可以使用 C++ 聚合初始化来构造实现接口的 class 的实例吗?

Can C++ aggregate initialization be used to construct an instance of a class which implements an interface?

我希望有人能告诉我为什么以下代码无法编译的技术细节,如果可能的话,解决方法。

我有一个名为 Foo 的现有结构,以及使用初始化列表创建 Foo 实例的代码。此代码编译并工作:

struct Foo {
    int id1;
    int id2;
};

int main()
{
    Foo f({1,2});

    return f.id1;
}

我希望 Foo 实现一个接口:

struct Interface {
    // All pure virtual methods, but this won't compile even if empty
};

struct Foo : public Interface{
    int id1;
    int id2;
};

int main()
{
    Foo f({1,2});

    return f.id1;
}

此代码不再编译,出现

中的错误
cannot convert argument 1 from 'initializer list' to 'const _Ty &'

(错误会根据您的具体编译器而变化。)

我找到了与聚合初始化相关的标准部分:

[dcl.init.aggr]/1 An aggregate is an array or a class (Clause 12) with 1.1 no user-provided, explicit, or inherited constructors (15.1), 1.2 no private or protected non-static data members (Clause 14), 1.3 no virtual functions (13.3), and 1.4 no virtual, private, or protected base classes (13.1).

虽然我实际上不确定这里是否发生了聚合初始化。有人可以解释发生的错误,如果可能的话,提供我可以对界面进行的更改吗?我有几个现有的结构需要这个接口,还有很多使用这种初始化形式的现有代码,我想尽可能少地重写它。谢谢!

你需要初始化基础 class 即使它是空的:

Foo f({{},1,2});

see it live on godbolt

在您所指部分的标准中,我们可以在 [dcl.init.aggr]p4.2 中看到这样的示例:

struct base1 { int b1, b2 = 42; };
struct base2 {
  base2() {
   b3 = 42;
 }
 int b3;
};

struct derived : base1, base2 {
 int d;
};

derived d1{{1, 2}, {}, 4};
derived d2{{}, {}, 4};

initializes d1.b1 with 1, d1.b2 with 2, d1.b3 with 42, d1.d with 4, and d2.b1 with 0, d2.b2 with 42, d2.b3 with 42, d2.d with 4. —end example]

另请参阅 [dcl.init.aggr]p2,其中解释了聚合的元素是什么:

The elements of an aggregate are:

-for an array, the array elements in increasing subscript order, or
-for a class, the direct base classes in declaration order, followed by the direct non-static data members ([class.mem]) that are not members of an anonymous union, in declaration order.

[dcl.init.aggr]p3说:

When an aggregate is initialized by an initializer list as specified in [dcl.init.list], the elements of the initializer list are taken as initializers for the elements of the aggregate. ...

注意,答案假定 C++17 或更高版本,因为 before C++17 an aggregate was not allowed to have a base class

@ShafikYaghmour 解释了为什么当 Interface 为空时,无法像以前那样进行聚合初始化。

但是如果 Interface 有虚函数,如问题中所建议的,从 Interface 派生的 class 将不是聚合 .因此,实现 Interface 并将数据成员保存为 Foo 的 class 必须实现一个构造函数。我看到的最简单的方法(根据数据成员的 "triviality",就速度而言可能不是最有效的)是这样的:

struct Interface {
   // All pure virtual methods, but this won't compile even if empty
   virtual void bar() =0;
   };

struct Foo_data{ //change the name of the aggregate
  int id1;
  int id2;
  };

struct Foo
  :Interface  //Foo has virtual function => Foo is not an aggregate
  ,Foo_data
  {
  Foo() =default;
  Foo(Foo_data data):Foo_data(std::move(data)){}//a constructor must be provided
  void bar() override {}
  };

int main(){
  Foo f({1,2});
  return f.id1;
  }