在初始化列表之外手动调用基类的构造函数 [英] Manually calling constructor of base class outside initialization list

问题描述

我有一个 Derived 类，其构造函数必须填充作为参数传递的 struct 的字段到 Base 类的构造函数。我希望能够命名要填充的结构的字段，以使我的代码适应未来（例如：抵抗 MyStruct ）。

I have a Derived class whose constructor has to populate the fields of a struct that is passed as an argument to the constructor of the Base class. I want to be able to name the fields of the struct that I am populating, to keep my code future-proof (i.e.: resistant to addition and/or reordering of the members of MyStruct).

请注意， struct MyStruct 具有默认值，因此无法直接在其中使用命名字段进行初始化初始化列表（例如： Base（{。a = a，.b = b}）不起作用）。另外，就我而言， Base 的副本构造函数也被删除。另外，我正在使用C ++ 11。

Note that struct MyStruct has default values, so it cannot be initialised with named fields directly in the initialization list (e.g.: Base({.a = a, .b = b}) does not work). Also, in my case, Base's copy constructor is deleted. Also, I am using C++ 11.

我想到的解决方案使用 placement new 运算符手动调用 Base 类的构造函数在 this 指向的内存上。为此，我还必须在 Base 类中添加受保护的默认构造函数。这种方法是否有任何弊端和/或有人可以建议一种更好的方法？

The solution I came up with uses the placement new operator to manually call the constructor of the Base class on the memory pointed to by this. To achieve this I also had to add a protected default constructor to my Base class. Are there any possible downsides to this approach and/or could anyone suggest a better method?

#include <iostream>

struct MyStruct
{
        int a = 0;
        int b = 1;
};

class Base
{
public:
        Base(MyStruct str){
                std::cout << "a: " << str.a << ", b: " << str.b << "\n";
        }
        Base(Base&&) = delete; // no copy constructor
protected:
        Base(){ // dummy, does exactly nothing.
                // it only exists to be called by
                // the derived class's constructor
        }
private:
        int amember;
};

class Derived : public Base
{
public:
        Derived(int a, int b)
        {
                MyStruct str;
                str.a = a;
                str.b = b;
                new (this) Base(str);
        }
private:
        int anothermember;
};

int main()
{
        MyStruct str;
        str.a = 10;
        str.b = 20;
        Base b(str);
        Derived d(10, 20);
        return 0;
}

编辑：添加提及不能复制Base，明确指出 Base :: Base（）完全不执行任何操作。

edit: added mention that Base cannot be copied, made explicit that Base::Base() does exactly nothing.

推荐答案

使用辅助函数代替例如

class Derived : public Base
{
public:
        Derived(int a, int b) : Base(make_mystruct(a, b)), anothermember(some_value) {}
private:
        int anothermember;
        static MyStruct make_mystruct(int a, int b) { return MyStruct(a, b); }
};

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