通过引用/指针返回静态局部变量

Question

我很困惑为什么静态unordered_map被清除如果我通过引用得到它但不是如果我通过指针得到它...(你可以在这里执行代码:http://cpp.sh/4ondg)

是因为当引用超出范围时,它的析构函数会被调用吗？如果是这样,那么第二个获取功能会得到什么？

class MyTestClass {
    public:
    static std::unordered_map<int, int>& getMap() {
        static std::unordered_map<int, int> map;
        return map;
    }
    static std::unordered_map<int, int>* getMapByPointer() {
        static std::unordered_map<int, int> map;
        return &map;
    }

};


int main()
{
    // By reference
    {
        auto theMap = MyTestClass::getMap();
        std::cout << theMap.size() << std::endl;
        theMap[5] = 3;
        std::cout << theMap.size() << std::endl;
    }
    {
        auto theMap = MyTestClass::getMap();
        std::cout << theMap.size() << std::endl;
        theMap[6] = 4;
        std::cout << theMap.size() << std::endl;
    }

    // By pointer
    {
        auto theMap = MyTestClass::getMapByPointer();
        std::cout << theMap->size() << std::endl;
        (*theMap)[5] = 3;
        std::cout << theMap->size() << std::endl;
    }
    {
        auto theMap = MyTestClass::getMapByPointer();
        std::cout << theMap->size() << std::endl;
        (*theMap)[6] = 4;
        std::cout << theMap->size() << std::endl;
    }
}

Answer 1

当你这样做

auto theMap = MyTestClass::getMap();

theMap推断的类型是std::unordered_map<int, int>- 不是参考.因此,函数调用返回的引用被复制到局部变量中theMap; 修改时theMap,您只修改此副本.

要存储引用,请将其声明为auto&:

auto& theMap = MyTestClass::getMap();

然后,您将按预期修改原始对象.