是否可以定义一组模板参数并有条件地专门化它

Question

struct TypeA {
    using data_t = int;
    enum { thread_create = pthread_create }; // ???
};

struct TypeB {
    using data_t = double;
    enum { thread_create = another_kind_of_thread_create }; // ???
};

template<typename T>
class Test {
public:
    void func() {
        T::thread_create(); // ???
    }
private:
    T::data_t a;
};

Test<TypeA> t1;
Test<TypeB> t2;

我想做的是用一个模板参数专门化模板 class Test。

如您所见，data_t 应该没有问题，但传递函数似乎并不容易。这段代码会产生一个错误：error: enumerator value for 'thread_create' must have integral or unscoped enumeration type.

是否可以传递这样的函数？

Answer 1

您描述的是一个非常普通的特征 class。您可能正在寻找以下方面的内容：

struct TypeA {
    using data_t = int;
    static void thread_create() { pthread_create() };
};

这将允许 T::thread_create(); 语法。

Answer 2

如果我没有理解错的话，你想根据专业化来切换线程创建函数和数据类型。如果是这样，为什么不这样呢？

#include <type_traits>

void pthread_create(void*, void*);
void another_kind_of_thread_create(void*, void*);
//my fakes. use your includes here

struct TypeA {
    using data_t = int;
    using thread_create_t = void (*)(void*, void*);
    static constexpr thread_create_t  thread_create{pthread_create};
};

struct TypeB {
    using data_t = double;
    using thread_create_t = void (*)(void*, void*);
    static constexpr thread_create_t thread_create{another_kind_of_thread_create};
};

template<typename T>
class Test : public T {
public:
    void func() {
        T::thread_create(this, &a); //whatever, put here just to match the prototype...
    }
private:
    typename T::data_t a;
};

Test<TypeA> t1;
Test<TypeB> t2;

static_assert(std::is_same<decltype(t1)::data_t, int>::value, "t1 should be int");
static_assert(std::is_same<decltype(t2)::data_t, double>::value, "t2 should be double");

demo

是否可以定义一组模板参数并有条件地专门化它

Is it possible to define a group of template parameters and specialize it conditionally

c++

templates

template-meta-programming

c++11