Re^3: ... for (@_) x= 2; (scalar assignment)

the list assignment returns the length of the list in scalar context, so

   DB<114> $l = ( @a = qw/a b/ )
 => 2
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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.

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Re^4: ... for (@_) x= 2; (scalar assignment) by rsFalse (Chaplain) on Dec 29, 2015 at 14:15 UTC
I tried compare three of my codes and their outputs. 'x=' seems doing the same in all three cases - it asks '@_' to be a scalar.	[reply]
Re^5: ... for (@_) x= 2; (scalar assignment) by LanX (Saint) on Dec 29, 2015 at 14:20 UTC
> I tried compare three of my codes sorry? > `x=` seems doing the same in all three cases - it asks `@_` to be a scalar. which is correct and documented. Anyway it might be an implementation flow that you sometimes don't get a warning or compilation error, but not a very serious one. I don't see the point discussing it further, since you have plenty of possibilities to achieve your goal in a much saner way. Cheers Rolf _{(addicted to the Perl Programming Language and ☆☆☆☆ :) Je suis Charlie!}	[reply] [d/l] [select]