Question

关于Is it possible to figure out the parameter type and return type of a lambda?的后续问题。我测试了代码并且它对我来说很好但只有当我从main()调用它时，当我从Do调用代码时代码无法编译时，如果你可以取消注释行，请告诉我在Do函数中编译代码

#include <tuple>
#include <iostream>
#include <typeinfo>
#include <cxxabi.h>
#include <memory>

template <class T>
std::string GetTypeName()
{
  // http://stackoverflow.com/questions/23266391/find-out-the-type-of-auto/23266701#23266701
  std::unique_ptr<char, void(*)(void*)> name{abi::__cxa_demangle(typeid(T).name(), 0, 0, nullptr), std::free};
  return name.get();
}

template <typename T>
struct function_traits
    : public function_traits<decltype(&T::operator())>
{};

template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const>
{
    enum { arity = sizeof...(Args) };

    typedef ReturnType result_type;

    template <size_t i>
    struct arg
    {
        typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
    };
};

template <typename FuncType>
void Do(FuncType func)
{
  typedef function_traits<decltype(func)> traits;
  //typename traits::arg<1>::type x = 0;
  std::cout << GetTypeName<traits>() << std::endl;
  //std::cout << GetTypeName<traits::arg<1>::type>() << std::endl;
}

int main()
{
  auto func = [] (int x, float y, double z) -> void { };
  Do(func);
  typedef function_traits<decltype(func)> traits;
  traits::arg<1>::type x = 0;
  std::cout << GetTypeName<traits>() << std::endl;
  std::cout << GetTypeName<traits::arg<1>::type>() << std::endl;
  return 0;
}

编译错误是

test.cpp: In function ‘void Do(FuncType)’:
test.cpp:37:20: error: non-template ‘arg’ used as template
   typename traits::arg<1>::type x = 0;
                    ^
test.cpp:37:20: note: use ‘traits::template arg’ to indicate that it is a template
test.cpp:37:20: error: declaration does not declare anything [-fpermissive]

我使用g++ (GCC) 5.3.1 20160406 (Red Hat 5.3.1-6)，编译命令为g++ -std=c++11 test.cpp -o test

在函数内部获取lambda参数类型无效

0 个答案: