Problem: That will return a list of characters, not a string.

But, when you print, it forces scalar context. The output of the print statement is the same. Same if you assigned result of sub to a scalar explicitly:

my $result = invert($string);
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is the same with or without the join.

.02

cLive ;-) - see above

Not quite. A list grep in a scalar context return the size of the list.

sub invert {
    #23456789012345678901234567890
    grep/[^\Q@_\E]/,map{chr}1..255
};

$inverted = invert( 'Just Another Perl Hacker!' );

print $inverted;
237
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237 + 18 (unique characters in the string) = 255.

---

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail

Huh? grep() in scalar context returns a number, not a join()ed list of elements that passed through. And print() doesn't "force scalar context" -- in fact, it invokes list context, which is the only reason why print invert($string) works.
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Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area)
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;