如何在类obj中使用enable_if？

Question

The code below is OK:

template <class T>
std::enable_if<std::is_atomic<T>::value, bool>
foo(T t) { return true; }

template <class T>
std::enable_if<tmp::is_sequence<T>::value, bool>
foo(T t) { return false; }

int main(void){
  foo(1);  // return true
  auto std::vector<int> a{2};
  foo(a);  // return false
}

But when I use a class to bundle them, it can not be compiled:

template <class T>
class test {
public:

std::enable_if<std::is_atomic<T>::value, bool>
foo(T t) { return true; }

std::enable_if<tmp::is_sequence<T>::value, bool>
foo(T t) { return false; }
};

int main(...) {
  test<int> obj;
  obj.foo(1);
  test<std::vector<int>> obj2;
  std::vector<int> tmp{2};
  obj2.foo(tmp);
}

clang++ print:

error: functions that differ only in their return type cannot be overloaded

So I write something to cheat to the compiler(add an S in second foo ):

template <class S>
std::enable_if<tmp::is_sequence<T>::value, bool>
foo(T t) { return false; }

It still can not work:

error: no type named 'type' in 'std::enable_if<false, bool>'

How can I make it work in a class?

Answer 1

You forgot to add ::type after enable_if : (see enable_if )

template <class T> std::enable_if<std::is_atomic<T>::value, bool>::type
foo(T t) { return true; }

Answer 2

Both member-functions should have different template parameter (following will work ok)

template <class T>
class test {
public:

template<typename U>
typename std::enable_if<std::is_atomic<U>::value, bool>::type
foo(U t) { return true; }

template<typename U>
typename std::enable_if<tmp::is_sequence<U>::value, bool>::type
foo(U t) { return false; }

};

Answer 3

如果您真的想做自己想做的事情，那么经典的成语就是引入一个带有默认值的伪SFINAE参数：

bool foo(T t, std::enable_if<..., void*> = nullptr) { ... }