1

归结起来的问题如下：

use std::marker::PhantomData;

struct WorldState<'a> {
    state: &'a f64,
}

trait CalculateWorldState<T> {
    fn state_value(&mut self, input: &T) -> f64;
}


trait LearningAlgorithm<T> {
    fn print_learning_information(&self, &T);
}

struct EvolutionaryAlgorithm<F, T>
where
    F: CalculateWorldState<T>,
{
    //I need this since I only use T as a method parameter, I do not save it anywhere
    //T are different ways to represent the current worldstate and are
    //short-lived (new ones generated every frame)
    _p_: PhantomData<T>,
    //I don't actually need this one in the real example since I have
    //an instatiated version of type CalculateWorldState saved in the
    //struct but I use phantomdata for simplicity of the example
    _p: PhantomData<F>,
}

impl<F, T> LearningAlgorithm<T> for EvolutionaryAlgorithm<F, T>
where
    F: CalculateWorldState<T>,
{
    fn print_learning_information(&self, input: &T) {
        println!("My learning goes splendid!");
        //do something with &T by calling the object of type
        //CalculateWorldState which we have saved somewhere, but do
        //not save the &T reference anywhere, just look at it
    }
}

struct WorldIsInGoodState {}

impl<'a> CalculateWorldState<WorldState<'a>> for WorldIsInGoodState {
    fn state_value(&mut self, input: &WorldState) -> f64 {
        100.
    }
}

fn main() {
    let mut a: Box<LearningAlgorithm<WorldState>> =
        Box::new(EvolutionaryAlgorithm::<WorldIsInGoodState, WorldState> {
            _p: PhantomData,
            _p_: PhantomData,
        });
    {
        let state = WorldState { state: &5. };
        a.print_learning_information(&state);
    }
}

游乐场。

上面的代码编译失败：

error[E0597]: borrowed value does not live long enough
  --> src/main.rs:59:5
   |
57 |         let state = WorldState { state: &5. };
   |                                          -- temporary value created here
58 |         a.print_learning_information(&state);
59 |     }
   |     ^ temporary value dropped here while still borrowed
60 | }
   | - temporary value needs to live until here

WorldState<'a>是一种寿命很短的数据类型（每帧一个），而LearningAlgorithm是一种寿命很长的数据类型（多个游戏）。但是我实现这个东西的方式，Rust 渴望相信，WorldState我传递给的每个人print_learning_information都必须比LearningAlgorithm.

我做错了什么？这还能怎么处理？

一些我不想做的事情：

  • WorldState包含正常状态（因为实际上它包含一些向量而不是 a并且我f64不想WorldState在传递每个玩家自己的世界视图时将它们复制到结构中）
  • 退出这个项目并开始一个新项目（你们都知道，投入一些时间后，你不想把所有的工作都扔掉）
4

1 回答 1

0

你的问题可以归结为

struct WorldState<'a> {
    state: &'a f64,
}

trait LearningAlgorithm<T> {
    fn print_learning_information(&self, &T);
}

struct EvolutionaryAlgorithm();

impl<T> LearningAlgorithm<T> for EvolutionaryAlgorithm
{
    fn print_learning_information(&self, input: &T) {
    }
}

fn main() {
    // scope a
    let mut a: Box<LearningAlgorithm<WorldState>> =
        Box::new(EvolutionaryAlgorithm());
    { // scope b
        let val = 5.;
        let state = WorldState { state: &val };
        a.print_learning_information(&state);
    }
}

请注意，这WorldState是一个类型构造函数，而不是具体类型。生命周期省略允许您在Box<LearningAlgorithm<WorldState>>没有明确指定生命周期参数的情况下编写WorldState，但这只是意味着编译器选择了一些适当的生命周期参数。

在这种情况下，为WorldStateis选择的生命周期scope a，因此是的a类型Box<LearningAlgorithm<WorldState<'scope_a>>>。因此，state应该有 type WorldState<'scope_a>，并且它包含的引用应该对 有效scope a，但引用指向的值只存在于scope b.

您需要对更高种类的类型的支持才能使您的示例按原样工作，但 Rust 不提供它。

WorldState最简单的解决方案是通过将引用替换为 来摆脱的生命周期参数Rc。也许有人会想出更好的解决方案。

于 2017-08-27T03:44:54.810 回答