Re: Find Element of array but use the next Element

Yet another way, using "reverse" and a faked-out "slice":

>perl -e "@arr = qw{ app Oracle EPDMCA Oracle EPZXC Oracle };
  print (map {$_ eq 'Oracle'? ($x):($x=$_)[2]} reverse @arr)"
[download]

I hope life isn't a big joke, because I don't get it.
-SNL

Comment on Re: Find Element of array but use the next Element Download Code

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Re^2: Find Element of array but use the next Element by choroba (Cardinal) on Jun 14, 2012 at 22:01 UTC
Not getting it. Care to explain?	[reply]
Re^3: Find Element of array but use the next Element by NetWallah (Canon) on Jun 14, 2012 at 23:19 UTC
Analyzing `map {$_ eq 'Oracle'? ($x):($x=$_)[2]} reverse @arr` [download] Ok - reading backwards, the first thing that happens, is @arr gets "reversed". Next we process each element of the reversed array. If 'Oracle' is found, we return "$x". You way well ask - what is $x, and where did it get a value? That happens if $_ does NOT match. In that case, what happens is: * $x gets assigned the current element * a temporary LIST is generated because of the () - this contains ($x) * We index index into the second element ($x)[2] which does not exist resulting in an EMPTY list That gets sent to "map" which ignores that. SO - the only elements that "map" returns is the $x value when the current element is 'Oracle'. In this case, $x contains the value of the previous element, and because we are 'reversed', in real life, it contains the 'next' element. There is a potential complaint that the order of the returned values is reversed. This is easily fixed by throwing a 'reverse' in front of the 'map'. I hope life isn't a big joke, because I don't get it. -SNL	[reply] [d/l]