我建议这个解决方案。它应该与 lambda 以及函数对象一起使用。它可以扩展以使其也适用于函数指针(只需通过底部提供的链接)
框架:
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) };
};
template<typename Functor, size_t NArgs>
struct count_arg : std::enable_if<function_traits<Functor>::arity==NArgs, int>
{};
Usage:
template<typename Functor>
typename count_arg<Functor, 1>::type Set(Functor f)
{
std::function<void(int)> fn = f;
std::cout << "f with one argument" << std::endl;
}
template<typename Functor>
typename count_arg<Functor, 2>::type Set(Functor f)
{
std::function<void(int, int)> fn = f;
std::cout << "f with two arguments" << std::endl;
}
int main() {
Set([](int a){});
Set([](int a, int b){});
return 0;
}
Output:
f with one argument
f with two arguments
我从该主题的已接受答案中获得了一些帮助:
- 是否可以计算出 lambda 的参数类型和返回类型? https://stackoverflow.com/questions/7943525/is-it-possible-to-figure-out-the-parameter-type-and-return-type-of-a-lambda
Visual Studio 2010 的解决方法
由于 Microsoft Visual Studio 2010 不支持可变参数模板,因此框架部分可以实现为:
template <typename T>
struct function_traits : public function_traits<decltype(&T::operator())>
{};
template <typename C, typename R, typename T0>
struct function_traits<R(C::*)(T0) const> { enum { arity = 1 }; };
template <typename C, typename R, typename T0, typename T1>
struct function_traits<R(C::*)(T0,T1) const> { enum { arity = 2 }; };
template <typename C, typename R, typename T0, typename T1, typename T2>
struct function_traits<R(C::*)(T0,T1,T2) const> { enum { arity = 3 }; };
//this is same as before
template<typename Functor, size_t NArgs, typename ReturnType=void>
struct count_arg : std::enable_if<function_traits<Functor>::arity==NArgs, ReturnType>
{};
EDIT
现在此代码支持任何返回类型。