原因是
我仍然有点困惑Fn-FnMut-FnOnce是如何与其他特征一起工作的。我想从下面的遍历树结构的函数中消除Copy约束。
pub fn for_each<F>(&mut self, mut f: F)
where
F: FnMut(&mut Tree<T>) + Copy,
{
self.children.iter_mut().for_each(|c| c.for_each(f));
f(self);
}
原因是我试图将一个修改外部变量的闭包传递给for_each，而Copy阻止了这种情况(E0277(。然而，当我去掉Copy时，我会得到以下错误消息：
error[E0507]: cannot move out of `f`, a captured variable in an `FnMut` closure
--> src/tree.rs:34:58
|
30 | pub fn for_each<F>(&mut self, mut f: F)
| ----- captured outer variable
...
34 | self.children.iter_mut().for_each(|c| c.for_each(f));
| ^ move occurs because `f` has type `F`, which does not implement the `Copy` trait
error[E0382]: borrow of moved value: `f`
--> src/tree.rs:35:9
|
30 | pub fn for_each<F>(&mut self, mut f: F)
| ----- move occurs because `f` has type `F`, which does not implement the `Copy` trait
...
34 | self.children.iter_mut().for_each(|c| c.for_each(f));
| --- - variable moved due to use in closure
| |
| value moved into closure here
35 | f(self);
| ^ value borrowed here after move
|
help: consider further restricting this bound
|
32 | F: FnMut(&mut Tree<T>) + Copy,
| ^^^^^^
error: aborting due to 2 previous errors
我该如何解决这个问题？如果这能让它更容易的话，我也愿意把它变成迭代器。
for_each拥有f的所有权。
这意味着，一旦调用c.for_each(f)，就会失去对f的所有权，因此以后无法使用f。为了解决这个问题，可以将for_each函数改为引用，这样就可以像一样传递它
pub fn for_each<F>(&mut self, f: &mut F)
where
F: FnMut(&mut Tree<T>),
{
// Only a reference is passed into for_each here
self.children.iter_mut().for_each(|c| c.for_each(f));
// f can be used again, as reference is reborrowed implicitly
f(self);
}
或者你也可以(如果可能的话(把呼叫转移到f，就像一样
pub fn for_each<F>(&mut self, mut f: F)
where
F: FnMut(&mut Tree<T>),
{
// f is used, but ownership is kept
f(self);
// ownership of f can now be transferred, as there is no further use of it
self.children.iter_mut().for_each(|c| c.for_each(f));
}