On Thu, 14 May 2020 at 13:58, Avi Kivity via Std-Proposals
<std-proposals_at_[hidden]> wrote:
> Coroutines capture parameters passed by value and copy them to the
> coroutine frame (all quotes from cppreference):
>
>
> > When a coroutine begins execution, it performs the following:
> >
> > allocates the coroutine state object using operator new (see below)
> > copies all function parameters to the coroutine state: by-value
> parameters are moved or copied, by-reference parameters remain
> references (and so may become dangling if the coroutine is resumed after
> the lifetime of referred object ends)
> >
>
> This allows a coroutine's parameters to outlive the point it is launched at:
>
>
> void foo() {
>
> call_some_coroutine(a_temporary_value());
>
> }
>
>
> Normally, a_temporary_value() would e destroyed at the end of the
> expression, which can be before the coroutine completes.
>
>
> With a lambda coroutine, we have a problem. This is because lambdas are
> captured by reference:

Says what?

> > If the coroutine is a non-static member function, such as task<void>
> my_class::method1(int x) const;, its Promise type is
> std::coroutine_traits<task<void>, const my_class&, int>::promise_type
>
>
> For a lambda, the non-static member function is
> synthetic_lambda_type::operator() const, and so the lambda would be
> captured as const synthetic_lambda_type&. Consider this call:

No. The lambda is the coroutine, not its operator().