Question

//test


#include <map>
#include <iostream>


using namespace std;

int main()
{
    map<int, string> Inventory;

    Inventory[1] = "Sword";
    Inventory[2] = "Armor";
    Inventory[3] = "Shield";

    map<int, string>::iterator iter;

    for (iter = Inventory.begin(); iter != Inventory.end(); ++iter)
        cout << (*iter).first << " - " << (*iter).second << "\n";

    iter == Inventory.find(2);

    Inventory.erase(iter);

    cout << "\n" << Inventory.count(2) << "\n";

    iter == Inventory.find(2);
    cout << "\n" << (*iter).first << " - " << (*iter).second.size() << "\n\n";


    if (Inventory.find(2) == Inventory.end())
        cout << "\nNot found.\n";

    for (iter = Inventory.begin(); iter != Inventory.end(); ++iter)
        cout << (*iter).first << " - " << (*iter).second << "\n";

    cout << "\n" << Inventory.size() << "\n";

    system("PAUSE");
    return EXIT_SUCCESS;

}

这是我运行程序时得到的结果：

1 - 剑

2 - 护甲

3 - 盾

1

2 - 1968772512

1 - 剑

3 - 盾

2

所以我有点困惑为什么键'2'没有被完全删除。

Inventory.count（2）返回1，这意味着密钥'2'仍然在某处的库存中挥之不去？

显然，在删除键'2'后，find（）仍会返回该键值的迭代器？

erase（）究竟是如何工作的？

Answer 1

iter == Inventory.find(2);

不是作业，我猜你打算这样做：

iter = Inventory.find(2);

对map :: erase（）真的很困惑

1 个答案: