On Thu, 7 Dec 2023, 05:15 Egor via Std-Proposals, <
std-proposals_at_[hidden]> wrote:

> Hello!
>
> std::optional<T> has a really greedy converting constructor. It accepts
> almost any type that T is constructible from, and constructs a T from
> it, even if that type is also convertible to optional<T>. Example:
> https://gcc.godbolt.org/z/ncsYehW19
>
> #include <iostream>
> #include <optional>
> #include <source_location>
>
> struct A
> {
> template <typename T>
> operator T()
> {
> std::cout <<
> std::source_location::current().function_name() << '\n';
> return {};
> }
> };
>
> int main()
> {
> std::optional<int> x(A{}); // (1) Calls `operator int`, not
> `operator optional<int>`.
> // std::optional<int> y = A{}; // (2) Ambiguous.
> x = A{}; // (3) Calls `operator int`, not `operator
> optional<int>`.
> }
>


What's the practical use case for this type's conversion operator? If it
can convert to both int and optional int, what does that mean? Why is it a
problem to convert to int first and then wrap that in an optional? Does
your type have a special way to construct an optional that is more
efficient than converting to an int and then setting optional<int> from
that?

"I find this confusing" is not very strong motivation.


> This is very confusing in my eyes. std::optional is first type I see in
> the wild that doesn't play nice with templated conversion operators.
>
> I suggest we make this constructor of optional not participate in
> overload resolution if the argument is convertible to optional<T>.
>
> This would make (2) valid, and make all three lines call A::operator
> optional<T>.
>
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>