![]() |
|
"be consistent" | |
PerlMonks |
Re^3: Perl Renameby Hugmeir (Sexton) |
on Jul 03, 2010 at 18:01 UTC ( [id://847922]=note: print w/replies, xml ) | Need Help?? |
I'm new to Perl myself, so this explanation might not only be insufficient but also wrong. That mandatory warning aside.. Map takes a list and returns a list. The best way I've found to understand map, is to remember that you have to read the list argument backwards; that is, the list it modifies goes at the end, and the list you get, at the start. In this case, both lists are @ARGV, which might complicate your understanding somewhat. Basically, it takes each element in @ARGV, transforms them with whatever it's in the block (in this case), and returns the transformed elements as a new list to be set in @ARGV again. So, let's take a look inside the block. I've never used File::DosGlob, but from skimming over the documentation, I assume it ports the glob function from *nix to a DOS shell/cmd. So we are using it to expand the wildcard, returning a list of elements that match. Then, my @g = File::DosGlob::glob($_) if /[*?]/;Checks whenever $_ has a wildcard in it, and if it does, calls the DosGlob glob function, which returns a list of files to be set in @g. The last line of the code tells the block which values send over to the new list being generated by map, like in your usual function. Here it's a ternary operator; if @g is true (has nonzero elements), evaluates @g, and thus sends that to map. If @g is false (there were no wildcards in $_), it evaluates whatever is after the :, which is $_, and sends that to the map. Here's hoping this helps/I didn't massively screw up.
In Section
Seekers of Perl Wisdom
|
|