Re: Re: Back to Remedial Perl for Me: map{} function

[mikfire]

Sorry, merlyn, but the second example *is* turning a list into another list. map is very capable of mapping 1-to-n relations, ie, every element on the right will produce more than one element on the right.

If I read your statement correctly ( and I may not have ), the first use of map isn't correct

my @first = qw/ one-1 two-2 three-3/;
my %hash  = map { split /-/ } @first; # Which works for me btw
print map { "$_ => $hash{$_}\n" } keys %hash;

But this usage is correct

my @first = qw/ one-1 two-2 three-3/;
my @second  = map { split /-/ } @first;
my %hash = @second;  # Legal code, isn't it?
print map { "$_ => $hash{$_}\n" } keys %hash;

How else is map supposed to make @first into @second except by performing an action on every element of @first?

I am not arguing that a foreach wouldn't be appropriate here. If that is what works, by all means use it. But if I am not supposed to use map when I want an action on every element of an array, doesn't it make more sense to say

@second = @first;

because that seems to be the only option you have left me with this statement.

Would you be kind enough to expand on your answer so I can figure out what I missed?

TIA
mikfire

[merlyn]

I wasn't commenting on either of those. I was commenting on the original poster's last example:

if ($access) { 
    foreach (@{$access}) { 
        $set{$_->[0]} = $_->[1]; 
    } 
}

That's just fine as it is. No need for a map. In fact, a map would be difficult here, unless we presume %set is already empty before we get here.

-- Randal L. Schwartz, Perl hacker

[mikfire]

Ahh. I got scope locked and assumed that you were referring to the map examples, not the foreach loop.

Thanks
mikfire

[$code or die]

Out of interest, why does this (both examples) print:

one => 1
three => 3
two => 2

and not:

one => 1
two => 2
three => 3

[davorg]

Because hashes are unordered.

--
<http://www.dave.org.uk>

"Perl makes the fun jobs fun
and the boring jobs bearable" - me

[Anonymous Monk]

"unsorted" would be more precise...the elements in the hash do have an order, but not not "in the order" that we might put them.

[$code or die]

That's what I thought, but it did the same order every time I ran it, so it seems that there is some order (i.e. it's not completely random). Was just wondering how perl decides to order it. Even Data::Dumping the hash gives the same order!