Re: Class::Struct question

No, but for reasons that have nothing to do with Class::Struct, and everything about Perl OO and operator precedence. Since '++' binds more tightly than '->', when you say:

$x->Counter++;
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According to B::Deparse, Perl translates that to:

++$x->Counter;
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Which should then be tokenized into something like:

(++$x) -> Counter
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Which is probably not at all what you want.

Without knowing your specific situation, I would suggest looking more carefully at your object structure. What you're trying to do indicates to me that you're playing too closely to the object's internals.

Update: Forget I said anything. See ikegami's response below.

"There is no shame in being self-taught, only in not trying to learn in the first place." -- Atrus, Myst: The Book of D'ni.

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Re^2: Class::Struct question by ikegami (Patriarch) on Sep 14, 2004 at 18:03 UTC
++' binds more tightly than '->' That's not true. '->' is just above '++' in precedence. `++$x->Counter;` is the same as `++($x->Counter);`. The problem is that Counter does not return an lvalue. See the other post I'm making in this thread for an example.	[reply] [d/l] [select]