C++ 实现类根据它使用的字符串构造另一个类的实例

Question

我需要实现一个抽象类、它的三个具体子类，该类的目标是创建这三个类实例之一和三个类的最后一个类执行器。要求是 c++98，而不是使用 if/elseif/else 来构造类实例，就像我在 Maker 类方法 make Form 中所做的那样。 if/elseif/else 我需要避免什么机制？

例如:

测试.h

#ifndef TEST_H
#define TEST_H
#include <iostream>

class Executor {
private:
    const std::string name;
public:
    Executor(const std::string &name = "") {};

    const std::string getname() const {return name;}
};

class BForm {
private:
    const std::string _name;
public:
    BForm(const std::string &name = "") : _name(name) {};
    virtual ~BForm() {};

    virtual void execute(const Executor &src) = 0;

    const std::string getname() {return _name;}
    virtual const std::string gettarget() = 0;
};

class Form1 : public BForm{
private:
    std::string _target;
public:
    Form1(const std::string &target = "") : BForm("form1"), _target(target) {};
    virtual ~Form1() {};

    virtual void execute(const Executor &src) {
        std::cout << src.getname() << " exec form1 target:" << _target << std::endl;
    }

    virtual const std::string gettarget() {return _target;}
};

class Form2 : public BForm {
private:
    std::string _target;
public:
    Form2(const std::string &target = "") : BForm("form2"), _target(target) {};
    virtual ~Form2() {};

    virtual void execute(const Executor &src) {
        std::cout << src.getname() << " exec form2 target:" << _target << std::endl;
    };
    virtual const std::string gettarget() {return _target;}
};

class Form3 : public BForm {
private:
    std::string _target;
public:
    Form3(const std::string &target = "") : BForm("form3"), _target(target) {};
    virtual ~Form3() {};

    virtual void execute(const Executor &src) {
        std::cout << src.getname() << " exec form3 target:" << _target << std::endl;
    };
    virtual const std::string gettarget() {return _target;}
};

class Maker {
public:
    BForm *makeForm(const std::string &name, const std::string &target)
    {
        /* need to avoid */
        if (name == "form1")
            return new Form1(target);
        else if (name == "form2")
            return new Form2(target);
        else
            return new Form3(target);
    }
};

#endif

main.cpp

#include "test.h"

int main() {
    Maker maker;
    BForm *form;
    Executor exec("executor");

    form = maker.makeForm("form1", "A");
    std::cout << form->getname() << " " << form->gettarget() << std::endl;
    form->execute(exec);
    delete form;
    return (0);
}

Answer 1

你可以定义一个指向函数的指针，然后使用从字符串到这个类型(指向函数的指针)的映射。然后使用带有索引器语法的参数来访问指向函数的正确指针。

这是一个例子:

#include <iostream>
#include <map>

// The class definitions with a virtual function hello() common to all
class Base { public: virtual void hello() = 0; };
class Derived1 : public Base { public: void hello() { std::cout << "Derived1"; } };
class Derived2 : public Base { public: void hello() { std::cout << "Derived2"; } };

// The object making functions
Base* Maker1() { return new Derived1; }
Base* Maker2() { return new Derived2; }

int main()
{
    // In C++98, without auto, it's worthwhile to typedef complicated types.
    // The first one is a function type returning a pointer to Base...
    typedef Base* MakerT();
    // ... the second one is a map type projecting strings to such function pointers
    typedef std::map<std::string, MakerT*> StrToMakerT;

    /// The actual map projecting strings to maker function pointers
    StrToMakerT strToMaker;
    // Fill the map
    strToMaker["D1"] = &Maker1;
    strToMaker["D2"] = &Maker2;

    // user input
    std::string choice;

    // as long as output works, input works, and the user didn't say "Q":
    while (std::cout << "Please input 'D1' or 'D2' or 'Q' for quit: " 
        && std::cin >> choice 
        && choice != "Q")
    {
        // Prevent adding new entries to the map foir unknown strings
        if (strToMaker.find(choice) != strToMaker.end())
        {
            // Simply look the function up again, the iterator type is too 
            // cumbersome to write in C++98
            Base* b = (*strToMaker[choice])();
            b->hello();
            std::cout << '\n';
            delete b;
        }
        else
        {
            std::cout << "Didn't find your choice, try again.\n";
        }
    }
    std::cout << "Thank you, good bye\n";
}