docs: Remove sub-headers from 公共方法 sections to follow SKILL.md规范
This commit is contained in:
@@ -18,8 +18,6 @@
|
||||
|
||||
## 公共方法
|
||||
|
||||
### 构造/析构
|
||||
|
||||
| 方法 | 描述 |
|
||||
|------|------|
|
||||
| `ResourceHandle() = default` | 默认构造空句柄 |
|
||||
@@ -27,40 +25,15 @@
|
||||
| `ResourceHandle(const ResourceHandle& other)` | 拷贝构造(自动增加引用) |
|
||||
| `ResourceHandle(ResourceHandle&& other) noexcept` | 移动构造 |
|
||||
| `~ResourceHandle()` | 析构函数(自动调用 Reset) |
|
||||
|
||||
### 赋值
|
||||
|
||||
| 方法 | 描述 |
|
||||
|------|------|
|
||||
| `ResourceHandle& operator=(const ResourceHandle& other)` | 拷贝赋值(自动管理引用) |
|
||||
| `ResourceHandle& operator=(ResourceHandle&& other) noexcept` | 移动赋值 |
|
||||
|
||||
### 资源访问
|
||||
|
||||
| 方法 | 描述 |
|
||||
|------|------|
|
||||
| `T* Get() const` | 获取裸指针 |
|
||||
| `T* operator->() const` | 通过指针访问资源成员 |
|
||||
| `T& operator*() const` | 解引用获取资源引用 |
|
||||
|
||||
### 状态查询
|
||||
|
||||
| 方法 | 描述 |
|
||||
|------|------|
|
||||
| `bool IsValid() const` | 检查句柄是否持有有效资源 |
|
||||
| `explicit operator bool() const` | 隐式布尔转换 |
|
||||
|
||||
### GUID 和类型
|
||||
|
||||
| 方法 | 描述 |
|
||||
|------|------|
|
||||
| `ResourceGUID GetGUID() const` | 获取资源的全局唯一标识符 |
|
||||
| `ResourceType GetResourceType() const` | 获取资源类型 |
|
||||
|
||||
### 资源释放
|
||||
|
||||
| 方法 | 描述 |
|
||||
|------|------|
|
||||
| `void Reset()` | 释放当前资源引用 |
|
||||
| `void Swap(ResourceHandle& other)` | 交换两个句柄的内容 |
|
||||
|
||||
|
||||
Reference in New Issue
Block a user