Re: 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($_); }` [download] Consider this to see the difference: `@old = (10..20); sub act_on { return 1..$_; }` [download] The first element becomes 10 elements of output, the second becomes 11, and so on. -- Randal L. Schwartz, Perl hacker	[reply] [d/l] [select]
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.	[reply] [d/l] [select]
Re: Re: Re: playing with map by merlyn (Sage) on Mar 09, 2001 at 23:21 UTC
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]
Re: Re: Re: Re: playing with map by mirod (Canon) on Mar 09, 2001 at 23:28 UTC
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`.	[reply]


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