c++ - 将类的每个实例添加到列表对象的构造函数

Question

C++ 和 OOP 新手。我正在尝试找出列表和迭代，因此我创建了以下示例代码。我创建了几个 Thing 对象，但我想确保在创建 Thing 时，其构造函数将其添加到列表“things”(在列表对象内)，以便我可以跟踪 Thing 的每个实例。然后，在 main() 的底部，我迭代事物列表。有没有更好的方法来执行此操作，或者您能否指出如何在我的 Thing 构造函数中执行此操作？谢谢!!

#include <iostream>
#include <list>

class Thing;

class Lists
{
public:
    std::list<Thing> things;
    Lists() {
        std::cout << "List object with list 'things' created" << std::endl;
    }
};

class Thing
{
public:
    int howMuch, pointer;
    Thing(int x, Lists* y)
    {
        howMuch = x;
        y->things.push_back(this);
    }
};

int main()
{

    //create the object that holds the list of things
    Lists lists;

    //make some objects, and pass a pointer of the lists to the constructor
    Thing thingA(123, &lists);
    Thing thingB(456, &lists);

    for (std::list<Thing>::iterator it = lists.things.begin(); it != lists.things.end(); ++it)
        std::cout << "test" << it->howMuch << std::endl;

    return 0;
}

Answer 1

您可以使用静态字段 _things 将创建的项目存储在 Thing 类本身内:

#include <iostream>
#include <list>

class Thing
{
    static std::list<Thing> _things;

public:
    int howMuch, pointer;
    Thing(int x) : howMuch(x)
    {
        _things.push_back(*this);
    }

    static std::list<Thing> getAllThings()
    {
        return _things;
    }
};


std::list<Thing> Thing::_things;

int main()
{
    Thing thingA(123);
    Thing thingB(456);

    auto allThings = Thing::getAllThings();

    for (auto it = allThings.begin(); it != allThings.end(); ++it)
        std::cout << "test " << it->howMuch << std::endl;

    return 0;
}