参考下面的代码，请问有人能想出如何适应这个代码吗？
这是我对解决方案的进展。它产生了相同的输出，但需要自动生成标记为
如何自动生成元组。
template <typename RET, typename... ARGS1, typename... ARGS2>
RET Mediator::change (Object* o, RET (Object::*f)(ARGS1...), ARGS2&&... args) {
const std::tuple<ARGS2...> t(args...);
for (Object* x : objects)
(x->*f)(std::get<0>(t), o->rating, std::get<1>(t), o->str);
}
这样我就不必每次更改ARGS2...时都重写不同版本。如果参数只包含4个参数，我不介意这样做，但是如果参数大于4，则需要进行泛化。ARGS1...中的类型应该是不同的类型，因此应该有一种方法使std::get<0>(t)、std::get<1>(t)等正确放置，以便不需要像上面那样手动执行（即使存在重复类型，它们也可以简单地放置在重复类型的第一个插槽中）。下面是完整的代码（上下文是每个订阅者对象更改Mediator时，其他订阅者对象也应相应更改）：
#include <iostream>
#include <string>
#include <vector>
#include <tuple>
struct Mediator {
std::vector<struct Object*> objects;
void registerObject (Object* o) {objects.emplace_back(o);}
template <typename RET, typename... ARGS1, typename... ARGS2>
RET change (Object*, RET (Object::*)(ARGS1...), ARGS2&&...);
};
struct Object {
int value;
double rating;
char letter;
std::string str;
Mediator& mediator;
Object (int v, double r, char l, const std::string& s, Mediator& m) :
value(v), rating(r), letter(l), str(s), mediator(m) {mediator.registerObject(this);}
virtual void adjust (int, double, char, const std::string&) = 0;
template <typename RET, typename... ARGS1, typename... ARGS2>
RET change (RET (Object::*f)(ARGS1...), ARGS2&&... args) {
return mediator.change(this, f, std::forward<ARGS2>(args)...);
}
};
struct A : Object {
using Object::Object;
virtual void adjust (int a, double b, char c, const std::string& s) override {
std::cout << "Type A adjusted using values " << a << ", " << b << ", " << c << ", and " << s << "." << std::endl;
}
};
struct B : Object {
using Object::Object;
virtual void adjust (int a, double b, char c, const std::string& s) override {
std::cout << "Type B adjusted using values " << a << ", " << b << ", " << c << ", and " << s << "." << std::endl;
}
};
struct C : Object {
using Object::Object;
virtual void adjust (int a, double b, char c, const std::string& s) override {
std::cout << "Type C adjusted using values " << a << ", " << b << ", " << c << ", and " << s << "." << std::endl;
}
};
template <typename RET, typename... ARGS1, typename... ARGS2>
RET Mediator::change (Object* o, RET (Object::*f)(ARGS1...), ARGS2&&... args) {
const std::tuple<ARGS2...> t(args...);
for (Object* x : objects)
(x->*f)(std::get<0>(t), o->rating, std::get<1>(t), o->str);
}
int main() {
Mediator mediator;
Object *a = new A(6, 1.2, 'a', "alan", mediator);
Object *b = new B(2, 6.5, 'b', "bob", mediator);
Object *c = new C(4, 0.8, 'c', "craig", mediator);
c->change (&Object::adjust, 8, 'k');
}
输出：
Type A adjusted using values 8, 0.8, k, and craig.
Type B adjusted using values 8, 0.8, k, and craig.
Type C adjusted using values 8, 0.8, k, and craig.
这是我对解决方案的进展。它产生了相同的输出，但需要自动生成标记为
// Here!的行。#include <iostream>
#include <string>
#include <vector>
#include <tuple>
template <std::size_t...> struct index_sequence {};
template <std::size_t N, std::size_t... Is>
struct make_index_sequence_helper : make_index_sequence_helper<N-1, N-1, Is...> {};
template <std::size_t... Is>
struct make_index_sequence_helper<0, Is...> {
using type = index_sequence<Is...>;
};
template <std::size_t N>
using make_index_sequence = typename make_index_sequence_helper<N>::type;
struct Mediator {
std::vector<struct Object*> objects;
void registerObject (Object* o) {objects.emplace_back(o);}
template <typename RET, typename... ARGS1, typename... ARGS2>
RET change (Object*, RET (Object::*)(ARGS1...), ARGS2&&...);
template <typename RET, typename... ARGS, std::size_t... Is>
RET changeHelper (RET (Object::*)(ARGS...), const std::tuple<ARGS...>&, index_sequence<Is...>);
};
struct Object {
int value;
double rating;
char letter;
std::string str;
Mediator& mediator;
Object (int v, double r, char l, const std::string& s, Mediator& m) :
value(v), rating(r), letter(l), str(s), mediator(m) {mediator.registerObject(this);}
virtual void adjust (int, double, char, const std::string&) = 0;
template <typename RET, typename... ARGS1, typename... ARGS2>
RET change (RET (Object::*f)(ARGS1...), ARGS2&&... args) {
return mediator.change(this, f, std::forward<ARGS2>(args)...);
}
};
struct A : Object {
using Object::Object;
virtual void adjust (int a, double b, char c, const std::string& s) override {
std::cout << "Type A adjusted using values " << a << ", " << b << ", " << c << ", and " << s << "." << std::endl;
}
};
struct B : Object {
using Object::Object;
virtual void adjust (int a, double b, char c, const std::string& s) override {
std::cout << "Type B adjusted using values " << a << ", " << b << ", " << c << ", and " << s << "." << std::endl;
}
};
struct C : Object {
using Object::Object;
virtual void adjust (int a, double b, char c, const std::string& s) override {
std::cout << "Type C adjusted using values " << a << ", " << b << ", " << c << ", and " << s << "." << std::endl;
}
};
template <typename RET, typename... ARGS1, typename... ARGS2>
RET Mediator::change (Object* o, RET (Object::*f)(ARGS1...), ARGS2&&... args) {
const std::tuple<ARGS2...> t(args...);
// Here!
const std::tuple<ARGS1...> tuple(std::get<0>(t), o->rating, std::get<1>(t), o->str);
changeHelper (f, tuple, make_index_sequence<sizeof...(ARGS1)>());
}
template <typename RET, typename... ARGS, std::size_t... Is>
RET Mediator::changeHelper (RET (Object::*f)(ARGS...),
const std::tuple<ARGS...>& tuple, index_sequence<Is...>) {
for (Object* x : objects)
(x->*f) (std::get<Is>(tuple)...);
}
int main() {
Mediator mediator;
Object *a = new A(6, 1.2, 'a', "alan", mediator);
Object *b = new B(2, 6.5, 'b', "bob", mediator);
Object *c = new C(4, 0.8, 'c', "craig", mediator);
c->change (&Object::adjust, 8, 'k');
}
如何自动生成元组。
const std::tuple<ARGS1...> tuple(std::get<0>(t), o->rating, std::get<1>(t), o->str);
使用类似于某物的东西
template <typename... ARGS1, typename... ARGS2>
std::tuple<ARGS1...> extractTuple (Object* o, ARGS2&&... args);
那么是否可以不需要为不同（如果ARGS1...很大，则可能为很多）的ARGS2...选择编写新版本的Mediator::change？我的当前想法是使用递归辅助方法，std::is_same，std::tuple_cat等。但我遇到了问题（我认为在检查类型期间我们正在解包ARGS1...中的ARGS2...）。
RET很可能是一个宏的名称。 - Cheers and hth. - Alf(char, double, int)。我们两个解决方案都无法使用c->change(&Object::foo，'z'，4);其中Object::foo的签名为(char, double, int)。 - prestokeys