std::result_of 语法混乱

Question

I have the following (CUDA) function:

__device__ auto foo(my_class& x, my_function_ptr_type y ) {
    return [gen](my_class& x) { return x.set_y(y); };
}

And I want typedef its return value's type. I've fiddling with the std::result_of syntax, and can't get it quite right. This won't work:

using foo_return_type = std::result_of<decltype(foo(my_class{}, my_function_ptr_type{nullptr}))>::type;

Nor this:

using foo_return_type = std::result_of<decltype(foo(my_class, my_function_ptr_type))>::type;

Nor this:

using foo_return_type = std::result_of<foo>::type;

What should I have as the template-argument to std::result_of ?

Notes:

• There's only one foo() in the namespace.
• No templates are involved (other than std::result_of ...)
• C++11 or C++14, take your pick (but note that this is CUDA, so theoretically that could be an issue).
• Compiler: NVCC 10.1

Answer 1

You need a function pointer type and the parameter types, then combine them in the format of F(ArgTypes...) . eg

using foo_return_type = std::result_of<decltype(&foo)(my_class&, my_function_ptr_type)>::type;

LIVE

---

You can also make your own type trait, if you don't stick to std::result_of . eg

template <typename F>
struct return_type_of_function {};
template <typename R, typename... Args>
struct return_type_of_function<R(Args...)> {
    using type = R;
};

then

using foo_return_type = return_type_of_function<decltype(foo)>::type;

LIVE