我有一个需要递归传递一个闭合参数的函数
use std::cell::RefCell;
use std::rc::Rc;
pub struct TreeNode {
val: i32,
left: Option<Rc<RefCell<TreeNode>>>,
right: Option<Rc<RefCell<TreeNode>>>,
}
pub fn pre_order<F>(root: Option<Rc<RefCell<TreeNode>>>, f: F)
where
F: FnOnce(i32) -> (),
{
helper(&root, f);
fn helper<F>(root: &Option<Rc<RefCell<TreeNode>>>, f: F)
where
F: FnOnce(i32),
{
match root {
Some(node) => {
f(node.borrow().val);
helper(&node.borrow().left, f);
helper(&node.borrow().right, f);
}
None => return,
}
}
}
这不起作用:
error[E0382]: use of moved value: `f`
--> src/lib.rs:23:45
|
22 | f(node.borrow().val);
| - value moved here
23 | helper(&node.borrow().left, f);
| ^ value used here after move
|
= note: move occurs because `f` has type `F`, which does not implement the `Copy` trait
error[E0382]: use of moved value: `f`
--> src/lib.rs:24:46
|
23 | helper(&node.borrow().left, f);
| - value moved here
24 | helper(&node.borrow().right, f);
| ^ value used here after move
|
= note: move occurs because `f` has type `F`, which does not implement the `Copy` trait
如果我尝试改变从f
f: F
的类型f: &F
我得到的编译器错误
error[E0507]: cannot move out of borrowed content
--> src/lib.rs:22:17
|
22 | f(node.borrow().val);
| ^ cannot move out of borrowed content
我怎样才能解决这个问题?
我打电话这样的功能:
let mut node = TreeNode::new(15);
node.left = Some(Rc::new(RefCell::new(TreeNode::new(9))));
let node_option = Some(Rc::new(RefCell::new(node)));
pre_order(node_option, |x| {
println!("{:?}", x);
});
FnOnce
是最最普通的功能限制。然而,这意味着你的代码必须为所有可能的功能,包括那些消耗他们的环境中工作。这就是为什么它被称为FnOnce
:你知道它的唯一的事情是,它可以至少一次把它叫做 - 但不一定更多。里面pre_order
我们只能假设是什么一切可能的F
如此:它可以被调用一次。
如果您更改为Fn
或FnMut
,排除消耗他们的环境封闭,你就可以调用它多次。 FnMut
是下一个最普通的功能特点,所以最好是接受,而不是Fn
,以确保您能接受的大多数功能:
pub fn pre_order<F>(root: Option<Rc<RefCell<TreeNode>>>, mut f: F)
where
F: FnMut(i32),
{
helper(&root, &mut f);
fn helper<F>(root: &Option<Rc<RefCell<TreeNode>>>, f: &mut F)
where
F: FnMut(i32),
{
match root {
Some(node) => {
f(node.borrow().val);
helper(&node.borrow().left, f);
helper(&node.borrow().right, f);
}
None => return,
}
}
}