'使用'声明为SFINAE

Question

我可以使用SFINAE(或其他技术)进行using声明,而私有派生自模板类吗？为了更好地理解,请参阅以下代

#include <iostream>

struct S1 {
    void f() { std::cout << "S1::f\n"; }
};

struct S2 {
    void f() { std::cout << "S2::f\n"; }
    void g() { std::cout << "S2::g\n"; }
};

template <class T>
struct D : private T {
    using T::f;
    // using T::g; // need this only if T provides g() function
};

int main() {
    D<S1>().f(); // ok. Prints 'S1::f'
    D<S2>().f(); // ok. Prints 'S2::f' 
    D<S2>().g(); // fail. But wants to be ok and prints 'S2::g'
    return 0;
}

我怎样才能达到预期的行为(如果可能的话)？

Answer 1

Bryan Chen的答案的变体看起来更丑,但是更容易扩展到多个检查,并且不需要复制在它们之间共享的代码,D<type-with-f>并且D<type-without-f>使用继承链,其中每个步骤检查一个额外的成员.唯一需要的重复是构造函数的继承,如果合适的话.

struct A {
  void f() { }
  void g() { }
  void i() { }
};

// The generic case. D<T, char[N]> simply provides what D<T, char[N+1]> provides.
template <typename T, typename U = char[1]>
struct D : D<T, char[sizeof(U) + 1]> {
  using D<T, char[sizeof(U) + 1]>::D;
};

// The end of the chain. This is where T gets inherited. It declares all of its own
// specialisations as its friends, so that they can access other members of T.
template <typename T>
struct D<T, char[6]> : private T {
  template <typename, typename>
  friend struct D;

  D(int) { }
  void fun() { }
};

// Check for T::f.
template <typename T>
struct D<T, char[2 + !sizeof(&T::f)]> : D<T, char[3]> {
  using D<T, char[3]>::D;
  using T::f;
};

// Check for T::g.
template <typename T>
struct D<T, char[3 + !sizeof(&T::g)]> : D<T, char[4]> {
  using D<T, char[4]>::D;
  using T::g;
};

// Check for T::h.
template <typename T>
struct D<T, char[4 + !sizeof(&T::h)]> : D<T, char[5]> {
  using D<T, char[5]>::D;
  using T::h;
};

// Check for T::i.
template <typename T>
struct D<T, char[5 + !sizeof(&T::i)]> : D<T, char[6]> {
  using D<T, char[6]>::D;
  using T::i;
};

int main() {
  D<A> d = 4; // ok: verify that constructors got inherited
  // A &a = d; // error: verify that inheritance of A is private
  d.f(); // ok: verify that f got inherited
  d.g(); // ok: verify that g got inherited
  // d.h(); // error: verify that h is not available
  d.i(); // ok: verify that i got inherited
  d.fun(); // ok: verify that the inheritance chain didn't get broken
}

注意:&T::f您可能希望代替检查,而不是检查std::declval<T>().f().前者无法处理重载函数.

Answer 2

decltype(void(&T::g))SFINAE 的C++ 部分模板特化和使用

#include <iostream>
#include <type_traits>

struct S1 {
    void f() { std::cout << "S1::f\n"; }
};

struct S2 {
    void f() { std::cout << "S2::f\n"; }
    void g() { std::cout << "S2::g\n"; }
};

template <class T, class V = void>
struct D : private T {
    using T::f;
};

template <class T>
struct D<T, decltype(void(&T::g))> : private T {
    using T::f;
    using T::g; // need this only if T provides g() function
};

int main() {
    D<S1>().f(); // ok. Prints 'S1::f'
    D<S2>().f(); // ok. Prints 'S2::f' 
    D<S2>().g(); // ok. Prints 'S2::g'
    return 0;
}

现场演示

编辑：

这是另一种更灵活的方法，但我不知道继承如何private virtual适用于实际用例。如果它可能导致任何问题，请告诉我（例如 UB）。

#include <iostream>
#include <type_traits>

struct S1 {
    void f() { std::cout << "S1::f\n"; }
};

struct S2 {
    void f() { std::cout << "S2::f\n"; }
    void g() { std::cout << "S2::g\n"; }
};

struct S3 {
    void g() { std::cout << "S3::g\n"; }
};

template <class T, class = void>
struct D_f {};

template <class T>
struct D_f<T, decltype(void(&T::f))> : private virtual T {
    using T::f;
};

template <class T, class = void>
struct D_g {};

template <class T>
struct D_g<T, decltype(void(&T::g))> : private virtual T {
    using T::g;
};

template <class T>
struct D : D_f<T>, D_g<T> {
};


int main() {
    D<S1>().f();
    D<S2>().f();
    D<S2>().g();
    D<S3>().g();
    return 0;
}

现场演示