在没有模板的构造函数中访问覆盖的字段 [英] Access overridden field in constructor without templates

问题描述

我无法从父C ++构造函数访问子类中重写的字段，并且我不能使用模板，因为上游项目不使用它们.

I can't access field overridden in child class from parent C++ constructor and I can not use templates, because upstream project doesn't use them.

这是正在工作的Python原型，我尝试在C ++中重新实现.该代码包含两个驱动程序类-一个子级和一个父级，并且父级在初始化期间打印该类的name.

This is working Python prototype that I try to reimplement in C++. The code contains two driver classes - one child and one parent, and parent prints name of the class during initialization.

class Driver(object):
  name = "Unknown"
  def __init__(self):
    print(self.name)

class SpecificDriver(Driver):
  name = "Specific"
  def __init__(self):
    super(SpecificDriver, self).__init__()

Driver()
SpecificDriver()

这将两个字符串打印到控制台

This prints two strings to console

Unknown
Specific

在C ++中看起来无法访问覆盖的name，因为此时name不存在-

Looks like in C++ can't access overridden name, because the name doesn't exist at this point - Call child method from parent constructor. So maybe there is another way to get driver name printed on initialization?

更新(2018):该问题的原标题是在C ++中初始化时在没有模板的情况下打印覆盖的子字段"，并且由于范围太广而被关闭.

UPDATE (2018): The original title for this question was "Print overridden child field during initialization in C++ without templates" and it was closed as too broad.

推荐答案

尽管您要求不使用模板来执行此操作，但这是从基类构造函数执行此操作的唯一方法.

Despite you're asking to do this without templates, it's the only way to do this from a base class constructor.

以下是应如何操作的示例:

Here's a sample how it should be done:

struct IDriver {
    // Public virtual API:
    virtual void func1() = 0;
    // ...
    virtual ~IDriver() {}
};

template<typename Derived>
class Driver : public IDriver {
public:
    Driver() {
         std::cout << "Driver" << std::endl;
         std::cout << static_cast<Derived*>(this)->name() << std::endl;
    }   
};

class SpecificDriver : public Driver<SpecificDriver> {
public:
    // Public virtual API:
    virtual void func1();
    std::string name() const { return "SpecificDriver"; }
    // or use typeid(SpecificDriver).name() if you prefer
};

int main() {
    SpecificDriver sd;
}

实时演示

至于您的评论:

是否可以像@tobspr方法中一样使用其他init()函数，但是将名称设为字段而不是函数调用?

Is it possible to use additional init() function as in @tobspr method, but making name a field instead of function call?

好吧，由于类名仍然是这些类的静态属性，因此您可以使用如下的static const字段:

Well, since the class name is a static property of these classes anyway, you can use a static const field like follows:

template<typename Derived>
class Driver : public IDriver {
public:
    Driver() {
         std::cout << name << std::endl;
         std::cout << Derived::name << std::endl;
    }   

private:
    static const std::string name;
};

template<typename Derived>
const std::string Driver<Derived>::name = "Driver";

class SpecificDriver : public Driver<SpecificDriver> {
public:
    static const std::string name;
};

const std::string SpecificDriver::name = "SpecificDriver";

int main() {
    SpecificDriver sd;
}

实时演示

甚至简化使用typeid():

#include <iostream>
#include <string>
#include <typeinfo>

template<typename Derived>
class Driver {
public:
    Driver() {
         std::cout << typeid(*this).name() << std::endl;
         std::cout << typeid(Derived).name() << std::endl;
    }  
};

class SpecificDriver : public Driver<SpecificDriver> {
};

int main() {
    SpecificDriver sd;
}

实时演示

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