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fix: 修复 space.RoomManager.AssumeControl 函数编译错误的问题

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kercylan98 2023-12-23 18:12:07 +08:00
parent f52d73e20e
commit 3f099e6f8e
1 changed files with 1 additions and 1 deletions
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game/space/room_manager.go
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@ -25,7 +25,7 @@ type RoomManager[EntityID comparable, RoomID comparable, Entity generic.IdR[Enti
// AssumeControl 将房间控制权交由 RoomManager 接管，返回 RoomController 实例 // AssumeControl 将房间控制权交由 RoomManager 接管，返回 RoomController 实例
// - 当任何房间需要被 RoomManager 管理时，都应该调用该方法获取到 RoomController 实例后进行操作 // - 当任何房间需要被 RoomManager 管理时，都应该调用该方法获取到 RoomController 实例后进行操作
// - 房间被接管后需要在释放房间控制权时调用 RoomController.Destroy 方法，否则将会导致 RoomManager 一直持有房间资源 // - 房间被接管后需要在释放房间控制权时调用 RoomController.Destroy 方法，否则将会导致 RoomManager 一直持有房间资源
func (rm *RoomManager[EntityID, RoomID, Entity, Room]) AssumeControl(room Room, options ...*RoomControllerOptions) *RoomController[EntityID, RoomID, Entity, Room] { func (rm *RoomManager[EntityID, RoomID, Entity, Room]) AssumeControl(room Room, options ...*RoomControllerOptions[EntityID, RoomID, Entity, Room]) *RoomController[EntityID, RoomID, Entity, Room] {
controller := newRoomController(rm, room, mergeRoomControllerOptions(options...)) controller := newRoomController(rm, room, mergeRoomControllerOptions(options...))
rm.OnRoomAssumeControlEvent(controller) rm.OnRoomAssumeControlEvent(controller)
return controller return controller