Pointer to array of base class, populate with deri

问题:

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

How do I:

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

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

or:

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

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

(or something similar)

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

回答1:

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.

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.

Fix this by changing the code to:

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

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

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.

回答2:

It is C++ use std::vector instead of simple array:

std::vector<BaseClass*> base;
base.push_back(new FirstDerivedClass());
base.push_back(new SecondDerivedClass());

As Kerrek SB noticed safest method is to use std::unique_ptr:

std::vector<std::unique_ptr<BaseClass> > base;
base.push_back( std_unique_ptr<BaseClass>(new FirstDerivedClass()) );
base.push_back( std_unique_ptr<BaseClass>(new SecondDerivedClass()) );

回答3:

If your BaseClass contains pure virtual methods, this will fail to compile :

BaseClass* base = new BaseClass[2];

If it doesn't, you are going to get memory leak.

In c++, this is done by using std::vector or std::array, with some kind of smart pointer. For example :

std::vector< std::shared_ptr< BaseClass > > arr( 2 );
arr[0].reset( new FirstDerivedClass() );
arr[1].reset( new SecondDerivedClass() );

回答4:

This was the answer (from Rubby)

BaseClass* Base[2];

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

回答5:

Define a pointer array , the pointer type is BaseClass. And assign the pointer to the derivedclass to the elements of the array. just like:

BaseClass* base [2];
base[0] = new FirstDerivedClass;
base[1] = new SecondDerivedClass;