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

From: language.lawyer_at <language.lawyer_at_[hidden]>
Date: Tue, 9 May 2023 18:02:41 +0500
> I read [expr.ref#2] which includes the following sentence:
>> *The expression E1->E2 is converted to the equivalent form (*(E1)).E2*
> My question is, does this mean that "every" expression that has the form
> *E1->E2* will be converted to *(*(E1)).E2. *I mean it is not clear to me
> in what situations/contexts does the above sentence hold?

Look at the previous sentence.

Received on 2023-05-09 13:02:48