Question

我有lib.rs：

pub mod genetics {
  use std;

  pub trait RandNewable {
    fn new() -> Self;
  }

  pub trait Gene : std::fmt::Debug + Copy + Clone + RandNewable {}

  #[derive(Debug, Copy, Clone)]
  pub struct TraderGene {
    cheat: bool,
  }

  impl RandNewable for TraderGene {
    fn new() -> TraderGene {
      TraderGene {
        cheat: true,
      }
    }
  }

  #[derive(Debug)]
  pub struct DNA<G: Gene> {
    genes: [G; 3],
  }

  impl <G: Gene> DNA<G> {
    pub fn create() -> DNA<G> {
      DNA {
        genes: [RandNewable::new(), RandNewable::new(), RandNewable::new()],
      }
    }

    pub fn mutate(&self) -> DNA<G> {
      let mut copy = self.genes;
      copy[0] = RandNewable::new();
      DNA {
        genes: copy,
      }
    }
  }
}

我正在尝试使用main.rs中的定义：

extern crate trafficgame;

use trafficgame::genetics::{DNA, TraderGene, Gene, RandNewable};

fn main() {
  let t1: DNA<TraderGene> = DNA::create();
  let t2 = t1.mutate();
}

但是我遇到两个错误：

the trait `trafficgame::genetics::Gene` is not implemented for the type `trafficgame::genetics::TraderGene

和

type `trafficgame::genetics::DNA<trafficgame::genetics::TraderGene>` does not implement any method in scope named `mutate`

TraderGene肯定会实现Gene特征，但似乎RandNewable的impl不在main范围内。另一个问题是，由于mutate未在特征中声明，我不知道在main中导入什么以引用mutate。我是否错误地组织了事情？

Answer 1

即使您的Gene特征没有方法，TraderGene也不会自动实现它，即使它实现了所有Gene的超级特征。您必须在impl模块中写一个空的genetics：

impl Gene for TraderGene {}

在不同的模块中使用无方法特征实现

1 个答案: