c++ - 派生函数图

Question

我有基类和派生类基础:

class Person{
public:
    Person(string name , int age ){
        this -> name = name;
        this -> age  = age;
    }
    virtual void getInfo(){
        cout << "Person " << name << " " << age;
    }

protected:
    string name;
    int age;
};

派生:

class Kid : public Person{
public:
    Kid(string name, int age):Person(name,age){};
    virtual void getInfo( ){
        cout << "Kid " << name << " " << age;
    }
};
 class Adult : public Person{
    public:
        Adult(string name, int age):Person(name,age){};
        virtual void getInfo( ){
            cout << "Adult " << name << " " << age;
        }
    };

当我做类似的事情时

map<string ,Person*> m;
Person *one;
Person *three;
Kid two("Mathew",15);
Adult four("JOhn",55);
three = &four;
one = &two;
m["first"] = one;
m["second"] = three;
for( auto &x : m )
   x.second-> getInfo();
return 0;

它很好地打印了它应该打印的信息//“Adult”代表成人类(class)，“Kid”代表 child 类(class)

但是，当我编辑类并将 map 移动到基类时。例如，创建 Add 方法。

 class Person{
    public:
        Person(string name , int age ){
            this -> name = name;
            this -> age  = age;
        }
        virtual void getInfo(){
            cout << "Person " << name << " " << age;
        }
        void add( string name , Person a){
            Person *one = &a;
            m[ name ] = one;
        }
        void print(){
            for( auto &x: m )
                 x.second -> getInfo()
        }
    protected:
        string name;
        int age;
       map< string , Person*> m;
    };



   Person one("John", 25);
   one.add("first",Kid("Mathew",15));
   one.add("second",Adult("Leo",55));
   one.print();

它抛出 seg fault ，为什么会这样？它的实现与 using of 方法基本相同。是什么导致段错误？有办法解决吗？

//编辑

我尝试使用 unique_ptr 重新定义 map 作为

map< string ,  unique_ptr<Person>> m;
   AddField (string name , Person   a ){

            m[name] = ( unique_ptr<Person> (a)); 
            return *this;
        }

或

  properties[name] = unique_ptr<Person> ( new Person( a ));

或

 AddField (string name , Person   a ){

        CData *one = unique_ptr<Person>(new Person(a));
        m[name] = one ;
        return *this;
    }

我没有使用唯一/共享 ptr 的经验。这扔了

‘std::unique_ptr’ to ‘Person*’

Answer 1

首先，让我们了解 map 实际存储的内容:

map< string , Person*> m;

这个映射将一个字符串绑定(bind)到一个Person*，这是一个指向一个人的内存地址。该映射实际上并不存储 person 实例，仅存储其内存地址。

现在让我们分析一下你的两种情况。

 map<string ,Person*> m;

// Allocating space for two memory addresses on the stack:
Person *one;
Person *three;

// Creating two instances on the stack:
Kid two("Mathew",15);
Adult four("JOhn",55);

// Setting the pointers to the memory addresses of the instances on the stack:
three = &four;
one = &two;

m["first"] = one;
m["second"] = three;
for( auto &x : m )
   x.second-> getInfo();
return 0;

// End of the function: now all objects are destroyed in reverse order.

实例 two 和 four 存在于堆栈中，并在作用域结束时被销毁(在 return 0; 之后)。获取它们的内存地址并将它们存储到 one 和 three 中就可以了，因为指针的生命周期将超过 two 和 four .

---

Person one("John", 25);

// Creating a Kid instance without a name (temporary).
// The Kid instance goes out of scope immediately and is destroyed:
one.add("first",Kid("Mathew",15));

// Creating an Adult instance without a name (temporary).
// The Adult instance goes out of scope immediately and is destroyed:
one.add("second",Adult("Leo",55));

one.print();

这里的问题是您正在创建的实例被销毁得太快了。您需要正确管理它们的生命周期，以确保您插入到 map 中的内存地址不会比指向的内存位置中的数据更长寿。

另一个主要问题是您按值接受add 的Person a 参数。这将创建传递实例的拷贝。

// Copy the passed instance:
void add( string name , Person a){
        // Take memory address of the copied instance:
        Person *one = &a;
        m[ name ] = one;

        // The copied instance is destroyed here!
    }

您应该将a 作为引用，以避免复制和object slicing :

void add( string name , Person& a){
        m[ name ] = &a;
    }

---

这是一个工作示例，在更正 a 的签名之后:

Kid k("Mathew",15);
Adult a("Leo",55);

// k and a will outlive one.

Person one("John", 25);
one.add("first", k);
one.add("second", a);

one.print();

// one gets destroyed.
// a gets destroyed.
// k gets destroyed.