Question

在Python或C ++中，类A可以将某些工作委派给类Say B的另一个实例，并在B中设置A的回调方法。我尝试在Rust中执行此操作，但到目前为止，我仍然一无所获，被Rust编译器击败。

这是我尝试过的代码，其余代码在本文末尾。

在A :: test中，我尝试使用闭包获取Fn（）特征对象作为回调。

// let b = B::new(self.finish)); // ideally but would not compile

// let test :Box<Fn(String)> = Box::new(move |msg| {self.finish(msg);}); // cannot infer an appropriate lifetime due to conflicting requirements
// let b = B::new(&test);

// let b = B::new(&Box::new( |msg| {A::finish(&self, msg);} )); // expected trait std::ops::Fn, found closure
// let b = B::new(&Box::new( |msg| {self.finish(msg);} )); // expected trait std::ops::Fn, found closure

什么都没有。有办法吗？

任何帮助将不胜感激！

还是我从根本上错了？ Rust是否要求其他方法在这里实施该想法？

这是我的测试代码

Play Ground Link

struct A {}

impl A {
    fn finish(&self, msg: String) {
        println!("{}", msg);
    }

    fn test(&self) {

        //let b = B::new(self.finish)); // would not compile

        // let test :Box<Fn(String)> = Box::new(move |msg| {self.finish(msg);}); // cannot infer an appropriate lifetime due to conflicting requirements
        // let b = B::new(&test);

        // let b = B::new(&Box::new( |msg| {A::finish(&self, msg);} )); // expected trait std::ops::Fn, found closure
        let b = B::new(&Box::new( |msg| {self.finish(msg);} )); // expected trait std::ops::Fn, found closure

        b.start("hi".to_string().clone());
    }
}

struct B<'b> {
    // cb:fn(msg:String),
    cb: &'b Box<Fn(String)>,
}

impl<'b> B<'b> {
    fn new(cb: &'b Box<Fn(String)>) -> B<'b> {
        B { cb: cb }
    }

    fn start(&self, msg: String) {
        (self.cb)(msg);
    }
}

fn main() {
    let a = A {};
    a.test();
}

Answer 1

是的，您可以将方法作为回调传递给您的结构，并从该结构的方法中调用它。并且您无需在传递引用时将封闭框装箱：

struct A {}

impl A {
    fn finish(&self, msg: String) {
        println!("{}", msg);
    }

    fn test(&self) {
        let fun = |msg: String| self.finish(msg);
        let b = B::new(&fun);
        b.start("hi".to_string().clone());
    }
}

struct B<'b> {
    cb: &'b Fn(String),
}

impl<'b> B<'b> {
    fn new(cb: &'b Fn(String)) -> B<'b> {
        B { cb }
    }

    fn start(&self, msg: String) {
        (self.cb)(msg);
    }
}

fn main() {
    let a = A {};
    a.test();
}

playground

将函数移到新结构时，此框很有用，但情况并非如此。

注意：由于您的函数名为start，我怀疑在您的实际用例中您想启动一个线程，在这种情况下，您可能应该看一下channels而不是回调。

如何将方法作为回调传递

1 个答案: