at 16:50 UTC ( #1053042=note: print w/replies, xml ) Need Help??


in reply to map sub to list?

chr defaults to using $_, so use: @a = map f( $_ ), @numbers;.

Or define f() to default to $_:

sub f(_){ $_[0]+1 };

Now my @a = map f, @numbers; works.


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Re^2: map sub to list?
by pldanutz (Acolyte) on Sep 09, 2013 at 17:15 UTC
    Thanks to everybody who replied! I think I've got it. When writing @a = map f( $_ ), @numbers; this is a case of map BLOCK LIST (as opposed to map EXPR LIST) right?
      No, using the comma makes it a case of EXPR, LIST.

        I can't seem to be able to reply to other replies (too much nesting probably). Perhaps I'll try the chatterbox or a separate question.

        I was trying to understand if the Perl compiler parses "f ($_)" in "map f ($_), @numbers" as a lambda. If it's not a block, but an expression, it must be either an arithmetic expression, a lambda expression etc. My question was from a syntactic/compiler point of view.

        And as to eager evaluation, obviously this is a counterexample, since map f($_) @nums doesn't immediately, eagerly evaluate f on the current $_, but instead passed a lambda to map()

        Where can I read about functions that have comma and non-comma parsings? This confuses me. Also, what expression is f ($_)? Is it a new function, i.e. a shorthand for sub f2 (_) { f ($_) }

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