C++中的部分class模板参数推导

Question

I have a templated class but only part of the template arguments can be deduced from the constructor.

Is there a way to provide the rest of the template arguments inside angle brackets when calling the constructor?

Assume we're using C++17.

template<typename T1, typename T2>
struct S
{
    T2 t2;

    S(const T2& _t2) : t2{_t2} {}

    void operator()(const T1& t1)
    {
        std::cout << t1 << ", " << t2 << '\n';
    }
};

int main()
{
    S<int, double> s {3.14};

    std::function<void(int)> func = s;
    
    func(42);

    // What I want:
    //S<int> s2 {3.14};     <- T1 is provided in the angle brackets, T2 is deduced
    //std::function<void(int)> func2 = s;
    //func2(42);
}

As far as I know we need to either provide all the template arguments in angle brackets or none of them and use CTAD. The problem is that I don't want to write all the template arguments (in my actual use case there's like 5-6 of them and they are quite verbose) but I also don't want to pass all the arguments in the constructor because some of them are not used to construct the object. I just need their types for the operator() method.

I cannot make the operator() method templated because I want to bind it to a std::function object and I cannot deduce the template parameter types during the bind. So that's why I need all the types in the wrapping class.

This partial template deduction exists for functions.

For example:

template<typename T1, typename T2>
void foo(const T2& t2)
{
    T1 t1{};
    std::cout << t1 << ", " << t2 << '\n';
}

int main()
{
    foo<int>(3.4); //T1 is explicitly int, T2 is deduced to be double
}

My current solution is to exploit this feature and construct the object through a function:

template<typename U1, typename U2>
S<U1, U2> construct_S(const U2& t2)
{
    return S<U1, U2>{t2};
}

int main()
{
    auto s2 = construct_S<int>(1.5);
    std::function<void(int)> func2 = s2;
    func2(23);
}

I find this solution clumsy because we're using an external function to construct the object.

I was wondering if there's a cleaner solution for doing this.

Maybe something with deduction guides? I'm not sure.

Answer 1

As mentioned in a comment, you can use a nested class such that the two parameters can be provided seperately (one explicitly the other deduced):

template<typename T1>
struct S {
    template <typename T2>
    struct impl {
        T2 t2;    
        impl(const T2& _t2) : t2{_t2} {}
    };

    template <typename T2>
    impl(const T2&) -> impl<T2>;
};


int main() {
    S<int>::impl<double> s {3.14};
    S<int>::impl s2 {3.14};    // <- T1 is provided in the angle brackets, T2 is deduced
}

I found this How to provide deduction guide for nested template class? . Though, the above code compiles without issues with both gcc and clang: https://godbolt.org/z/MMaPYGbe1 .

If refactoring the class template is not an option, the helper function is a common and clean solution. The standard library has many make_xxx functions, some of them were only needed before CTAD was a thing.

Answer 2

The simplest way is to provide factory function which will take over deduction of types:

template<typename T1, typename T2>
auto makeS(T2 x) -> S<T1, T2>
{
    return S<T1, T2>{x};
}

https://godbolt.org/z/4cPTdv7e3

Answer 3

template<typename T2>
struct S
{
    T2 t2;

    S(const T2& _t2) : t2{_t2} {}
    template<typename T1>
    void operator()(const T1& t1)
    {
        std::cout << t1 << ", " << t2 << '\n';
    }
};

int main()
{
    S s {3.14}; // T1 not provided
    std::function<void(int)> func2 = s; // T1 deduced by std function
    func2(42); // works
}

I just removed T1 and made it a template argument of operator() and everything works.

Live example .