How can I derive an output type for the template function?

问题

I am confused with my template-based multiplication function with two inputs and a type that needs to be derived. How should I derive a type of a function?

template<typename T, typename U>
DERIVED_TYPE multiply(T t, U u) {
    return t * u;
}

Well, I know that auto or decltype(auto) does the work quite well, but I would like the other way if I can, studying is one reason. For example, with Eigen,

DERIVED_TYPE multiply(Matrix<int, 2, 3> t, Matrix<double, 3, 4> u) {
    return t * u;
}

the DERIVED_TYPE should be Matrix<double, 2, 4> of course.

回答1:

In this particular case:

DERIVED_TYPE multiply(Matrix<int, 2, 3> t, Matrix<double, 3, 4> u) {
    return t * u;
}

you can use std::common_type to derive the common type of double and int:

Matrix< std::common_type<int,double> , 2 , 4>

In general:

template <typename T1, typename T2, size_t m,size_t n,size_t p>
Matrix< std::common_type<T1,T2>, m,p> multiply( Matrix<T1,m,n>, Matrix<T2,n,p>) {
    //...
}

回答2:

In C++11, you can use the operation itself to define the return type via the decltype specifier in conjunction with the auto specifier

template<typename T, typename U>
auto multiply(T const &t, U const &u) -> decltype(t * u)
{
    return t * u;
}

In C++14 you can omit the trailing return type in the function declaration

template<typename T, typename U>
auto multiply(T const &t, U const &u)
{
    return t * u;
}

In C++20 you can omit the template specifier and use the abbreviated function template syntax

auto multiply(auto const &t, auto const &u)
{
    return t * u;
}

Btw, I have made the arguments const & to avoid copies of large types.

来源：https://stackoverflow.com/questions/54635703/how-can-i-derive-an-output-type-for-the-template-function

标签

c++

templates

c++17