返回一个递归闭包，它将环境可变变量从函数移入其中？

Question

从问题的公认答案来看，递归闭包的方法不适合我：

Is it possible to make a recursive closure in Rust?

我的闭包需要从函数中返回，并且需要将变量从环境中移动到其中，并且可以对其进行变异。

然后我找到了一种似乎更适合我的方式：

Anonymous recursion with closures

use std::cell::RefCell;

fn main() {
    let id = &(|a| a) as &Fn(u64) -> u64;
    let (fib, fib_p): (_, RefCell<&Fn(u64) -> u64>);
    fib_p = RefCell::new(id);

    fib = |n: u64| {
        if n < 2 {
            n
        } else {
            (fib_p.borrow())(n - 2) + (fib_p.borrow())(n - 1)
        }
    };

    *fib_p.borrow_mut() = &fib;

    println!("{}", fib(10));        
}

上面的代码运行良好。

但是我的闭包需要从函数中返回，所以不能作为引用防止悬空引用，而且我们从编译时也不知道闭包的大小，所以我使用了智能指针@987654324 @ 为了它。下面的代码会报错：

use std::cell::RefCell;

fn main() {
    let id: Box<Fn(u64) -> u64> = Box::new(|a| a);
    let (fib, fib_p): (Box<Fn(u64) -> u64>, RefCell<&Box<Fn(u64) -> u64>>);
    fib_p = RefCell::new(&id);

    fib = Box::new(|n: u64| {
        if n < 2 {
            n
        } else {
            (fib_p.borrow())(n - 2) + (fib_p.borrow())(n - 1)
        }
    });

    *fib_p.borrow_mut() = &fib;

    println!("{}", fib(10));        
}

error[E0597]: `fib_p` does not live long enough
  --> src/main.rs:12:15
   |
8  |     fib = Box::new(|n: u64| {
   |                    -------- capture occurs here
...
12 |             (&fib_p.borrow())(n - 2) + (&fib_p.borrow())(n - 1)
   |               ^^^^^ borrowed value does not live long enough
...
19 | }
   | - borrowed value dropped before borrower
   |
   = note: values in a scope are dropped in the opposite order they are created

Answer 1

您可以将递归上下文包装在非递归闭包中：

pub fn fib() -> Box<Fn(u64) -> u64> {
    Box::new(|n: u64| {
        let id = &(|a| a) as &Fn(u64) -> u64;
        let (fib, fib_p): (_, RefCell<&Fn(u64) -> u64>);
        fib_p = RefCell::new(id);

        fib = |n: u64| {
            if n < 2 {
                n
            } else {
                (fib_p.borrow())(n - 2) + (fib_p.borrow())(n - 1)
            }
        };

        *fib_p.borrow_mut() = &fib;

        fib(n)
    })
}