基类对象的 C++ vector

Question

如果我有一个基类和一个派生类，如果我想将多个基类和/或派生类组合在一个容器中，我可以创建一个基类指针 vector 。

例子:

class base
{
}

class derived : public base
{
}

std::vector<base*> group;

但是有没有可能做到以下几点呢？

std::vector<base> group;

即:没有指针，需要 newing 和 deleteing？

编辑:2015 年我还不知道多态性，如果你在 2022 年阅读这个问题，那么值得搜索一些相关信息。

Answer 1

是的，您可以使用 vector<base> ＆编译器不会为此用法引发任何错误。但是，vector<base> 的问题是它未能实现 polymorphism .见下文:-

#include <iostream>
#include <vector>
using namespace std;

class base
{
    int x, id;
    static int i;
    public:
    base()
    {
        id = ++i;
        cout << "Base constructed: " << id << "\n";
    }
    base (const base &b)
    {
        id = ++i;
        cout << "Base copy constructed: " << id << "\n";
    }
    virtual int& getx()
    {
        cout << "Base getx() called\n";
        return x;
    }
    virtual ~base()
    {
        cout << "Base destroyed: " << id << "\n";
    }
};
int base :: i = 0; 

class derived : public base
{
    int x, id;
    static int j;
    public:
    derived()
    {
        id = ++j;
        cout << "Derived constructed: " << id << "\n";
    }
    derived (const derived& d)
    {
        id = ++j;
        cout << "Derived copy constructed: " << id << "\n";
    }
    virtual int& getx()
    {
        cout << "Derived getx() called\n";
        return x;
    }
    virtual ~derived()
    {
        cout << "Derived destroyed: " << id << "\n";
    }
};
int derived :: j = 0;

int main()
{
    vector<base> v;
    v.emplace_back(derived());
    v[0].getx() = 7;
    cout << "\n\n";
    for (int i=0; i<v.size(); ++i)
    cout << v[i].getx() <<"\n";
    cout << "\n\n";
    return 0;
}
/* Output :-
Base constructed: 1
Derived constructed: 1
Base copy constructed: 2
Derived destroyed: 1
Base destroyed: 1
Base getx() called


Base getx() called
7


Base destroyed: 2
*/

可以清楚的看到虽然对象是derived的copy constructor 都不是的 derived被称为 nor getx()一样的。所以使用vector<base>的目的实现多态是失败的。因此你不应该使用 vector<base> & 而更喜欢 vectors的 smart pointers或原始指针。