Can I force a user to provide a value for a member when using designated initializers?

Question

I have created an options struct, intended to be used via designated initializer:

struct FooOptions {
    bool enableReticulation;
};

void Foo(FooOptions&& opts);

Foo(FooOptions{.enableReticulation = true});

Unfortunately, because bool has a default constructor, it's also valid to do this:

Foo(FooOptions{});

but I don't want this; I want users of Foo to explicitly decide to enable reticulation or not. I can achieve this with a runtime error this way:

struct FooOptions {
    bool enableReticulation = []() -> bool { 
        assert(false, "No value provided for enableReticulation");
    }();
};

But I would prefer to do this with a compile time error. Is there any way to do that? I am OK with changing bool to SomeWrapper<bool> if necessary, as long I can mostly initialize SomeWrapper<T> as if it were T, but without a default initializer.

Answer 1

Way to handle classes and non-classes types, thank to SFINAE:

template<typename T, typename Enabler = void> class TWrapper;

template<typename T>
class TWrapper<T, std::enable_if_t<std::is_class<T>::value>> : public T {
public:
    TWrapper()=delete;

    using T::T;
};

template<typename T>
class TWrapper<T, std::enable_if_t<!std::is_class<T>::value>>
{
public:
    TWrapper()=delete;

    T value;
    TWrapper(T arg) : value(arg) {}
    operator T() const { return value; }
};

Demo