the list assignment returns the length of the list in scalar context, so 
   DB<114> $l = ( @a = qw/a b/ )
 => 2

[download]

So it's not the same code!

The difference between your code and my example is that there is no explicit variable (like $l) used here, but somehow it's still possible to change it.

Maybe it's a side effect of for ? I dunno.

I'd prefer a Can't modify warning here too!

But it's still a very peculiar construction, so I'm not surprised if this edge case wasn't covered.

HTH!

Cheers Rolf
(addicted to the Perl Programming Language and ☆☆☆☆ :)
Je suis Charlie!

update

More insights:

... it's not the temporary scalar which is modified but the resulting list in brackets

  DB<121> print "$_\n" for ($a=666) x= 2
666
666
 => ""

  DB<122> $a
 => 666

[download]

According to the already cited documentation of "combined assignments" this shouldn't be possible, b/c its a list operation.

In reply to Re^3: ... for (@_) x= 2; (scalar assignment) by LanX
in thread ... for (@_) x= 2; by rsFalse

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