But scalar is setting scalar context, I believe that is its sole purpose in life. The sort function can look to see if the context it is being used is scalar or list via the function wantarray. It seems that sort returns undef in a scalar context: 
@a=('a', 'b', 'c');
$a = sort(@a);
print $a;
Use of uninitialized value at - line 3.

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To put sort into a list context there is no clean way to do it, (sort(@a)) doesnt do it in this case. But a snip from the perlfunc man pages for scalar says: 
There is no equivalent operator to force an expression to
be interpolated in list context because it's in practice never
needed.  If you really wanted to do so, however, you could use
the construction @{[ (some expression) ]}, but usually a simple
(some expression) suffices.

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So this does work with -w (but it is a bit of extra work): 
@a=('a', 'b', 'c');
$a = @{[sort(@a)]};
print $a;

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In reply to Re: scalar doesn't work values returned by a function? by perlmonkey
in thread scalar doesn't work values returned by a function? by ZZamboni

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