成员初始化器不会为非静态数据成员或基类命名

问题描述

我很难在google上找到匹配。

I'm having a hard time finding hits on google for this.

struct a {
    float m_x;
    float m_z;
public:
    a(float x): m_x(x) {}
};

class b : public a {
    b(float z): m_z(z) {}
};

在clang 3.2上：

On clang 3.2:

error: member initializer 'm_z' does not name a non-static data member or base class
    b(float z): m_z(z) {}

推荐答案

不能直接从初始化列表初始化基类成员。这是因为初始化顺序是以这种方式进行的。

No you cannot initialize base class members from initializer list directly. This is because order of initialization proceeds in this way

C ++标准n3337 § 12.6.2 / 10

C++ Standard n3337 § 12.6.2/10

- 首先，只有最多的
派生类（1.8）的构造函数，虚拟基类按照
的顺序初始化，它们出现在以深度优先的从左到右遍历的直接
acyclic基类的图，其中从左到右是在导出类
base-specifier-list中的基类的
出现的顺序。

— First, and only for the constructor of the most derived class (1.8), virtual base classes are initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where "left-to-right" is the order of appearance of the base classes in the derived class base-specifier-list.

- 然后，直接基类在
声明顺序中初始化，因为它们出现在base-specifier-list
（不考虑mem初始化器的顺序）。

— Then, direct base classes are initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).

- 然后，非静态
数据成员按照在
类定义中声明的顺序初始化 （再次不考虑
mem初始化器的顺序）。

— Then, non-static data members are initialized in the order they were declared in the class definition (again regardless of the order of the mem-initializers).

- 最后，执行
构造函数体的复合语句。

[注意：声明顺序要求确保base和
成员子对象以与
初始化相反的顺序被销毁。 - end note]

[ Note: The declaration order is mandated to ensure that base and member subobjects are destroyed in the reverse order of initialization. — end note ]

因此，您可以在基类中指定一个构造函数派生类（），或者您可以分配给派生类中的基类成员class

So you can specify a constructor in a base class (it can be protected) and use that one in initialization list of derived class (should be preferred) or you can assign to a base class member in derived class ctor body (less efficient - you are assigning to default initialized member).

在前一种情况下，你可以这样写：

In the former case you might write it this way:

struct A {
    float m_x;
    float m_z;
    A(){}
protected:
    A(float x): m_x(x) {}
};

class B : public A {
public:
    B(float z) : A(z) {}
    // alternatively
    // B(float z) {
    //     m_x = z;
    // }
};

int main(){
    B b(1);
    return 0;
}

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