问题描述
我有一个指针State *state;
,它是其他状态继承的类。
而我做state = new GameState;
如果我将对象状态指向的内容从GameState更改为其他内容,是否需要以某种方式释放内存?
解决方法
如果您的意思是State* state
不拥有该对象,则需要执行以下操作:
// create states
State* game_state = new GameState();
State* menu_state = new MenuState();
// set current state
State* state = game_state;
// ... do somethings ...
// delete states
delete game_state;
game_state = nullptr;
delete menu_state;
menu_state = nullptr;
state = nullptr;
如果要编写现代代码,请使用:
// create states
auto game_state = make_unique<GameState>();
auto menu_state = make_unique<MenuState>();
// set current state
State* state = game_state.get();
// ... do somethings ...
// no need to delete states
state = nullptr;
您还可以结合使用shared_ptr和weak_ptr。
如果不使用智能指针,则需要在拥有该对象的任何对象上调用delete
。
如果State* state
确实拥有对象,那么您需要这样做:
// create current state
State* state = new GameState();
// ... do somethings ...
// switch state
delete state;
state = new MenuState();
// ... do somethings ...
// delete state
delete state;
state = nullptr;
并带有智能指针:
// create current state
unique_ptr<State> state = make_unique<GameState>();
// ... do somethings ...
// switch state (old state automatically deleted)
state = make_unique<MenuState>();
// ... do somethings ...
// delete state
state = nullptr;
,
这是一个简单的规则:如果您写new
,请为其写一个delete
。如果在堆上分配内存,则必须释放它,否则会发生内存泄漏,这就是没有内置垃圾收集功能的语言的本质。
因此,如果您写:
State *state;
state = new GameState;
您还必须这样做:
delete state;
此外,考虑使用Smart pointers,它们更安全,如果您想编写现代 C ++,则应该认真研究它们。