如何在不重复代码的情况下在子类中获取相同的方法代码？

Question

我有以下代码：

#include <exception>

class Exception : public std::exception {
private:
    const char* MESSAGE = "Exception"

public:
    inline virtual const char* what() const throw() {
        return this->MESSAGE;
    }
};

class ShoulderROMException : public Exception {
private:
    typedef Exception super;
    const char* MESSAGE = "ShoulderROM exception";

protected:
    static const int MAX_MESSAGE_LENGTH = 200;
    mutable char composedMessage[ShoulderROMException::MAX_MESSAGE_LENGTH];

public:
    virtual const char* what() const throw() {
        strcpy(this->composedMessage, super::what());
        strcat(this->composedMessage, " -> ");
        strcat(this->composedMessage, this->MESSAGE);
        return this->composedMessage;
    }
};

class KinectInitFailedException : public ShoulderROMException {
private:
    typedef ShoulderROMException super;
    const char* MESSAGE = "Kinect initialization failed."

public:
    virtual const char* what() const throw() {
        strcpy(this->composedMessage, super::what());
        strcat(this->composedMessage, " -> ");
        strcat(this->composedMessage, this->MESSAGE);
        return this->composedMessage;
    }
};

这会生成如下所示的日志条目： Exception -> ShoulderROM exception -> Kinect initialization failed. 这正是我想要的，但我想避免明显的代码重复，而且似乎找不到（优雅的）方法来做到这一点。

如果有人能帮助我，那就太好了。 :)

此致，星际宝贝

Answer 1

通过一个通用的class实现它。我会像这样重写你的代码：

class Exception : public std::exception {
    static const char* MESSAGE = "Exception"
    static const int MAX_MESSAGE_LENGTH = 200;
    mutable char composedMessage[MAX_MESSAGE_LENGTH];

public:
    virtual const char* name() const throw() {
        return MESSAGE;
    }

    virtual const char* what() const throw() {
        strcpy(this->composedMessage, name());
        strcat(this->composedMessage, " -> ");
        strcat(this->composedMessage, this->MESSAGE);
        return this->composedMessage;
    }
};

class ShoulderROMException : public Exception {
    static const char* MESSAGE = "ShoulderROM exception";
public:    
    virtual const char* name() const throw() {
        return MESSAGE;
    }
};

class KinectInitFailedException : public ShoulderROMException {
    static const char* MESSAGE = "Kinect initialization failed."
public:
    virtual const char* name() const throw() {
        return MESSAGE;
    }
};

如果您不希望 Exception class 中有太多实现，请添加另一个 ShoulderROMException 和 KinectInitFailedException 都将继承的实现。

您的代码还有其他问题：MESSAGE 成员应该是 static，并且您处理字符串的方式不是很 C++ish.我还要补充一点，内联虚函数没有任何意义。

Answer 2

感谢您的帮助。它给了我灵感。借助一位同学的一些额外想法，我想出了这个非常有效的解决方案。 :)

#include <exception>

class Exception :
    public std::exception {
private:
    static const std::string MESSAGE = "Exception";

protected:
    std::string composedMessage;

public:
    Exception() :
    composedMessage(this->MESSAGE) {
    }

    virtual const char* what() const throw() {
        return this->composedMessage.c_str();
    }
};

class ShoulderROMException :
    public Exception {
private:
    static const std::string MESSAGE = "ShoulderROM exception";

public:
    ShoulderROMException() {
        this->appendMessage(this->MESSAGE);
    }

    virtual void appendMessage(std::string message) {
        this->composedMessage += " -> ";
        this->composedMessage += message;
    }
};

class KinectInitFailedException :
    public ShoulderROMException {
private:
    static const std::string MESSAGE = "Kinect initialization failed.";

public:
    KinectInitFailedException() {
        this->appendMessage(this->MESSAGE);
    }
};

我从错误的角度看问题：自上而下而不是自下而上。 ^^

感谢您的帮助和最诚挚的问候，星际宝贝

如何在不重复代码的情况下在子类中获取相同的方法代码？

How can I get the same method code in a subclass without code duplication?

c++

exception

code-duplication

duplicates

generalization