指向基类的数组，填充派生类 [英] Pointer to array of base class, populate with derived class

问题描述

如果我有一个基类，只有虚拟方法，并从2派生类
基类，与实现这些虚拟的方法。

If I have a base class, with only virtual methods and 2 derived classes from the base class, with those virtual methods implemented.

我如何：

 // causes C2259
 BaseClass* base = new BaseClass[2];

 BaseClass[0] = new FirstDerivedClass;
 BaseClass[1] = new SecondDerivedClass;

或

// causes "base is being used without being initialized"
BaseClass* base;
// causes CC59 again
BaseClass* base = new BaseClass;

base[0] = FirstDerivedClass();
base[1] = SecondDerivedClass();

（或类似的东西）

(or something similar)

......这样我就可以通过 DerivedClass 访问的BaseClass 取值方法，
但指针和指针 DerivedClass S'

...so that I can access the BaseClasss methods through the DerivedClass, but by pointer and the pointer is an array of DerivedClasss?

推荐答案

您数组类型错误的：它存储的BaseClass 对象的实例的代替的指针的给他们。由于的BaseClass 似乎是抽象的，编译器会抱怨它不能缺省方式构造的实例来填补你的阵列。

Your array is of the wrong type: it stores BaseClass object instances instead of pointers to them. Since BaseClass seems to be abstract, the compiler complains that it cannot default-construct instances to fill your array.

即使的BaseClass 是不是抽象的，使用数组多态是<一个href=\"https://www.securecoding.cert.org/confluence/display/cplusplus/ARR39-CPP.+Do+not+treat+arrays+polymorphically\">big没有没有在C ++中，所以你应该在任何情况下做不同的事情。

Even if BaseClass were not abstract, using arrays polymorphically is a big no-no in C++ so you should do things differently in any case.

通过改变code来解决这个问题：

Fix this by changing the code to:

BaseClass** base = new BaseClass*[2];

BaseClass[0] = new FirstDerivedClass;
BaseClass[1] = new SecondDerivedClass;

这是说，大部分的时间是preferable使用的的std ::矢量不是纯数组和智能指针（如的std :: shared_ptr的），而不是愚蠢的指针。使用这些工具，而不是手工编写code的会照顾问题的主机透明地在一个非常小的运行成本。

That said, most of the time it is preferable to use std::vector instead of plain arrays and smart pointers (such as std::shared_ptr) instead of dumb pointers. Using these tools instead of manually writing code will take care of a host of issues transparently at an extremely small runtime cost.

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