复制对象

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以下代码尝试复制对象并保留原始类型。 不幸的是它不工作（每个复制的对象将成为 Super ，而不是与它的原来的类相同）。

请注意， copySuper（const Super& givenSuper）不应该知道任何关于 Super 。

有可能做这样的副本吗？或者我必须更改 copySuper ？

的定义

#include < string> #include< iostream> class Super { public： Super（）{}; virtual〜Super（）{}; virtual std :: string toString（）const { returnI'm Super！ } }; class Special：public Super { public： Special（）{}; virtual〜Special（）{}; virtual std :: string toString（）const { returnI'm Special！ } }; Super * copySuper（const Super& givenSuper） { Super * superCopy（new Super（givenSuper））; return superCopy; } int main（） {特殊; std :: cout<< special.toString（）<< std :: endl; std :: cout<< ---<< std :: endl; Super * specialCopy = copySuper（special）; std :: cout<< specialCopy-> toString（）<< std :: endl; return 0; } //所需输出： //＃我是特殊！ //＃--- //＃我是特别的！ // //实际输出： //＃我是Sepcial！ //＃--- //＃我是超级！

解决方案

尝试：

class Super { public： Super（）; // regular ctor Super（const Super& _rhs）; //复制构造函数 virtual Super * clone（）const {return（new Super（* this））;}; }; // eo class Super class特殊：public Super { public： Special（）：Super（）{}; Special（const Special& _rhs）：Super（_rhs）{}; virtual Special * clone（）const {return（new Special（* this））;}; }; // eo class Special

注意，我们已经实现了一个clone

<$ c $

c> Super * s = new Super（）; Super * s2 = s-> clone（）; // copy of s Special * a = new Special（）; Special * b = a-> clone（）; //复制

编辑：正如其他评论员指出的， * this ，而不是此。

EDIT2：我真的不应该邮局这么快，当在中间工作。用于协变返回类型的Special :: clone（）的修改返回类型。

The following code tries to copy an object and keep the original type. Unfortunately it does not work (every copied object will become a Super instead of being of the same class as its original).

Please note that copySuper(const Super& givenSuper) should not know anything about the subclasses of Super.

Is it possible to do such a copy? Or do I have to change the definition of copySuper ?

#include <string> #include <iostream> class Super { public: Super() {}; virtual ~Super() {}; virtual std::string toString() const { return "I'm Super!"; } }; class Special : public Super { public: Special() {}; virtual ~Special() {}; virtual std::string toString() const { return "I'm Special!"; } }; Super* copySuper(const Super& givenSuper) { Super* superCopy( new Super(givenSuper) ); return superCopy; } int main() { Special special; std::cout << special.toString() << std::endl; std::cout << "---" << std::endl; Super* specialCopy = copySuper(special); std::cout << specialCopy->toString() << std::endl; return 0; } //Desired Output: // # I'm Special! // # --- // # I'm Special! // //Actual Output: // # I'm Sepcial! // # --- // # I'm Super!

解决方案

Try this:

class Super { public: Super();// regular ctor Super(const Super& _rhs); // copy constructor virtual Super* clone() const {return(new Super(*this));}; }; // eo class Super class Special : public Super { public: Special() : Super() {}; Special(const Special& _rhs) : Super(_rhs){}; virtual Special* clone() const {return(new Special(*this));}; }; // eo class Special

Note that we have implemented a clone() function that Special (and any other derivative of Super) overrides to create the correct copy.

e.g:

Super* s = new Super(); Super* s2 = s->clone(); // copy of s Special* a = new Special(); Special* b = a->clone(); // copy of a

EDIT: As other commentator pointed out, *this, not this. That'll teach me to type quickly.

EDIT2: Another correction.

EDIT3: I really should not post so quickly when in the middle of work. Modified return-type of Special::clone() for covariant return-types.