rust - 在结构中存储函数特征对象的生命周期不够长

Question

我正在尝试将两个函数存储在这样的结构中:

struct Example<'a> {
    func1: &'a Fn(f64) -> f64,
    func2: &'a Fn(f64) -> f64,
}

impl<'a> Example<'a> {

    fn new(f1: &'a Fn(f64) -> f64, f2: &'a Fn(f64) -> f64) -> Example<'a> {
        Example {
            func1: f1,
            func2: f2,
        }
    }

    fn test_func(self, x: f64, y: f64) -> f64 {
        (self.func1)(x) + (self.func2)(y)
   }
}


fn main() {
    fn first(x: f64) -> f64 {
        x
    }
    fn second(x: f64) -> f64 {
        x
    }
    let test = Example::new(&first, &second);
    test.test_func(1.0, 2.0);
}

Rust playground

生成以下错误:

borrowed value does not live long enough
    let test = Example::new(&first, &second);
                            ^~~~~
note: reference must be valid for the block suffix following statement 2 at 28:45...
    let test = Example::new(&first, &second);
    test.test_func(1.0, 2.0);
}
note: ...but borrowed value is only valid for the statement at 28:4
    let test = Example::new(&first, &second);
    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
help: consider using a `let` binding to increase its lifetime
    let test = Example::new(&first, &second);
    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
error: borrowed value does not live long enough
    let test = Example::new(&first, &second);
                                    ^~~~~~
note: reference must be valid for the block suffix following statement 2 at 28:45...
    let test = Example::new(&first, &second);
    test.test_func(1.0, 2.0);
}
note: ...but borrowed value is only valid for the statement at 28:4
    let test = Example::new(&first, &second);
    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
help: consider using a `let` binding to increase its lifetime
    let test = Example::new(&first, &second);
    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

我对这是一个生命周期问题感到困惑，因为函数和结构体在 main 中应该共享相同的生命周期。也对为什么编译器的建议似乎与我已有的建议相同感到困惑。

Answer 1

按照编译器的建议(考虑使用 let 绑定(bind)来增加其生命周期)允许代码编译:

fn main() {
    fn first(x: f64) -> f64 {
        x
    }
    fn second(x: f64) -> f64 {
        x
    }

    let x = &first;
    let y = &second;
    let test = Example::new(x, y);

    println!("{}", test.test_func(1.0, 2.0));
}

同样

let x = first;
let y = second;
let test = Example::new(&x, &y);

我相信(但不知道如何证明)问题是必须创建一些东西来将函数指针 (fn(f64) -> f64) 适配到特征对象中(&Fn(f64) -> f64)。但是，在 new 调用中创建它意味着它会在语句返回时立即被删除。

这类似于文字引用的情况，它有同样的错误:

struct Foo<'a>(&'a u8);

fn main() {
    let f = Foo(&4);
    println!("{}", f.0);
}