递归可变参数函数模板

Question

我想编写一个类方法,它接受一个模板参数包,但是零参数,并对类型进行"迭代":

struct Bar {
    template <typename T, typename... Ts>
    void foo() {
        // something with T that involves Bar's members
        foo<Ts...>();
    }
};

实现这个的首选方法是什么？

Answer 1

您可以使用以下内容：

struct Bar {
    template <typename... Ts>
    void foo() {
        int dummy[] = {0 /*Manage case where Ts is empty*/,
                       (bar<Ts>(), void() /* To avoid overload `operator,` */, 0)...};
        (void) dummy; // suppress warning for unused variable.
    }

    template <typename T>
    void bar()
    {
        // something with T that involves Bar's members
    }

};

在C++17中，可以用Folding表达式来简化：

struct Bar {
    template <typename... Ts>
    void foo() {
        (static_cast<void>(bar<Ts>()), ...);
    }

    template <typename T>
    void bar()
    {
        // something with T that involves Bar's members
    }

};