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From: Doug Gregor (dgregor_at_[hidden])
Date: 2007-01-08 11:48:58


On Jan 8, 2007, at 11:32 AM, Tobias Schwinger wrote:
> Doug Gregor wrote:
>> TR1 says...
>>
>> Given an rvalue f of type F and values t1, t2, ..., tN of types T1,
>> T2, ..., TN, respectively, the type member
>> is the result type of the expression f(t1, t2, ...,tN). The values
>> ti are lvalues when the corresponding type Ti is
>> a reference type, and rvalues otherwise.
>
> Sorry - it seems I missed that last part of the sentence, somehow
> (maybe
> by looking at the wrong version?).

It's likely that you were looking at an older draft of TR1. We had
originally said that everything was an lvalue. We later realized that
making them rvalues by default, with references signifying lvalues,
gave us a more functional result_of that was actually easier to deal
with. The final TR1 draft and the C++0x draft have the revised
wording I quoted.

        Cheers,
        Doug


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