c++ - 带有模板参数的 decltype

Question

test1.cpp 编译，但 test2.cpp 不编译。两者之间的唯一区别是我在 test1.cpp 中的类声明中定义了 Handle::add_it，但在 test2.cpp 中定义了类声明.

test1.cpp:g++ test1.cpp -o test1 -std=c++11

#include <iostream>

template<typename B>
class Handle
{
public:
        decltype(B.operator(int)) add_it(int x)
        {
                return b(x);
        }

        B b;
};

struct Add1
{
        int operator()(int x)
        {
                return x + 1;
        }
};

int main()
{
        Handle<Add1> h;
        std::cout << h.add_it(5) << std::endl;
}

test2.cpp:g++ test2.cpp -o test2 -std=c++11

#include <iostream>

template<typename B>
class Handle
{
public:
        decltype(B.operator(int)) add_it(int x);

        B b;
};

template<typename B>
decltype(B.operator(int)) Handle<B>::add_it(int x)
{
        return b(x);
}

struct Add1
{
        int operator()(int x)
        {
                return x + 1;
        }
};

int main()
{
        Handle<Add1> h;
        std::cout << h.add_it(5) << std::endl;
}

错误

test2.cpp:13:11: error: expected primary-expression before ‘.’ token
 decltype(B.operator(int))
           ^
test2.cpp:13:20: error: expected type-specifier before ‘(’ token
 decltype(B.operator(int))
                    ^
test2.cpp:13:21: error: expected primary-expression before ‘int’
 decltype(B.operator(int))

Answer 1

您可以使用 std::declval 修改它:

template<typename B>
class Handle
{
public:
  decltype(std::declval<B>()(int())) add_it(int x) {
    return b(x);
  }

  B b;
};

Live Demo

或者在类的定义之外:

template<typename B>
class Handle {
public:
  decltype(std::declval<B>()(int())) add_it(int x);
  B b;
};

template<typename B>
decltype(std::declval<B>()(int())) Handle<B>::add_it(int x) {
  return b(x);
}

Live Demo