部分专门化lambda的类模板

Question

I have template type that store information about function or member function (like return type, number or parameters and so on).

template<class R, class... FuncParams>
struct SFuncInfo
{
    using Signature = R(FuncParams...);
    using Ret = R;

    static constexpr size_t numParams = sizeof...(FuncParams);
};

// member
template<class T, class Ret, class... Params>
struct SFuncInfo<Ret(T::*)(Params...)> : SFuncInfo<Ret, Params...> 
{
    static constexpr bool isMemberFunction = true;
};

// function
template<class R, class... FuncParams>
struct SFuncInfo<R(FuncParams...)> : SFuncInfo<R, FuncParams...> 
{
    static constexpr bool isMemberFunction = false;
};

This is how it can be used:

int func(const char* str) { return 1; }

struct MyType
{
    bool memFunc(int val, float fl) { return true; }
};

int main() 
{
    static_assert(!SFuncInfo<decltype(func)>::isMemberFunction, "");
    static_assert(std::is_same<SFuncInfo<decltype(func)>::Ret, int>::value, "");

    static_assert(SFuncInfo<decltype(&MyType::memFunc)>::isMemberFunction, "");
    static_assert(std::is_same<SFuncInfo<decltype(&MyType::memFunc)>::Ret, bool>::value, "");
}

This code compiles. But I want also to handle cases with lambdas. Something like this:

auto lambda = [](int, bool) -> float { return 3.14f; };

static_assert(SFuncInfo<decltype(lambda)>::isMemberFunction, "");
static_assert(std::is_same<SFuncInfo<decltype(lambda)>::Ret, float>::value, "");

I tried different option. Few of the are listed below.

template<class T>
struct SFuncInfo<T, decltype(T())>
{
    static constexpr bool isMemberFunction = true;
};

template<class T>
struct SFuncInfo<T, decltype(&std::decay<decltype(std::declval<T>())>::type::operator())>
{
    static constexpr bool isMemberFunction = true;
};

It doesn't resolve to any of these specialization.

By the way, the code below also compiles:

auto lambda = [](int, bool) -> float { return 3.14f; };

using LambdaType = std::decay<decltype(std::declval<decltype(lambda)>())>::type;
using CallOperator = decltype(&LambdaType::operator());
static_assert(std::is_same<SFuncInfo<CallOperator>::Ret, float>::value, "");
static_assert(SFuncInfo<CallOperator>::isMemberFunction, "");

Here is LIVE DEMO is someone wants to play around.

Does anyone has a good solution for this?

Answer 1

Adding this single partial specialization works on my side:

template<class Lambda>
struct SFuncInfo<Lambda> : SFuncInfo<decltype(&Lambda::operator())> { };

Answer 2

The solution would be to make an overload that is only available to callable object types and inherit from SFuncInfo with the operator() type.

template<typename T>
struct SFuncInfo<T, decltype(void(&T::operator()))> : SFuncInfo<decltype(&T::operator())> {};
//     constraint ----------------^

However to support this I separated the specializations and the metadata classes, splitting them into SFuncInfo and SFuncInfoBase :

template<class R, class... FuncParams>
struct SFuncInfoBase
{
    using Signature = R(FuncParams...);
    using Ret = R;

    static constexpr size_t numParams = sizeof...(FuncParams);
};

template<class T, typename = void>
struct SFuncInfo;

// member
template<class T, class Ret, class... Params>
struct SFuncInfo<Ret(T::*)(Params...)const> : SFuncInfo<Ret(T::*)(Params...)> {};

template<class T, class Ret, class... Params>
struct SFuncInfo<Ret(T::*)(Params...)> : SFuncInfoBase<Ret, Params...> 
{
    static constexpr bool isMemberFunction = true;
};

Live example