使用基类的静态函数而不指定参数以避免歧义

Question

我的一些基类获得了大量的参数。 现在我想指定要使用的静态函数：

template <typename... Types>
struct SBase {
    static void func() {
    }
};

struct A : public SBase<int> {
};

struct B : public A, public SBase<int, double, short,
    unsigned int, float, unsigned char, long, unsigned long> {

    // using SBase::func; // Not possible.
    // Horrible, but works.
    using SBase<int, double, short,
    unsigned int, float, unsigned char, long, unsigned long>::func;
};

您可以看到，我需要两次编写模板参数，这会导致代码重复。

有没有办法摆脱它？

Answer 1

你可以使用typedef：

typedef SBase<int, double, short, unsigned int, float, unsigned char,
      long, unsigned long> B_SBase;

struct B : public A, public B_SBase {
    using B_SBase::func;
};

Answer 2

如果B已经是一个模板（在我的代码中大多是这种情况），那么你可以像这样使用：

template <typename MyBase = SBase<int, double, short,
                                  unsigned int, float, unsigned char,
                                  long, unsigned long> >
struct B : public A, public MyBase {
  using MyBase::func;
};

但是，如果不是，我不可能知道不重复基类或使用typedef SBase<...> Bs_Base污染命名空间。 但如果你聪明，你只需要写两次：

struct B : public A, public SBase<int, double, short,
    unsigned int, float, unsigned char, long, unsigned long> {
  typedef SBase<int, double, short, unsigned int, float,
                unsigned char, long, unsigned long> MyBase;
};
static_assert(std::is_base_of<B::MyBase, B>::value, "");

Answer 3

使B成为一个类模板。 在那里，没有重复：

template<typename... Types>
struct B : public A, public SBase<Types...> {
  using SBase<Types...>::func;
};

typedef B<int, double, short, unsigned int, float, unsigned char, long, unsigned long> BB;

void foo ()
{
  BB::func();
}