c++ - 指针和地址

Question

我只是稍微思考一下，然后我看到了一些对我来说没有意义的东西。该应用程序几乎要求您输入一个类的名称，如果它在 map 中找到它，它就会创建该类的一个实例并打印出该对象的地址。

我第一次输入 Derived1，它生成 1 个地址，例如 (0x00001)，第二次我输入相同的名称，它显示另一个地址 (0x00002)，第三次它再次使用相同的地址。

为什么我只得到第二个对象的新地址而没有得到其余对象的新地址？

这绝不是问题，我只是好奇。

#include <iostream>
#include <memory>
#include <map>
#include <string>
#include <vector>
#include <typeinfo>

class Base
{
protected:
    int id;
public:
    Base(int _id)
        :id(_id)
    {
        std::cout << "Im the base class, and this is my address: " << this << std::endl;
    }
    virtual void PrintMyAddress() const = 0;
    void PrintId()
    {
        std::cout << "This is my ID: " << this->id << std::endl;
    }
    virtual ~Base(){}
};

class Derived1 : public Base
{
public:
    Derived1(int _id) : Base(_id){}
    void PrintMyAddress() const override
    {
        std::cout << "I'm derived 1, this is my address: " << this << std::endl;
    }
    ~Derived1(){}
};

class Derived2 : public Base
{
public:
    Derived2(int _id) : Base(_id){}
    virtual void PrintMyAddress() const override
    {
        std::cout << "I'm derived 2, this is my address: " << this << std::endl;
    }

    virtual ~Derived2(){}
};

class Derived3 : public Derived2
{
public:
    Derived3(int _id) : Derived2(_id){}
    void PrintMyAddress() const override
    {
        std::cout << "I'm derived 3, this is my address: " << this << std::endl;
    }

    ~Derived3(){}
};

class Generate
{
private:
    typedef std::unique_ptr<Base> (Generate::*SomeFunction)(int);
    std::map<std::string, SomeFunction> listFunction;

public:
    Generate()
    {
        this->MapClasses();
    }

    template <typename T>
    std::unique_ptr<Base> CreateDerived(int _id)
    {
        std::unique_ptr<Base> tmp(new T(_id));
        return std::move(tmp);
    }

    void MapClasses()
    {
        this->listFunction["Derived1"] = &Generate::CreateDerived<Derived1>;
        this->listFunction["Derived2"] = &Generate::CreateDerived<Derived2>;
        this->listFunction["Derived3"] = &Generate::CreateDerived<Derived3>;
    }

    std::unique_ptr<Base> Class(std::string _name)
    {
        if (this->listFunction.find(_name) != this->listFunction.end())
        {
            return std::move((this->*this->listFunction[_name])(rand() % 100));
        }
        else
        {
            std::unique_ptr<Base> itsnull;
            return std::move(itsnull);
        }
    }

    std::map<std::string, SomeFunction>& GetFunctionList()
    {
        return this->listFunction;
    }
};


int main()
{
    Generate gen;
    bool run = true;
    while (run)
    {
        std::cout << std::endl << "What class do you want to generate? Type in the name of the class: " << std::endl << std::endl;
        std::string className;
        std::cin >> className;
        if (className == "exit")
        {
            run = false;
            continue;
        }

        auto genclass = gen.Class(className);
        if (genclass.get())
        {
            genclass->PrintMyAddress();
            genclass->PrintId();
            std::cout << "This is my TypeID: " << typeid(*genclass).name() << std::endl;
        }
        else
        {
            std::cout << "Class couldn't be created because it doesn't exist..." << std::endl;
        }
    }

    std::cin.ignore(std::numeric_limits<std::streamsize>::max(),'\n');
    return 0;
}

这是一个类似的示例，但始终显示相同的地址:

for (int i = 0; i < 10; ++i)
{
    int* test = new int;
    std::cout << test << std::endl;
    delete test;
}

编译器是否在第二次创建后优化了一切？

Answer 1

您的代码基本上创建了一个新对象，然后在每个循环中将其删除。

对象的地址取决于堆把它放在哪里。堆可以自由地重用最近释放的空间(如果不这样做，那将是愚蠢的)，因此您很可能会得到重复的地址。