Rust：使用泛型特征作为特征参数

Question

如何在 Rust 中使用相关的泛型类型？

这是我得到的（只有第一行给我带来了麻烦）：

impl<G, GS> TreeNode<GS> where G: Game, GS: GameState<G>{
    pub fn expand(&mut self, game: &G){
        if !self.expanded{
            let child_states = self.data.generate_children(game);
            for state in child_states{
                self.add_child_with_value(state);
            }
        }
    }
}

GameState 是Game 的通用特征，self.data 实现了这种类型的GameState<Game>。编译器告诉我

error[E0207]: the type parameter `G` is not constrained by the impl trait, self type, or predicates
  --> src/mcts.rs:42:6
   |
42 | impl<G, GS> TreeNode<GS> where G: Game, GS: GameState<G>{
   |      ^ unconstrained type parameter

error: aborting due to previous error

但在我看来，我在 expand 函数和 G 需要属于 GS 的事实中都限制了 G。非常感谢任何帮助。

编辑： 以下是目前更多的定义

trait GameState<G: Game>: std::marker::Sized + Debug{
    fn generate_children(&self, game: &G) -> Vec<Self>;
    fn get_initial_state(game: &G) -> Self;
}

trait Game{}

struct TreeNode<S> where S: Sized{
    parent: *mut TreeNode<S>,
    expanded: bool,
    pub children: Vec<TreeNode<S>>,
    pub data: S,
    pub n: u32
}

impl<S> TreeNode<S>{
    pub fn new(data: S) -> Self{
        TreeNode {
            parent: null_mut(),
            expanded: false,
            children: vec![],
            data,
            n: 0
        }
    }

    pub fn add_child(&mut self, mut node: TreeNode<S>){
        node.parent = self;
        self.children.push(node);
    }

    pub fn add_child_with_value(&mut self, val: S){
        let new_node = TreeNode::new(val);
        self.add_child(new_node);
    }

    pub fn parent(&self) -> &Self{
        unsafe{
            &*self.parent
        }
    }

}

Answer 1

impl<G, GS> TreeNode<GS> where G: Game, GS: GameState<G>{
    // ...
}

问题在于G 不受约束，因此在此块中可能存在多个（可能是冲突的）实现，因为GS 可能为多个G 实现GameState<G>。参数G 不明确。

---

如果您想让GameState<G> 能够为多个G 实现，您应该将约束从impl 块移到方法中：

// note: G is now a type parameter of the method, not the impl block, which is fine
impl<GS> TreeNode<GS> {
    pub fn expand<G>(&mut self, game: &G) where G: Game, GS: GameState<G> {
        if !self.expanded{
            let child_states = self.data.generate_children(game);
            for state in child_states{
                self.add_child_with_value(state);
            }
        }
    }
}

---

如果您只想为单个G 实现GameState，则应将G 设为GameState 的关联类型，而不是泛型类型参数：

trait GameState: std::marker::Sized + Debug {
    type G: Game;
    fn generate_children(&self, game: &Self::G) -> Vec<Self>;
    fn get_initial_state(game: &Self::G) -> Self;
}

// note: now G is given by the GameState implementation instead of
//       being a free type parameter
impl<GS> TreeNode<GS> where GS: GameState {
    pub fn expand(&mut self, game: &GS::G){
        if !self.expanded{
            let child_states = self.data.generate_children(game);
            for state in child_states{
                self.add_child_with_value(state);
            }
        }
    }
}

Answer 2

不能根据TreeNode<GS>的类型确定G的具体类型；只有在调用expand 时才知道。请注意，expand 可以使用 不同 类型为 G 调用两次。

您可以通过约束方法的类型参数而不是整个实现块来表达这一点：

impl<GS> TreeNode<GS> {
    pub fn expand<G>(&mut self, game: &G)
    where
        G: Game,
        GS: GameState<G>,
    {
        if !self.expanded {
            let child_states = self.data.generate_children(game);
            for state in child_states {
                self.add_child_with_value(state);
            }
        }
    }
}

---

如果不能用不同的Gs 调用expand，那么这是您的建模问题。解决此问题的另一种方法是确保所有TreeNodes 都知道G 的类型。例如：

struct TreeNode<G, S>
where
    S: Sized,
{
    parent: *mut TreeNode<G, S>,
    expanded: bool,
    pub children: Vec<TreeNode<G, S>>,
    pub data: S,
    pub n: u32,
}

然后，一旦考虑了额外的类型参数，您的原始实现块应该可以按照编写的方式工作。