I know one could check the existence of a particular method using expression SFINAE in C++11 as follows.
What I can't find though, is an example to do the same, checking method arguments as well. In particular I would like to match a method that takes a const parameter.
#include <iostream>
struct A
{
void method() const
{
return;
}
};
template <typename T, typename = std::string>
struct hasMethod
: std::false_type
{
};
template <typename T>
struct hasMethod<T, decltype(std::declval<T>().method())>
: std::true_type
{ };
int main() {
std::cout << hasMethod<A>::value << std::endl;
}
In reality I would like the hasMethod:: to match
void method(const Type& t) const
{
return;
}
What is the syntax to pass to decltype?
I have tried:
struct hasMethod<T, decltype(std::declval<T>().method(const int&))>
: std::true_type
but it obviously doesn't work.
Related
I'm aware that my code should work in c++14 but I have to replicate this behaviour in c++11, I havent been able to make a equivalent init() could anyone help?
enum MyEnum {
BANANA, APPLE, PINEAPPLE, ORANGE, FRUIT_AMOUNT
};
template<MyEnum>
struct Fruit{
virtual ~Fruit(){}
virtual void dostuff(){}
};
template <>
struct Fruit<ORANGE>{
void dostuff(){ cout<<"Hey apple!"<<endl;
}
constexpr array< Fruit*, FRUIT_AMOUNT > init(){
array< Fruit*, FRUIT_AMOUNT > myArray;
for(int i =0; i < FRUIT_AMOUNT; i++)
myArray[i] = new Fruit< (MyEnum) i >();
return myArray;
}
array<Fruit*, FRUIT_AMOUNT> myPrettyFruits = init();
I'm aware that my code should work in c++14
Ehmm... not at all.
Your code has a lot of problems.
Some of them, in no particular order
(1) You can't write
array<Fruit*, FRUIT_AMOUNT>
because Fruit isn't a type; it's a template class.
So, by example, Fruit<BANANA> is a type, and you can write
std::array<Fruit<BANANA> *, FRUIT_AMOUNT>
but you can't have a pointer to Fruit because Fruit (without explicating a template argument) isn't a type.
A possible solution for this problem is make all Fruit types inheriting from a common base; by example
struct FruitBase
{ };
template <MyEnum>
struct Fruit : public FruitBase
{
virtual ~Fruit () {}
virtual void dostuff () {}
};
This way you can have an array of FruitBase pointers
std::array<FruitBase *, FRUIT_AMOUNT>
So you can put in the array pointer to Fruit<Something> types that are also FruitBase pointers.
(2) You can't have
Fruit< (MyEnum) i >
where i is a run-time known variable, because a template argument must be known compile time.
A possible C++14 solution is use std::make_index_sequence to get a sequence of template (so compile-time known) std::size_t values.
I suggest something as follows
template <std::size_t ... Is>
constexpr std::array<FruitBase *, FRUIT_AMOUNT>
init_helper (std::index_sequence<Is...> const &)
{ return { { (FruitBase*)(new Fruit<(MyEnum)Is>()) ... } }; }
constexpr std::array<FruitBase *, FRUIT_AMOUNT> init ()
{ return init_helper(std::make_index_sequence<FRUIT_AMOUNT>{}); }
Observe that both functions are a single return statement; and this is necessary for a constexpr C++11 function.
Unfortunately std::index_sequence and std::make_index_sequence are available only starting from C++14. But it's possible make a C++11 substitute for they.
(3) new Something{} can't be executed compile-time
So you can define init_helper() constexpr but it's a fake constexpr function (so also init() is a fake constexpr function) because can't be executed compile time.
So you can write
std::array<FruitBase *, FRUIT_AMOUNT> myPrettyFruits = init();
but myPrettyFruits is initialized run-time.
If you try to initialize it compile-time
constexpr std::array<FruitBase *, FRUIT_AMOUNT> myPrettyFruits = init();
you get a compilation error.
The following is a full compiling C++11 example, with a std::index_sequence/std::make_index_sequence replacement, that works only run-time
#include <array>
#include <iostream>
template <std::size_t...>
struct indexSequence
{ using type = indexSequence; };
template <typename, typename>
struct concatSequences;
template <std::size_t... S1, std::size_t... S2>
struct concatSequences<indexSequence<S1...>, indexSequence<S2...>>
: public indexSequence<S1..., ( sizeof...(S1) + S2 )...>
{ };
template <std::size_t N>
struct makeIndexSequenceH
: public concatSequences<
typename makeIndexSequenceH<(N>>1)>::type,
typename makeIndexSequenceH<N-(N>>1)>::type>::type
{ };
template<>
struct makeIndexSequenceH<0> : public indexSequence<>
{ };
template<>
struct makeIndexSequenceH<1> : public indexSequence<0>
{ };
template <std::size_t N>
using makeIndexSequence = typename makeIndexSequenceH<N>::type;
enum MyEnum
{ BANANA, APPLE, PINEAPPLE, ORANGE, FRUIT_AMOUNT };
struct FruitBase
{ };
template <MyEnum>
struct Fruit : public FruitBase
{
virtual ~Fruit () {}
virtual void dostuff () {}
};
template <>
struct Fruit<ORANGE> : public FruitBase
{ void dostuff () { std::cout << "Hey apple!" << std::endl; } };
// fake constexpr function: new can't be executed compile-time
template <std::size_t ... Is>
constexpr std::array<FruitBase *, FRUIT_AMOUNT>
init_helper (indexSequence<Is...> const &)
{ return { { (FruitBase*)(new Fruit<(MyEnum)Is>()) ... } }; }
// fake constexpr: init_helper() can't be executed compile-time
constexpr std::array<FruitBase *, FRUIT_AMOUNT> init ()
{ return init_helper(makeIndexSequence<FRUIT_AMOUNT>{}); }
int main ()
{
// compile (executed run-time)
std::array<FruitBase *, FRUIT_AMOUNT> myPrettyFruits = init();
// compilation error (init() can't be executed compile-time)
//constexpr std::array<FruitBase *, FRUIT_AMOUNT> myPrettyFruits = init();
}
I have the following scenario:
struct AP;
struct B
{
B() : m(2) {}
int m;
};
struct A : private B
{
A() : B(), n(1) {}
private:
int n;
friend AP;
};
struct AP
{
AP(A& a) : a_(a) {}
template<typename T>
struct A_B {
using type = typename std::enable_if< std::is_base_of< typename std::remove_reference<T>::type,
A >::value,
T >::type;
};
template<typename T>
operator typename A_B<T>::type()
{
return static_cast<T>(a_);
}
template<typename T>
typename A_B<T>::type get()
{
return static_cast<T>(a_);
}
int& n() { return a_.n; }
private:
A& a_;
};
int main()
{
A a;
AP ap(a);
ap.n() = 7;
const B& b = ap.get<const B&>();
//const B& b = ap; candidate template ignored: couldn't infer template argument 'T'
//auto b = static_cast<const B&>(ap); candidate template ignored: couldn't infer template argument 'T'
std::cout<<b.m;
}
The commented lines wouldn't compile. Clang++ notes that "candidate template ignored: couldn't infer template argument 'T'"
Why am I not able to get a reference to A's base with the cast operator?
I think the code would look much nicer that way.
The answer that you posted works, but is overkill unless you really want a static_assert message.
Classic templating works just fine in this instance because A is already convertible to B:
struct AP
{
AP(A& a) : a_(a) {}
template<typename T>
operator T()
{
return a_;
}
template<typename T>
T get()
{
return a_;
}
int& n() { return a_.n; }
private:
A& a_;
};
Demo
I found the answer here: http://www.mersenneforum.org/showthread.php?t=18076
This is the key: "when you want the compiler to deduce argument types, those types must not be dependent types"
With this it compiles:
template<typename T>
operator T()
{
static_assert(std::is_base_of< typename std::remove_reference<T>::type,A >::value,
"You may cast AP only to A's base classes.");
return static_cast<T>(a_);
}
The following is a mockup code that I wrote to experiment with trailing return types in a CRTP setup.
#include <iostream>
#include <memory>
#include <utility>
using namespace std;
struct t_aspect{
struct t_param1 {};
};
// Generic Selector
template <typename t_detail>
struct Select;
template <>
struct Select<t_aspect::t_param1> {
using typeof = t_aspect::t_param1;
};
//Base CRTP class
template<typename dclas>
class CrtpB
{
public:
template<typename T1>
auto func1() -> // What should be here?
{
return(static_cast<dclas*>(this)->func1<T1>());
}
};
//Derived CRTP class
class CrtpD : public CrtpB<CrtpD>
{
private:
uint32_t param1 = 0;
private:
auto func1(const t_aspect::t_param1&) -> uint32_t
{
return(param1);
}
public:
static auto create() -> unique_ptr<CrtpB>
{
return(unique_ptr<CrtpD>(new CrtpD));
}
template<typename T1>
auto func1() -> decltype(func1(typename Select<T1>::typeof()))
{
return(func1(typename Select<T1>::typeof()));
}
};
int main()
{
auto crtp = CrtpD::create();
auto parm = crtp->func1<t_aspect::t_param1>();
return 0;
}
I would like some help in deciphering what should be the trailing return type of func1 in CrtpB.
I have tried using
decltype(static_cast<dclas*>(this)->func1<T1>())
but this does not work. I have also tried writing a helper function based on a solution found in Inferring return type of templated member functions in CRTP.
template <typename D, typename T>
struct Helpr {
typedef decltype(static_cast<D*>(0)->func1<T>()) type;
};
But this does not work either.
dclas is an incomplete type when the base class is instantiated. You need to do two things to make this work:
Defer the checking of the type of func1<T1>() until the type is complete
Use the template keyword on the dependent expression so that the template definition is parsed correctly:
We can do this by adding a layer of indirection:
namespace detail {
template <class T, class Func1Arg>
struct func1_t {
using type = decltype(std::declval<T>().template func1<Func1Arg>());
};
};
Then you use this trait as the trailing return type:
template<typename T1>
auto func1() -> typename detail::func1_t<dclas,T1>::type
{
return(static_cast<dclas*>(this)->template func1<T1>());
}
I'm not sure of the title, because I'm not sure the issue comes from the "copyablility" of my container.
I tryied quite everything but I can't get rid of this error.
Here is a simplified version of my code (please do not challenge the class design, I really would like to keep the end-used syntax in the BOOST_FOREACH):
template <typename T>
class MyContainer
{
public:
typedef typename std::vector<T>::iterator iterator;
typedef typename std::vector<T>::const_iterator const_iterator;
MyContainer(std::vector<T>& vec, boost::mutex& mutex) :
m_vector(vec),
m_lock(mutex)
{
}
iterator begin() { return m_vector.begin(); }
const_iterator begin() const { return m_vector.begin(); }
iterator end() { return m_vector.end(); }
const_iterator end() const { return m_vector.end(); }
private:
std::vector<T>& m_vector;
boost::lock_guard<boost::mutex> m_lock;
};
template <typename T>
struct GetContainer
{
GetContainer(std::vector<T>& vec, boost::mutex& mutex) :
m_vector(vec),
m_mutex(mutex)
{
}
MyContainer<T> Get()
{
return MyContainer<T>(m_vector, m_mutex);
}
std::vector<T>& m_vector;
boost::mutex& m_mutex;
};
int main()
{
std::vector<int> v;
v.push_back(1);
v.push_back(2);
boost::mutex m;
GetContainer<int> getter(v, m);
BOOST_FOREACH(int i, getter.Get())
{
std::cout << i << std::endl;
}
return 0;
}
The compiler complains about not having a copy constructor for MyContainer::MyContainer(const MyContainer&).
I also have :
error: no matching function for call to ‘MyContainer::MyContainer(boost::foreach_detail_::rvalue_probe >::value_type)’
I follow the extensibility tips:
http://www.boost.org/doc/libs/1_58_0/doc/html/foreach/extensibility.html#foreach.extensibility.making__literal_boost_foreach__literal__work_with_non_copyable_sequence_types
But, making
MyContainer<T> : private boost::noncopyable
doesn't solve the issue.
Nor defining the function
boost_foreach_is_noncopyable
or specializing the template struct
is_noncopyable
for MyContainer (in fact, how would I specialize this template for a template type ?)
Last "tip":
If I remove the mutex and the lock from everywhere (I just pass the vector to GetContainer and to MyContainer), it works.
But it doesn't work if I make
MyContainer<T> : private boost::noncopyable
(I expected it should, so I'm not sure my problem is with BOOST_FOREACH, but maybe because I return a copy of MyContainer with my getter ?)
I thank you if you read me until here, and thanks in advance for help.
Seems to be a limitation of BOOST_FOREACH with move-only types. I didn't find a way around it¹ (except for the - ugly - obvious approach to put the lock_guard in a shared_ptr).
You didn't specify a c++03 requirement, though, so you can make it work without BOOST_FOREACH by replacing lock_guard with unique_lock.
Here's my take on things in c++11 (note how generic it is):
Live On Coliru
#include <boost/thread.hpp>
#include <boost/range.hpp>
namespace detail {
template <typename R, typename M>
struct RangeLock {
RangeLock(R&r, M& m) : _r(r), _l(m) {}
RangeLock(RangeLock&&) = default;
using iterator = typename boost::range_iterator<R>::type;
iterator begin() { using std::begin; return begin(_r); }
iterator end () { using std::end; return end (_r); }
using const_iterator = typename boost::range_iterator<R const>::type;
const_iterator begin() const { using std::begin; return begin(_r); }
const_iterator end () const { using std::end; return end (_r); }
private:
R& _r;
boost::unique_lock<M> _l;
};
}
template <typename R, typename M>
detail::RangeLock<R,M> make_range_lock(R& r, M& mx) { return {r,mx}; }
template <typename R, typename M>
detail::RangeLock<R const,M> make_range_lock(R const& r, M& mx) { return {r,mx}; }
#include <vector>
#include <map>
int main() {
boost::mutex mx;
std::vector<int> const vec { 1, 2 };
std::map<int, std::string> const map { { 1, "one" }, { 2, "two" } };
for(int i : make_range_lock(vec, mx))
std::cout << i << std::endl;
for(auto& p : make_range_lock(map, mx))
std::cout << p.second << std::endl;
for(auto& p : make_range_lock(boost::make_iterator_range(map.equal_range(1)), mx))
std::cout << p.second << std::endl;
}
Prints
1
2
one
two
one
¹ not even using all the approaches from Using BOOST_FOREACH with a constant intrusive list
I post my answer if it can help...
With C++03, I finally provide a copy constructor to be able to use the class with BOOST_FOREACH.
So the issue is moved to another topic: make the class copied in a logic and suitable way.
In my case, I "share the lock and the vector", the user shouldn't use this copy itself if he doesn't want to do bugs, but in BOOST_FOREACH it's okay:
I change the mutex to a recursive_mutex
I change the lock to an unique_lock
and:
MyContainer(const MyContainer& other) :
m_vector(other.vec),
m_lock(*other.m_lock.mutex())
{
}
With C++11
Thanks to Chris Glover on the boost mailling list, a C++11 solution:
You can't do what you are trying to do in C++03. To accomplish it, you
need C++11 move semantics to be able to move the MyContainer out of the Get
function. Even without using BOOST_FOREACH, the following code fails;
GetContainer<int> getter(v, m);
MyContainer<int> c = getter.Get(); // <-- Error.
Here's an example with the necessary changes; I changed the scoped_lock to
a unique_lock and added a move constructor.
template <typename T>
class MyContainer
{
public:
[...]
MyContainer(MyContainer&& other)
: m_vector(other.m_vector)
{
m_lock = std::move(other.m_lock);
other.m_vector = nullptr;
}
Using variadic template arguments together with a simple template argument I have experienced some strange behaviour of is_base_of when it was instantiated from a binded functor.
Here is the code:
template <class T, class Index>
class Base{};
template<typename T>
struct Checker {
typedef int result_type;
// Returns 1 if a given T type is descendant of Base<T,First>
template<typename First, typename ...Args>
result_type operator()(First&& first, Args&&... params)
{
return check(std::is_base_of<Base<T,First>, T>(),
std::forward<First>(first),
std::forward<Args>(params)...);
}
template<typename ...Args>
result_type check(const std::true_type&, Args&&... params)
{
return 1;
}
template<typename ...Args>
result_type check(const std::false_type&, Args&&... params)
{
return 0;
}
};
struct A {};
struct B : Base<B,int> {};
int main()
{
Checker<A> ch1;
std::cout<<ch1(3.14)<<std::endl;
Checker<B> ch2;
std::cout<<ch2(1 ,3.14)<<std::endl; // output is 1
std::cout<<std::bind(ch2, 1, 3.14)()<<std::endl; // output is 0 but it should be 1 !
return 0;
}
The program output is:
0
1
0
But I would expect:
0
1
1
Am I using the variadic templates in a wrong way? Is there any other (correct) way to get the first type of a variadic type list like Args? Why this is a problem only when it is used with the bind expression?
Note, if I am modifing the Base template to have only one template parameter, then the bind expression works:
template <class T>
class Base{};
template<typename T>
struct Checker {
typedef int result_type;
// Returns 1 if a given T type is descendant of Base<T>
template<typename ...Args>
result_type operator()(Args&&... params)
{
return check(std::is_base_of<Base<T>, T>(),
std::forward<Args>(params)...);
}
template<typename ...Args>
result_type check(const std::true_type&, Args&&... params)
{
return 1;
}
template<typename ...Args>
result_type check(const std::false_type&, Args&&... params)
{
return 0;
}
};
struct A {};
struct B : Base<B> {};
int main()
{
Checker<A> ch1;
std::cout<<ch1(3.14)<<std::endl;
Checker<B> ch2;
std::cout<<ch2(3.14)<<std::endl; // output is 1
std::cout<<std::bind(ch2, 3.14)()<<std::endl; // output is 1 this time!
return 0;
}
You're not getting the expected output because the data-type of First in your Checker function object when called after std::bind() is of type int&, not int.
Therefore std::is_base_of<Base<B,int&>, B> does not instantiate to a std::true_type for the call to Checker::check.
The problem is that std::bind is creating an object that internally stores the arguments for the function you are passing to it. Therefore there is a named l-value as a non-static data-member of the object returned by std::bind that is holding the value you passed as an argument to be bound to your function. When that non-static data-member is then passed to the r-value reference at the time you call the operator() of the functor, it's passed as an l-value reference, since it is no longer a temporary object. You would have a similar problem if you did something like:
int x = 1;
Checker<B> ch2;
std::cout<<ch2(x, 3.14)<<std::endl;
The named-value x is an l-value, and would be passed to the first argument in your operator() method as an l-value reference, not as temporary, since first is a r-value reference. Therefore your type would end up again as an int& and not an int, and you'd print a value of 0.
To fix this problem, you can do something like:
template<typename First, typename ...Args>
result_type operator()(First&& first, Args&&... params)
{
if (std::is_reference<First>::value)
{
return check(std::is_base_of<Base<T, typename std::remove_reference<First>::type>, T>(),
std::forward<First>(first),
std::forward<Args>(params)...);
}
else
{
return check(std::is_base_of<Base<T,First>, T>(),
std::forward<First>(first),
std::forward<Args>(params)...);
}
}
This will strip off the reference-type of the object and give you the results you want.
Unfortunately std::is_reference did not give me the expected result on a more complicated issue.
So finally I choosed providing the reference and const-reference overloads:
template<typename First, typename ...Args>
result_type operator()(First& first, Args&&... params)
{
return check(std::is_base_of<Base<T,First>, T>(),
first,
std::forward<Args>(params)...);
}
template<typename First, typename ...Args>
result_type operator()(const First& first, Args&&... params)
{
return check(std::is_base_of<Base<T,First>, T>(),
first,
std::forward<Args>(params)...);
}