This pattern is usually used to deduce a function template argument from the type its result is assigned to.

Here's my usecase. I have a function that reads a field from a config file:

    template <typename T> T getConfigField(std::string_view name);

    int x = getConfigField<int>("field_x");
    std::optional<int> y = getConfigField<std::optional<int>>("field_y");

And let's say I don't want to repeat the type name twice. (In this specific case I could use auto, but it's not always an option, e.g. when passing getConfigField<..>(..) as a function argument or when initializing a non-static data member with it.)

One solution is to add a second non-template overload of getConfigField, which returns a proxy class with a templated conversion operator, like the one I've shown in the previous message:

    struct Proxy {
        std::string_view name;
        template <typename T> operator T() const {return getConfigField<T>(name);}
    };
    Proxy getConfigField(std::string_view name) {return Proxy{name};}

    int x = getConfigField("field_x"); // T is deduced as int
    void foo(int x);
    foo(getConfigField("field_x"));

This works correctly for most types, std::optional is only outlier I've seen.

> Why is it a problem to convert to int first and then wrap that in an optional?

Because converting to int directly doesn't give you the chance to return a null optional.

In this example, imagine that getConfigField<std::optional<int>>("foo") returns null if given a non-existing field, while getConfigField<int>("foo") throws. But if I do std::optional<int> x( getConfigField("foo") ); for an non-existing field, it unexpectedly throws too, since T=int.

On Thu, 7 Dec 2023, 05:15 Egor via Std-Proposals, <std-proposals@lists.isocpp.org> 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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