返回递归闭包的函数签名

Question

我正在尝试实现一个 returns 递归闭包的函数，尽管我不确定如何在函数签名中表达它。这是 Python

中有效实现的示例代码

def counter(state):
    def handler(msg):
        if msg == 'inc':
            print state
            return counter(state + 1)

        if msg == 'dec':
            print state
            return counter(state - 1)
    return handler

c = counter(1)
for x in range(1000000):
    c = c('inc')

和 Rust 的伪代码。

enum Msg {
    Inc,
    Dec
}

fn counter(state: Int) -> ? {
    move |msg| match msg {
        Msg::Inc => counter(state + 1),
        Msg::Dec => counter(state - 1),
    }
}

Answer 1

因为 Rust 支持递归类型，你只需要在一个单独的结构中编码递归：

enum Msg { 
    Inc,
    Dec,
}

// in this particular example Fn(Msg) -> F should work as well
struct F(Box<FnMut(Msg) -> F>);

fn counter(state: i32) -> F {
    F(Box::new(move |msg| match msg {
        Msg::Inc => {
            println!("{}", state);
            counter(state + 1)
        }
        Msg::Dec => {
            println!("{}", state);
            counter(state - 1)
        }
    }))
}

fn main() {
    let mut c = counter(1);
    for _ in 0..1000 {
        c = c.0(Msg::Inc);
    }
}

不幸的是，我们不能在这里取消装箱 - 因为未装箱的闭包具有不可命名的类型，我们需要将它们装箱到特征对象中以便能够在结构声明中命名它们。

返回递归闭包的函数签名

Function signature for returning a recursive closure

recursion

continuations

closures

types

rust