http://www.perlmonks.org?node_id=814297


in reply to accessing the result of a match as an array

@matches = "hello awesome" =~ /(el).*(om)/; print pop @awesome;
but this is clearly unacceptable.

Particularly since you match into @matches, but pop @awesome. :-) As

prototype "CORE::pop"; # => ;\@
indicates, if pop takes any argument at all, then it must be an array (not a list)—in common parlance, it must start with a @ sigil. (In fact, its suitability for feeding to pop is one of the ways that people usually advise to distinguish an array from a list.) The reason for this, in turn, is (I think—but this is a difficult point, so I make it small so as to avoid giving offence if I am mistaken) exactly the one you have hit on—a list is not a data structure, in the sense that it is essentially internal to the interpreter and not meant for the user to be slinging around via such tactics as naming it, or, especially, mutating it (which is what pop does).

Since you clearly have no need for the array/list (because you don't want to name it), it's not important that you actually mutate it; so why not just index into it?

print ( ( "hello awesome" =~ /(el).*(om)/ )[-1] );
(Since I always forget them, I'll point out that the parentheses are necessary to avoid parsing as ( print ("hello awesome" =~ /(el).*(om)/) )[-1], which is, as perl will gently remind you, a syntax error.)

UPDATE: By the way, [doc://perldata] and [doc://scalar] become, respectively, perldata and scalar.
UPDATE 2: On further thought, I realise that I've never understood this error:

Type of arg 1 to pop must be array (not list assignment) at -e line 1, + near "/(el).*(om)/)"
That is, I understand that there is a difference between an array and a list assignment; but, since a list assignment returns the array being assigned to:
$ perl -e '( my @a = (1) ) = (2); print "@a\n";' 2
I don't understand why pop complains. I suspect it's a parsing problem; maybe one of the internals experts could be enticed to say for sure?