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C++17 结构化绑定

在聊结构化绑定之前,有几个面试问题,看看你会不会?

  • 如何使用结构化绑定访问自定义类的私有成员?

  • 如何使用结构化绑定修改自定义类的成员呢?

这几个题目估计没几个人能答上来,题目与答案网上也都没有,基于这些问题,我们引出本文的内容:如何为自定义类型支持结构化绑定,里面有哪些坑?

注:懒人版,答案与代码已放星球。

为了让本文的内容更加丝滑,从示例出发。

1.自定义类型

如果给定一个类:

struct Foo {
  Foo(int64_t id, std::string data) : id_(id), data_(std::move(data)) {}
  void Print() { std::cout << "id = " << id_ << ", data = " << data_ << std::endl; }

 private:
  int64_t id_;
  std::string data_;
};

要让下面这个代码运行起来:

int main() {
  Foo f{12, "hello world"};
  const auto& [id, data] = f;
  std::cout << "id = " << id << ", data = " << data << std::endl;
  f.Print();
}

如果直接运行会编译:

 cannot decompose inaccessible member 'Foo::id_' of 'const Foo'
   42 |   const auto& [id, data] = f;

那么怎么做呢?

2.支持结构化绑定

这个问题我们使用两个步骤:

1.特化 std::tuple_size 和 std::tuple_element

namespace std {
  template <>
  struct tuple_size<Foo> : std::integral_constant<std::size_t, 2> {};

  template <>
  struct tuple_element<0, Foo> {
    using type = int64_t;
  };

  template <>
  struct tuple_element<1, Foo> {
    using type = std::string;
  };
}

2.为上面的结构体添加友元get函数,然后实现即可:

template <std::size_t Index>
decltype(auto) get(const Foo& f);

template <>
decltype(auto) get<0>(const Foo& f) {
  return f.id_;
}

template <>
decltype(auto) get<1>(const Foo& f) {
  return f.data_;
}

此时便可以正常的工作了,得到输出:

id = 12, data = hello world
id = 12, data = hello world

最后,抛出几个问题:

  • 上面提供的是只读访问,如果在访问的时候修改const auto &会发生什么?

例如:

auto& [id, data] = f;
  • 如何支持对类成员的修改访问?

即,让这个语句能够正常运行

auto& [id, data] = f;

好了,本节完,欢迎留言区交流。