rust - 为什么调用 FnOnce 关闭是一个 Action ？

Question

我正在尝试将一个闭包传递给一个函数，该函数随后会在该函数的范围内改变传递给它的某些内容。根据我目前对 Rust 的理解，它应该看起来像这样:

pub fn call_something(callback: &FnOnce(&mut Vec<i32>)) {
    let mut my_vec = vec![0, 1, 2, 3, 4];
    callback(&mut my_vec);
}

这会导致这些错误:

error[E0161]: cannot move a value of type dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>): the size of dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>) cannot be statically determined
 --> src/lib.rs:3:5
  |
3 |     callback(&mut my_vec);
  |     ^^^^^^^^

error[E0507]: cannot move out of borrowed content
 --> src/lib.rs:3:5
  |
3 |     callback(&mut my_vec);
  |     ^^^^^^^^ cannot move out of borrowed content

为什么调用 FnOnce 是一个 Action ？我在这里缺少什么？

Answer 1

Why is calling a FnOnce a move?

因为那是 the definition of what makes a closure FnOnce :

extern "rust-call" fn call_once(self, args: Args) -> Self::Output
//                              ^^^^

对比 FnMut和 Fn :

extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output
//                             ^^^^^^^^^

extern "rust-call" fn call(&self, args: Args) -> Self::Output
//                         ^^^^^

另见:

• When does a closure implement Fn, FnMut and FnOnce?
• "cannot move a value of type FnOnce" when moving a boxed function
• Cannot move out of borrowed content

---

你可能想要

pub fn call_something(callback: impl FnOnce(&mut Vec<i32>))

或

pub fn call_something<F>(callback: F)
where
    F: FnOnce(&mut Vec<i32>),

这些是相同的。它们都拥有闭包的所有权，这意味着您可以调用闭包并在流程中使用它。