in reply to playing with map

I have written lots and lots of Perl code without ever really using the map command very much. But I often see it in other's code. So, my impression is that the code

@new = map { act_on($_); } @old;

is functionally equivalent to

for (0..$#old) { $new[$_] = act_on( $old[$_] ); }

Is this a correct view of things?

CJW

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Re: Re: playing with map
by merlyn (Sage) on Mar 09, 2001 at 23:18 UTC
    No, that presumes a 1-1 mapping. Some are, some aren't. It's more like this: 
    @new = ();
for (@old) {
  push @new, act_on($_);
}

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    Consider this to see the difference: 
    @old = (10..20);
sub act_on {
  return 1..$_;
}

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    The first element becomes 10 elements of output, the second becomes 11, and so on.

    -- Randal L. Schwartz, Perl hacker

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Re: Re: playing with map
by mirod (Canon) on Mar 09, 2001 at 23:20 UTC

    Not exactly, more like:

    for (0..$#old) { push @new, act_on( $old[$_] ); }

    usually written as:

    foreach @old { push @new, act_on( $_); }

    If act_on returns a list then all members are pushed into @old.

    Update: darn! merlyn is right, you need to initialize @new to () to get the exact equivalent statement.

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      If act_on returns a list then all members are pushed into @old.
      In a list context, act_on must return a list! And map (and the map-equivalents) are providing list context.

      -- Randal L. Schwartz, Perl hacker

[reply]

        OK, I was not clear. I should have written "if act_on returns a list with more than one element" then the original code is not equivalent to a map (and I can play with fonts too ;--). If act_on returns a scalar or a 1 element list then the original code behaves the same way as a map.

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