指针功能的深层复制 [英] Deep copy of a pointer function

问题描述

我有一个类（myClass），该类的指针可以像这样充当私有成员：

I have a class (myClass) that has a pointer to function as private member like this:

void (gi::myClass::*ptrFunction) (int  , int  , vector<int> & , vector<int> & , vector<int> &);

在运行时（带有新对象）创建myClass的对象（* myClass ptrObject）时，此指针在构造函数中使用函数的地址初始化。
之后，使用上面创建的相同对象，我调用了另一个函数。
在此函数中，我必须创建调用对象的深层副本：

When you create at runtime (with new) a object (a *myClass ptrObject) of myClass this pointer is inizialited in the constructor with the address of a function. After with the same object created above I call another function. In this function I must create a deep copy of the calling object:

gi::myClass *ptrObject2 = new myClass(....);

上面我称一个构造函数执行复制，但是它不初始化成员* ptrFunction * ptrObject2
我的问题是：
如何将指针函数从* ptrObject复制到ptrObject2？
（我的编译器是C ++ 11）

Above I call a constructor that do the copy but it don't initialize the member *ptrFunction of *ptrObject2 My question is: How do I do a deep copy of the pointer function from the *ptrObject to ptrObject2? (My compiler is C++11)

推荐答案

没有深拷贝函数指针。您没有复制任何数据，只是将引用复制到v表中。这样：

There is no deep-copy of a function pointer. You are not copying any data, just a reference into the v-table. As such:

ptrObject2->ptrFunction = ptrObject->ptrFunction

很好。

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