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Re: [std-proposals] Why [expr.ref] says that E1->E2 is converted to (*(E1)).E2

From: Jason McKesson <jmckesson_at_[hidden]>
Date: Wed, 10 May 2023 11:18:02 -0400
On Wed, May 10, 2023 at 1:06 AM Sean Mayard via Std-Proposals
<std-proposals_at_[hidden]> wrote:
>>
>> In terms of syntax and evaluation order, yes (as [expr.pre] says).
>
>
> Okay so does this mean that the sentence quoted below from [expr.ref#1] also applies to overloaded operators since it talks about "evaluation"?
>
>> The postfix expression before the dot or arrow is evaluated;63 the result of that evaluation, together with the id-expression, determines the result of the entire postfix expression.
>
>
> That is, I want to know if the above quoted sentence from expr.ref#1 applies to overloaded operators since it talks about "evaluation" and [expr.pre] said that Overloaded operators obey the rules for syntax and evaluation order specified in [expr.compound].

"Evaluation order" and "evaluation" are not the same thing.

Received on 2023-05-10 15:18:14