When it's the same as

P:\test>perl -e"'fred' => 1;"
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A hash element pair in a void context?

There is no "hash" involved here. It's merely a list.

perl -MO=Deparse -e "'moo', 'oom', 'zoop'"
'???', '???', '???';
-e syntax OK
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I've been reading to much Perl 6 stuff where Pairs are distinct entities. Your right. It's just a 2 element list in a void context which is being optimised away. Hence the '???'.

---

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail
"Memory, processor, disk in that order on the hardware side. Algorithm, algoritm, algorithm on the code side." - tachyon

I suspect that use is stringified by a fat comma before it is recognised as a keyword, I haven't yet figured out how to test that theory.

Yes, that's exactly what's happening:


% perl -le 'print for f(); sub f {use => strict}'
use
strict
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Perhaps I'm just being thick, and there is something much more subtle going on here that I am missing, but isn't that exactly the documented difference between ',' and '=>'? From perldoc perlop (emphasis mine):

The => digraph is mostly just a synonym for the comma operator. It's useful for documenting arguments that come in pairs. As of release 5.001, it also forces any word to the left of it to be interpreted as a string.

So the statement 'use => strict' is a 2-element list in void context, whereas using a comma allows 'use' to be interpreted as a keyword, thus the syntax error...

I agree. When the string is made explicit, it does not change the result:

perl -MO=Deparse -e "'use', 'strict'"
'???', '???';
-e syntax OK

perl -MO=Deparse -e "'use'=> 'strict'"
'???', '???';
-e syntax OK
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