The return type of a lambda expression is automatically deduced. You don't have to use the auto keyword unless you specify a trailing-return-type. The trailing-return-type resembles the return-type part of an ordinary function or member function.
The type of a lambda expression is unspecified. But they are generally mere syntactic sugar for functors. A lambda is translated directly into a functor.
Funny, I've just written a function_traits
implementation based on Specializing a template on a lambda in C++0x which can give the parameter types. The trick, as described in the answer in that question, is to use the decltype
of the lambda's operator()
.
template <typename T>
struct function_traits
: public function_traits<decltype(&T::operator())>
{};
// For generic types, directly use the result of the signature of its '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.
};
};
// test code below:
int main()
{
auto lambda = [](int i) { return long(i*10); };
typedef function_traits<decltype(lambda)> traits;
static_assert(std::is_same<long, traits::result_type>::value, "err");
static_assert(std::is_same<int, traits::arg<0>::type>::value, "err");
return 0;
}
Note that this solution does not work for generic lambda like [](auto x) {}
.
Though I'm not sure this is strictly standard conforming, ideone compiled the following code:
template< class > struct mem_type;
template< class C, class T > struct mem_type< T C::* > {
typedef T type;
};
template< class T > struct lambda_func_type {
typedef typename mem_type< decltype( &T::operator() ) >::type type;
};
int main() {
auto l = [](int i) { return long(i); };
typedef lambda_func_type< decltype(l) >::type T;
static_assert( std::is_same< T, long( int )const >::value, "" );
}
However, this provides only the function type, so the result and parameter
types have to be extracted from it.
If you can use boost::function_traits
, result_type
and arg1_type
will meet the purpose.
Since ideone seems not to provide boost in C++11 mode, I couldn't post
the actual code, sorry.
The specialization method shown in @KennyTMs answer can be extended to cover all cases, including variadic and mutable lambdas:
template <typename T>
struct closure_traits : closure_traits<decltype(&T::operator())> {};
#define REM_CTOR(...) __VA_ARGS__
#define SPEC(cv, var, is_var) \
template <typename C, typename R, typename... Args> \
struct closure_traits<R (C::*) (Args... REM_CTOR var) cv> \
{ \
using arity = std::integral_constant<std::size_t, sizeof...(Args) >; \
using is_variadic = std::integral_constant<bool, is_var>; \
using is_const = std::is_const<int cv>; \
\
using result_type = R; \
\
template <std::size_t i> \
using arg = typename std::tuple_element<i, std::tuple<Args...>>::type; \
};
SPEC(const, (,...), 1)
SPEC(const, (), 0)
SPEC(, (,...), 1)
SPEC(, (), 0)
Demo.
Note that the arity is not adjusted for variadic operator()
s. Instead one can also consider is_variadic
.
The answer provided by @KennyTMs works great, however if a lambda has no parameters, using the index arg<0> does not compile. If anyone else was having this problem, I have a simple solution (simpler than using SFINAE related solutions, that is).
Just add void to the end of the tuple in the arg struct after the variadic argument types. i.e.
template <size_t i>
struct arg
{
typedef typename std::tuple_element<i, std::tuple<Args...,void>>::type type;
};
since the arity isn't dependent on the actual number of template parameters, the actual won't be incorrect, and if it's 0 then at least arg<0> will still exist and you can do with it what you will. If you already plan to not exceed the index arg<arity-1>
then it shouldn't interfere with your current implementation.
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