如何使用Rust中的引用将FnMut闭包传递给函数？ [英] How can I pass a FnMut closure to a function using a reference in Rust?

问题描述

我已经学习了如何将闭包参数传​​递给函数，因此可以两次调用 closure ：

I've learned how to pass a closure argument to a function so I can call closure twice:

let closure = || println!("hello");
fn call<F>(f: &F)
where
    F: Fn(),
{
    f();
}
call(&closure);
call(&closure);

当我使用 FnMut 时：

let mut string: String = "hello".to_owned();
let change_string = || string.push_str(" world");
fn call<F>(mut f: &mut F)
where
    F: FnMut(),
{
    f();
}
call(&change_string);
call(&change_string);

会出现错误：

error[E0308]: mismatched types
  --> src/main.rs:10:10
   |
10 |     call(&change_string);
   |          ^^^^^^^^^^^^^^ types differ in mutability
   |
   = note: expected type `&mut _`
              found type `&[closure@src/main.rs:3:25: 3:53 string:_]`

我该如何解决？

推荐答案

错误消息所述：

expected type `&mut _`
   found type `&[closure@src/main.rs:3:25: 3:53 string:_]`

期望对某物（& mut _ ）的可变引用，但您提供的是不变的引用闭包（& ... ）。采取可变的参考：

It is expecting a mutable reference to something (&mut _), but you are providing an immutable reference to a closure (&...). Take a mutable reference:

call(&mut change_string);

哪个会导致下一个错误：

Which leads to the next error:

error: cannot borrow immutable local variable `change_string` as mutable
 --> src/main.rs:9:15
  |
3 |     let change_string = || string.push_str(" world");
  |         ------------- use `mut change_string` here to make mutable
...
9 |     call(&mut change_string);
  |               ^^^^^^^^^^^^^ cannot borrow mutably

获取可变引用值本身是可变的：

Taking a mutable reference requires that the value itself be mutable:

let mut change_string = || string.push_str(" world");

---

在这种情况下，您不需要完全采用& mut F ，因为实现了 FnMut 用于可变引用 FnMut 。也就是说，这可行：

---

In this case, you don't need to take a &mut F at all, as FnMut is implemented for mutable references to FnMut. That is, this works:

fn call(mut f: impl FnMut()) {
    f();
}

call(&mut change_string);
call(&mut change_string);

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