c++ - 使用单例时内存泄漏

Question

我有一个实现单例设计模式的类。我知道有些人认为这不是个好主意，但它有很大帮助，

无论如何 - 我有内存泄漏，vlagrind 将我指向这些行:

_singleton = new Manager(); //Manager::instance() (Manager.cpp:18)

和

Manager::Manager() : _file(new ofstream), _tasks(new map<int, Task *>()),
        _idState(new map<int, int>()), _closing(false), _pending(false),
        _lock(new pthread_mutex_t), _endLock(new pthread_mutex_t),  _cond(new pthread_cond_t),
        _flushCond(new map<int, pthread_cond_t *>()), _attr(new pthread_attr_t) {
//The last line is line 25 in Manager::Manager

现在在 Manager 的析构函数中我不能显式删除它，因为它会创建一个愚蠢的循环(因为删除 _singleton 时将调用析构函数导致无限循环)。我该如何摆脱这种泄漏？谢谢!

附言这是 Valgrind 的输出:

==17823== 512 bytes in 1 blocks are definitely lost in loss record 2 of 2
==17823==    at 0x4C27297: operator new(unsigned long) (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==17823==    by 0x40151E: Manager::Manager() (Manager.cpp:25)
==17823==    by 0x4014DB: Manager::instance() (Manager.cpp:18)
==17823==    by 0x406475: initdevice(char*) (outputdevice.cpp:66)
==17823==    by 0x4061D5: main (driver.cpp:21)
==17823== 
==17823== LEAK SUMMARY:
==17823==    definitely lost: 512 bytes in 1 blocks
=    =17823==    indirectly lost: 0 bytes in 0 blocks
==17823==      possibly lost: 288 bytes in 1 blocks
==17823==    still reachable: 0 bytes in 0 blocks
==17823==         suppressed: 0 bytes in 0 blocks

另外:这是我创建管理器的代码:

Manager.h:
class Manager {
public:
    ~Manager();
    static Manager * instance();
private:
    Manager();
    static Manager * _singleton;
};

和实现:

Manager.cpp:
Manager * Manager::_singleton = 0;

Manager * Manager::instance() {
    if (!_singleton)
        _singleton = new Manager();
    return _singleton;
}

Answer 1

在 C++ 中实现单例的一种常见方法是使实例成为静态函数 std::unique_ptr<T>在实例 getter 内部，而不是类静态变量。这确保了在程序完成时调用析构函数，并允许您创建一个可以多态访问的实例，例如通过指向抽象基类的指针。

Scott Meyers 在 his "More Effective C++" book 中对此主题进行了很好的讨论。 .