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

您可以使用 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;
};