Inspired by another question I tried to find a way to deduce the type
of an overload member function given the actual argument used to call
that function. Here is what I have so far:
#include <type_traits>
template<typename F, typename Arg>
struct mem_fun_type {
// perform overload resolution here
typedef decltype(std::declval<F>()(std::declval<Arg>())) result_type;
typedef decltype(static_cast<result_type (F::*)(Arg)>(&F::operator())) type;
};
struct foo {};
struct takes_two
{
void operator()(int);
void operator()(foo);
};
struct take_one {
void operator()(float);
};
int main()
{
static_assert(std::is_same<mem_fun_type<take_one, float>::type,
void (take_one::*)(float)>::value, "Zonk");
static_assert(std::is_same<mem_fun_type<takes_two, double>::type,
void (takes_two::*)(float)>::value, "Zonk");
return 0;
}
As long as the template parameter Arg matches the actual type the
static_cast will succeed, but this is only the most trivial case of
overload resolution (exact match). Is it possible to perform the
complete overload resolution process in template metaprogramming?
This is purely hypothetical and not intended for real-world use.
This is closest I came to it so far: define function returning tables of different sizes and your result is sizeof(select(...)) receiving pointer to function you want to match. To ensure that the code will compile even if function does not exist in given class, you can use separate check has_function.
Result of overload resolution is in select<has_function<T>::value, T>::value.
With this code you can even "resolve" data members, not just functions, it's only a question of making right parameter for select function.
However there is one deficiency here - overload resolution is not on function parameters, but on function type. Meaning none of usual parameter type conversions takes place.
// Verify the name is valid
template <typename T>
struct has_function
{
struct F {int function;};
struct D : T, F {};
template <typename U, U> struct same_;
template <typename C> static char(&select_(same_<int F::*, &C::function>*))[1];
template <typename> static char(&select_(...))[2];
enum {value = sizeof(select_<D>(0)) == 2};
};
// Values to report overload results
enum type { none=1 , function_sz_size_t , function_sz , function_string };
template <bool, typename R> struct select;
template <typename R> struct select<false, R>
{
enum {value = none};
};
template <typename R> struct select<true, R>
{
// Define your overloads here, they don't have to be templates.
template <typename Ret, typename Arg> static char(&select_(Ret (R::*)(const char*, Arg)))[function_sz_size_t];
template <typename Ret, typename Arg> static char(&select_(Ret (R::*)(Arg)))[function_sz];
template <typename Ret> static char(&select_(Ret (R::*)(std::string)))[function_string];
template <typename Ret> static char(&select_(Ret (R::*)(std::string&&)))[function_string];
template <typename Ret> static char(&select_(Ret (R::*)(const std::string&)))[function_string];
static char(&select_(...))[none];
enum {value = sizeof(select_(&R::function))};
};
Related
Recently I designed meta-types and the possible operations that would allow compile-time type concatenations:
#include <tuple>
template<template<typename...> typename T>
struct MetaTypeTag
{};
/*variable template helper*/
template<template<typename...> typename T>
constexpr MetaTypeTag<T> meta_type_tag = {};
template<typename T>
struct TypeTag
{};
/*comparison*/
template<typename T>
constexpr bool operator==(TypeTag<T>, TypeTag<T>) { return true; }
template<typename T, typename U>
constexpr bool operator==(TypeTag<T>, TypeTag<U>) { return false; }
/*variable template helper*/
template<typename T>
constexpr TypeTag<T> type_tag = {};
template<template<typename...> typename T, typename... Ts>
constexpr TypeTag<T<Ts...>> combine(MetaTypeTag<T>, TypeTag<Ts>...)
{
return {};
}
int main()
{
constexpr auto combined_tag = combine(meta_type_tag<std::tuple>, type_tag<int>, type_tag<float>);
static_assert(combined_tag == type_tag<std::tuple<int, float>>, "");
}
The std::tuple without template arguments cannot be used as a type, but may still appear in the template template parameter.
Now if we try to go one step further, the question is whether there is any way to unify struct MetaTypeTag and struct TypeTag, since they are both empty classes with one template parameter, or at least it could be possible to use the same variable template type_tag but redirect to a different class depending on the type category? So I would imagine something like this:
template<???>
constexpr auto type_tag = ????{};
//use with 'incomplete type'
type_tag<std::tuple> //MetaTypeTag<std::tuple>
//use with regular type
type_tag<int> //TypeTag<int>
I tried all possible ways - redefinition, explicit specialization, partial specialization, optional template parameters, conditional using alias, but none worked. I had hoped C++17's template<auto> would help, but it turns out that one is for non-type only.
the question is whether there is any way to unify struct MetaTypeTag and struct TypeTag, since they are both empty classes with one template parameter
I don't thinks so.
The best I can imagine to simplify a little (very a little) your code is define a couple of overloaded constexpr function, say getTag()
template <typename T>
auto constexpr getTag ()
{ return TypeTag<T>{}; }
template <template <typename ...> typename T>
auto constexpr getTag ()
{ return MetaTypeTag<T>{}; }
so you can call getTag<T>() where T is either a type or a template.
So you can call combine() as follows
constexpr auto combined_tag
= combine(getTag<std::tuple>(), getTag<int>(), getTag<float>());
But I don't think is a great improvement.
I want to create a templated class or function, that receives a lambda, and puts it internally in std::function<>
Lambda could have any number of input parameters [](int a, float b, ...) std::function<> should correspond to the lambda's operator()'s type
template <typename T>
void getLambda(T t) {
// typedef lambda_traits::ret_type RetType; ??
// typedef lambda_traits::param_tuple --> somehow back to parameter pack Args...
std::function<RetType(Args...)> fun(t);
}
int main() {
int x = 0;
getLambda([&x](int a, float b, Person c){});
}
So I need to somehow extract the Return Type and Parameter Pack
Answer here suggests to use partial spec on lambda's :: operator()
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>
// we specialize for pointers to member function
{
enum { arity = sizeof...(Args) };
// arity is the number of arguments.
typedef ReturnType result_type;
template <size_t i>
struct arg
{
typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
// the i-th argument is equivalent to the i-th tuple element of a tuple
// composed of those arguments.
};
};
But I need a way to convert tuple<> back to parameters pack, to create a proper std::function<> instantiation
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>
// we specialize for pointers to member function
{
using result_type = ReturnType;
using arg_tuple = std::tuple<Args...>;
static constexpr auto arity = sizeof...(Args);
};
template <class F, std::size_t ... Is, class T>
auto lambda_to_func_impl(F f, std::index_sequence<Is...>, T) {
return std::function<typename T::result_type(std::tuple_element_t<Is, typename T::arg_tuple>...)>(f);
}
template <class F>
auto lambda_to_func(F f) {
using traits = function_traits<F>;
return lambda_to_func_impl(f, std::make_index_sequence<traits::arity>{}, traits{});
}
The code above should do what you want. The main idea, as you can see, is to create an integer pack. This is the non-type template equivalent of variadics. I don't know of any technique by which you can use such a pack without calling another function, so typically in these situations with tuples you'll see a nested "impl" function that does all the work. Once you have the integer pack, you expand it while accessing the tuple (works for getting the values too).
On a stylistic note: use using, not typename, especially in template heavy code as the former can alias templates too. And don't use that enum trick to store a static value without it using space; compilers will optimize this out anyhow and just using a static constexpr integer is much clearer.
How can I deduce statically if an argument is a C++ function object (functor)?
template <typename F>
void test(F f) {}
I tried is_function<F>::value, but this doesn't work. It also seems there is no is_functor trait, so perhaps it's not possible. I appear to be only looking for a specific member function, in this case the function call operator: F::operator().
It is possible to create such a trait, with two restrictions:
For the compiler, a free function is something fundamentally different from a class functor that overloads operator(). Thus we have to treat both cases seperately when implementing. This is not a problem for usage though, we can hide this implementation detail from the user.
We need to know the signature of the function you want to call. This is usually not a problem, and it does have the nice side effect that our trait is able to handle overloads pretty natively.
Step one: Free functions
Let's start with free functions, because they are little easier to detect. Our task is, when given a function pointer, to determine whether the signature of that function pointer matches the signature passed as the second template argument. To be able to compare those, we either need to get a grasp of the underlying function signature, or create a function pointer of our signature. I arbitrarily chose the latter:
// build R (*)(Args...) from R (Args...)
// compile error if signature is not a valid function signature
template <typename, typename>
struct build_free_function;
template <typename F, typename R, typename ... Args>
struct build_free_function<F, R (Args...)>
{ using type = R (*)(Args...); };
Now all that's left to do is to compare and we are done with the free function part:
// determine whether a free function pointer F has signature S
template <typename F, typename S>
struct is_function_with_signature
{
// check whether F and the function pointer of S are of the same
// type
static bool constexpr value = std::is_same<
F, typename build_free_function<F, S>::type
>::value;
};
Step two: Class functors
This one is a little more involved. We could easily detect with SFINAE whether a class defines an operator():
template <typename T>
struct defines_functor_operator
{
typedef char (& yes)[1];
typedef char (& no)[2];
// we need a template here to enable SFINAE
template <typename U>
static yes deduce(char (*)[sizeof(&U::operator())]);
// fallback
template <typename> static no deduce(...);
static bool constexpr value = sizeof(deduce<T>(0)) == sizeof(yes);
};
but that does not tell us whether one exists for our desired function signature! Luckily, we can use a trick here: pointers are valid template parameters. Thus we can first use the member function pointer of our desired signature, and check whether &T::operator() is of that type:
template <typename T, T> struct check;
Now check<void (C::*)() const, &C::operator()> will only be a valid template instantiation if C does indeed have a void C::operator()() const. But to do this we first have to combine C and the signature to a member function pointer. As we already have seen, we need to worry about two extra cases we did not have to care about for free functions: const and volatile functions. Besides that it's pretty much the same:
// build R (C::*)(Args...) from R (Args...)
// R (C::*)(Args...) const from R (Args...) const
// R (C::*)(Args...) volatile from R (Args...) volatile
// compile error if signature is not a valid member function signature
template <typename, typename>
struct build_class_function;
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...)>
{ using type = R (C::*)(Args...); };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) const>
{ using type = R (C::*)(Args...) const; };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) volatile>
{ using type = R (C::*)(Args...) volatile; };
Putting that and our findings concerning the check helper struct together, we get our check metafunction for functor objects:
// determine whether a class C has an operator() with signature S
template <typename C, typename S>
struct is_functor_with_signature
{
typedef char (& yes)[1];
typedef char (& no)[2];
// helper struct to determine that C::operator() does indeed have
// the desired signature; &C::operator() is only of type
// R (C::*)(Args...) if this is true
template <typename T, T> struct check;
// T is needed to enable SFINAE
template <typename T> static yes deduce(check<
typename build_class_function<C, S>::type, &T::operator()> *);
// fallback if check helper could not be built
template <typename> static no deduce(...);
static bool constexpr value = sizeof(deduce<C>(0)) == sizeof(yes);
};
Step three: Putting the pieces together
We are almost done. Now we only need to decide when to use our free function, and when the class functor metafunctions. Luckily, C++11 provides us with a std::is_class trait that we can use for this. So all we have to do is specialize on a boolean parameter:
// C is a class, delegate to is_functor_with_signature
template <typename C, typename S, bool>
struct is_callable_impl
: std::integral_constant<
bool, is_functor_with_signature<C, S>::value
>
{};
// F is not a class, delegate to is_function_with_signature
template <typename F, typename S>
struct is_callable_impl<F, S, false>
: std::integral_constant<
bool, is_function_with_signature<F, S>::value
>
{};
So we can finally add the last piece of the puzzle, being our actual is_callable trait:
// Determine whether type Callable is callable with signature Signature.
// Compliant with functors, i.e. classes that declare operator(); and free
// function pointers: R (*)(Args...), but not R (Args...)!
template <typename Callable, typename Signature>
struct is_callable
: is_callable_impl<
Callable, Signature,
std::is_class<Callable>::value
>
{};
Now we clean up our code, put implementation details into anonymous namespaces so they are not acessible outside of our file, and have a nice is_callable.hpp to use in our project.
Full code
namespace // implementation detail
{
// build R (*)(Args...) from R (Args...)
// compile error if signature is not a valid function signature
template <typename, typename>
struct build_free_function;
template <typename F, typename R, typename ... Args>
struct build_free_function<F, R (Args...)>
{ using type = R (*)(Args...); };
// build R (C::*)(Args...) from R (Args...)
// R (C::*)(Args...) const from R (Args...) const
// R (C::*)(Args...) volatile from R (Args...) volatile
// compile error if signature is not a valid member function signature
template <typename, typename>
struct build_class_function;
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...)>
{ using type = R (C::*)(Args...); };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) const>
{ using type = R (C::*)(Args...) const; };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) volatile>
{ using type = R (C::*)(Args...) volatile; };
// determine whether a class C has an operator() with signature S
template <typename C, typename S>
struct is_functor_with_signature
{
typedef char (& yes)[1];
typedef char (& no)[2];
// helper struct to determine that C::operator() does indeed have
// the desired signature; &C::operator() is only of type
// R (C::*)(Args...) if this is true
template <typename T, T> struct check;
// T is needed to enable SFINAE
template <typename T> static yes deduce(check<
typename build_class_function<C, S>::type, &T::operator()> *);
// fallback if check helper could not be built
template <typename> static no deduce(...);
static bool constexpr value = sizeof(deduce<C>(0)) == sizeof(yes);
};
// determine whether a free function pointer F has signature S
template <typename F, typename S>
struct is_function_with_signature
{
// check whether F and the function pointer of S are of the same
// type
static bool constexpr value = std::is_same<
F, typename build_free_function<F, S>::type
>::value;
};
// C is a class, delegate to is_functor_with_signature
template <typename C, typename S, bool>
struct is_callable_impl
: std::integral_constant<
bool, is_functor_with_signature<C, S>::value
>
{};
// F is not a class, delegate to is_function_with_signature
template <typename F, typename S>
struct is_callable_impl<F, S, false>
: std::integral_constant<
bool, is_function_with_signature<F, S>::value
>
{};
}
// Determine whether type Callable is callable with signature Signature.
// Compliant with functors, i.e. classes that declare operator(); and free
// function pointers: R (*)(Args...), but not R (Args...)!
template <typename Callable, typename Signature>
struct is_callable
: is_callable_impl<
Callable, Signature,
std::is_class<Callable>::value
>
{};
Ideone example with some tests
http://ideone.com/7PWdiv
Although this does not work for overloaded functions, for all other cases (free functions, classes implementing operator(), and lambdas) this short solutions works in C++11:
template <typename T, typename Signature>
struct is_callable: std::is_convertible<T,std::function<Signature>> { };
Note: std::is_invocable is available since C++17.
You can use the following concept in c++20
template<typename F>
concept FunctionObject = requires (F) {
&F::operator();
};
Is an is_functor C++ trait class possible?
Yes, it is possible to manually implement a validation for functors.
I tried is_function::value, but this doesn't work.
You are on the right path, it is possible to implement using std::function
Remember that std::function also accepts functions, function pointers and functor instances in its constructor.
Example:
struct Test {
public:
bool operator()(int){
return true;
}
};
void test(int){
}
void example(std::function<void(int)> ex){
cout << "Called" << endl;
};
int main()
{
example(test);
example(&test);
example(Test{});
}
That said, the logic that will be used to validate if a class has a function call overload operator (functor) is similar to the code above.
That is, if the std::function<void(int)> accepts an instance of the class Test{} means the class has a functor, otherwise it doesn't.
Example of an possible solution
Here is the source code:
//F: Test class
//Args...: The arguments, ex: int or none
template <typename F, typename... Args>
struct is_functor :
is_constructible <
function<void(Args ...)>, F
>
{};
Example usage:
is_functor<Test, int> -> True result
is_functor -> False result
Info about std::is_constructible
Is a trait class that identifies whether T is a constructible type with the set of argument types specified by Arg.
For this class, a constructible type is a type that can be constructed using a particular set of arguments.
is_constructible inherits from integral_constant as being either true_type or false_type, depending on whether T is constructible with the list of arguments Args.
In short, it checks if a given class has a constructor, for example:
struct Test2 {
Test(bool, int){}
};
std::is_constructible<Test2, bool, int> -> True result
std::is_constructible<Test2, float> -> False result
Example of implementation:
template <typename, typename, typename ...Args>
struct is_constructible_impl : false_type {};
template <typename T, typename ...Args>
struct is_constructible_impl <
void_t<decltype(T(std::declval<Args>()...))>, T, Args...
> : true_type {};
template <typename T, typename ...Args>
struct is_constructible : is_constructible_impl<void_t<>, T, Args...> {};
Final explanation
In the implementation of is_functor it was checked if std::function<void(int)> accepts an instance of Test{}, which is true.
References:
How is std::is_constructible<T, Args> implemented?
Can std::is_invocable be emulated within C++11?
https://replit.com/#LUCASP6/RowdyAlphanumericCode#main.cpp
template<typename T, typename Sign>
struct is_functor
{
typedef char yes[1];
typedef char no [2];
template <typename U, U> struct type_check;
template <typename _1> static yes &chk(type_check<Sign, &_1::operator()>*);
template <typename > static no &chk(...);
static bool const value = sizeof(chk<T>(nullptr)) == sizeof(yes);
};
Altered from this answer.
It could be used like...
template<typename T>
typename std::enable_if<is_functor<T, void(T::*)()>::value>::type func()
{
}
The following code shows an SFINAE implementation to check whether a type (basically a class) contains a member function member_func at compile time.
#define CHECKER(func_name,class_name) sizeof(class_name<T>::template func_name<T>(0)) == 1
#include <iostream>
struct A
{
void member_func();
};
struct B {};
template<typename T>struct Check_If_T_Is_Class_Type
{
template<typename C> static char func (char C::*p);
template<typename C> static long func (...);
enum{val = CHECKER(func,Check_If_T_Is_Class_Type)};
};
//APPROACH 1
template <typename T>struct TypeHasMemberFunc
{
template <typename C, C> struct TypeCheck;
template <typename C> struct Prototype_Holder {typedef void (C::*fptr)();};
template <typename C> static char func(TypeCheck
<
typename Prototype_Holder<C>::fptr,
&C::member_func
>*);
template <typename C> static long func(...);
enum {value = CHECKER(func,TypeHasMemberFunc)};
};
//APPROACH 2
template <typename T>struct has_member_func
{
template<class C> static char func(char (*)[sizeof(&C::member_func)]);
template<class C> static long func(...);
enum{value = CHECKER(func,has_member_func)};
};
int main(){
if(Check_If_T_Is_Class_Type<A>::val)
std::cout<<TypeHasMemberFunc<A>::value; //using APPROACH 1
if(Check_If_T_Is_Class_Type<B>::val)
std::cout<<has_member_func<B>::value; //using APPROACH 2
}
However my question is which approach would you prefer (APPROACH 1 or APPROACH 2) and why?
Do you find any inconsistency in the given approaches? If yes please let me know.
P.S : Assume sizeof(char)!= sizeof(long)
Second approach doesn't check function type (return type or arguments types) and does work with all types, not only class types.
I would personally prefer the second approach to play with as it is shorter and easier to understand. But GCC won't compile it, so you have to use something like that for GCC:
namespace detail
{
template<class C> char func(char (*)[sizeof(&C::member_func)]);
template<class C> long func(...);
}
template <typename T>struct has_member_func
{
enum{value = (sizeof(detail::func<T>(0)) == 1)};
};
Also, It would be nice to get rid of CHECKER macro. It makes your code extremely less readable.
Anyway, I would refrain from using such C++ hacks in production code (except you are a Boost team member :-)
Such stuff is error-prone, hard to support, hardly portable between compilers but the principal point is that I do not remember any real life task demanded such hard-code C++.
EDIT: completed and corrected.
Another approach, using ambiguity from inheritance, probably functionally equivalent to your Approach 2. I remember having problems with Approach 1 (it compiles with G++ 4.4.5 though) because the name resolution triggered an error and not a substitution failure. I had to resort to:
template <class T>
struct has_foo
{
struct fallback { void foo(...); };
struct D : T, fallback { };
template <typename U, U> struct K;
// Will be ambiguous for U = D iff T has a foo member function.
// It doesn't trigger an error, since D will always have at least one
// foo member function.
template <class U> static char (&test(K<void (fallback::*)(...), &U::foo>*))[1];
template <class U> static char (&test(...))[2];
static const bool value = sizeof(test<D>(0)) == 2;
};
This works when T is a class, so you may want to add your layer for checking whether T is a class type.
Note that any foo member function will be detected. If you want to check whether the detected foo function can be called with given arguments, you have to do another layer of SFINAE:
// Check whether foo can be called with an argument of type Arg
// and yields an element of type Res.
// If you need Res = void, this code does not work.
template <class T, typename Arg, typename Res>
struct check_foo
{
struct flag {};
struct D : T { using T::foo; flag foo(...); };
template <typename U>
static char (&test(U))[1];
template <typename> static char (&test(...))[2];
static Arg f();
static const bool value = sizeof(test<Arg>( ((D*)0)->foo(f()) )) == 1;
};
I've written a traits class that lets me extract information about the arguments and type of a function or function object in C++0x (tested with gcc 4.5.0). The general case handles function objects:
template <typename F>
struct function_traits {
template <typename R, typename... A>
struct _internal { };
template <typename R, typename... A>
struct _internal<R (F::*)(A...)> {
// ...
};
typedef typename _internal<decltype(&F::operator())>::<<nested types go here>>;
};
Then I have a specialization for plain functions at global scope:
template <typename R, typename... A>
struct function_traits<R (*)(A...)> {
// ...
};
This works fine, I can pass a function into the template or a function object and it works properly:
template <typename F>
void foo(F f) {
typename function_traits<F>::whatever ...;
}
int f(int x) { ... }
foo(f);
What if, instead of passing a function or function object into foo, I want to pass a lambda expression?
foo([](int x) { ... });
The problem here is that neither specialization of function_traits<> applies. The C++0x draft says that the type of the expression is a "unique, unnamed, non-union class type". Demangling the result of calling typeid(...).name() on the expression gives me what appears to be gcc's internal naming convention for the lambda, main::{lambda(int)#1}, not something that syntactically represents a C++ typename.
In short, is there anything I can put into the template here:
template <typename R, typename... A>
struct function_traits<????> { ... }
that will allow this traits class to accept a lambda expression?
I think it is possible to specialize traits for lambdas and do pattern matching on the signature of the unnamed functor. Here is the code that works on g++ 4.5. Although it works, the pattern matching on lambda appears to be working contrary to the intuition. I've comments inline.
struct X
{
float operator () (float i) { return i*2; }
// If the following is enabled, program fails to compile
// mostly because of ambiguity reasons.
//double operator () (float i, double d) { return d*f; }
};
template <typename T>
struct function_traits // matches when T=X or T=lambda
// As expected, lambda creates a "unique, unnamed, non-union class type"
// so it matches here
{
// Here is what you are looking for. The type of the member operator()
// of the lambda is taken and mapped again on function_traits.
typedef typename function_traits<decltype(&T::operator())>::return_type return_type;
};
// matches for X::operator() but not of lambda::operator()
template <typename R, typename C, typename... A>
struct function_traits<R (C::*)(A...)>
{
typedef R return_type;
};
// I initially thought the above defined member function specialization of
// the trait will match lambdas::operator() because a lambda is a functor.
// It does not, however. Instead, it matches the one below.
// I wonder why? implementation defined?
template <typename R, typename... A>
struct function_traits<R (*)(A...)> // matches for lambda::operator()
{
typedef R return_type;
};
template <typename F>
typename function_traits<F>::return_type
foo(F f)
{
return f(10);
}
template <typename F>
typename function_traits<F>::return_type
bar(F f)
{
return f(5.0f, 100, 0.34);
}
int f(int x) { return x + x; }
int main(void)
{
foo(f);
foo(X());
bar([](float f, int l, double d){ return f+l+d; });
}
The void_t trick can help. How does `void_t` work?
Unless you have C++17, you'll need to include the definition of void_t:
template<typename... Ts> struct make_void { typedef void type;};
template<typename... Ts> using void_t = typename make_void<Ts...>::type;
Add an extra template argument to the original template, defaulted to void:
template <typename T, typename = void>
struct function_traits;
The traits object for simple functions is the same as you already have:
template <typename R, typename... A>
struct function_traits<R (*)(A...)>
{
using return_type = R;
using class_type = void;
using args_type = std:: tuple< A... >;
};
For non-const methods:
template <typename R, typename... A>
struct function_traits<R (C::*)(A...)>
{
using return_type = R;
using class_type = void;
using args_type = std:: tuple< A... >;
};
Don't forget const methods:
template <typename R, typename C, typename... A>
struct function_traits<R (C::*)(A...) const> // const
{
using return_type = R;
using class_type = C;
using args_type = std:: tuple< A... >;
};
Finally, the important trait. Given a class type, including lambda types, we want to forward from T to decltype(&T::operator()). We want to ensure that this trait is only available for types T for which ::operator() is available, and this is what void_t does for us. To enforce this constraint, we need to put &T::operator() into the trait signature somewhere, hence template <typename T> struct function_traits<T, void_t< decltype(&T::operator())
template <typename T>
struct function_traits<T, void_t< decltype(&T::operator()) > >
: public function_traits< decltype(&T::operator()) >
{
};
The operator() method in (non-mutable, non-generic) lambdas is const, which explains why we need the const template above.
But ultimately this is very restrictive. This won't work with generic lambdas, or objects with templated operator(). If you reconsider your design, you find find a different approach that is more flexible.
By delegating some of the work to a series of function templates instead of a class template, you can extract the relevant info.
First though, I should say that the relevant method is a const method, for a lambda (for a non-capturing, non-generic, non-mutable lambda). So you will not be able to tell the difference between a true lambda and this:
struct {
int operator() (int) const { return 7; }
} object_of_unnamed_name_and_with_suitable_method;
Therefore, I must assume that you don't want "special treatment" for lambdas, and you don't want to test if a type is a lambda type, and that instead you want to simply extract the return type, and the type of all arguments, for any object which is simple enough. By "simple enough" I mean, for example, that the operator() method is not itself a template. And, for bonus information, a boolean to tell us if an operator() method was present and used, as opposed to a plain old function.
// First, a convenient struct in which to store all the results:
template<bool is_method_, bool is_const_method_, typename C, typename R, typename ...Args>
struct function_traits_results {
constexpr static bool is_method = is_method_;
constexpr static bool is_const_method = is_const_method_;
typedef C class_type; // void for plain functions. Otherwise,
// the functor/lambda type
typedef R return_type;
typedef tuple<Args...> args_type_as_tuple;
};
// This will extract all the details from a method-signature:
template<typename>
struct intermediate_step;
template<typename R, typename C, typename ...Args>
struct intermediate_step<R (C::*) (Args...)> // non-const methods
: public function_traits_results<true, false, C, R, Args...>
{
};
template<typename R, typename C, typename ...Args>
struct intermediate_step<R (C::*) (Args...) const> // const methods
: public function_traits_results<true, true, C, R, Args...>
{
};
// These next two overloads do the initial task of separating
// plain function pointers for functors with ::operator()
template<typename R, typename ...Args>
function_traits_results<false, false, void, R, Args...>
function_traits_helper(R (*) (Args...) );
template<typename F, typename ..., typename MemberType = decltype(&F::operator()) >
intermediate_step<MemberType>
function_traits_helper(F);
// Finally, the actual `function_traits` struct, that delegates
// everything to the helper
template <typename T>
struct function_traits : public decltype(function_traits_helper( declval<T>() ) )
{
};