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lvalue rvalue

 
 
Ioannis Vranos
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      07-16-2004
Ali Cehreli wrote:

> They are destroyed at the end of the complete statement where they are
> created. The complete statement may have many other function
> calls. The destruction of the temporary must wait untill all of those
> calls are completed.
>
> This is incorrect. The temporary lives longer than the function call
> as I described above. The destruction must wait until the statement
> completes.



Yes you are right.






Regards,

Ioannis Vranos

http://www23.brinkster.com/noicys
 
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JKop
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      07-17-2004
Andrey Tarasevich posted:

> but that's just the way it is in C++.



The only valid explanantion.


-JKop
 
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JKop
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      10-18-2004

I've just had a quick read over this thread.

We can summarize temporaries with the following:

non-const (they're *not* const)
r-value (they're *not* an l-value)

(strange how it is!)

With has the following repercussions (for want of a better word):

A temporary cannot be bound to a non-const reference ( not because it's
const, because it *isn't*, but because it's an r-value, ie. you can only
bind a non-const reference to an l-value ).

So, where you have:

AnyType& blah = Anything...

then "Anything" must be:

non-const
l-value


And where you have:

AnyType const & blah = Anything...

Then "Anything" can quite literally be anything!


-JKop
 
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