如何处理仿制药？

我想使用泛型来计算因子，但在main中遇到错误。如何处理仿制药？

我全码：

pub trait Body { 
    fn new() -> Self; 

    fn fact(&self, x: usize) -> usize { 
     match x { 
      1 => 1, 
      _ => x * self.fact(x - 1), 
     } 
    } 
} 

#[derive(Clone, Debug)] 
pub struct RecursiveCall<T: Body> { 
    level: usize, 
    indicator: String, 
    n_repeat: usize, 
    body: T, 
} 

impl<T> RecursiveCall<T> 
    where T: Body 
{ 
    fn new(n_repeat: usize) -> RecursiveCall<T> { 
     RecursiveCall { 
      level: 0, 
      indicator: "- ".to_string(), 
      n_repeat: n_repeat, 
      body: <T as Body>::new(), 
     } 
    } 

    fn pre_trace(&self, fname: &str, arg: &usize) { 
     let args: String = arg.to_string(); 
     println!("{}", 
       (vec![self.indicator.as_str(); self.level]).join("") + 
       self.level.to_string().as_str() + ":" + fname + "(" + 
       args.as_str() + ")"); 
    } 

    fn post_trace(&self, fname: &str, arg: &usize, ret: &usize) { 
     println!("{}", 
       (vec![self.indicator.as_str(); self.level]).join("") + 
       self.level.to_string().as_str() + ":" + fname + "=" + 
       ret.to_string().as_str()); 
    } 

    fn print_trace(&mut self) { 
     &self.pre_trace("fact", &self.n_repeat); 
     self.level += 1; 
     let ret = &self.body.fact(self.n_repeat); 
     self.level -= 1; 
     &self.post_trace("fact", &self.n_repeat, ret); 

     println!("Difference={}", &ret.to_string().as_str()); 
    } 
} 

type B = Body; 
fn main() { 
    let t = RecursiveCall::<B>::new(); 
}

这个错误发生在main()：

error: no associated item named `new` found for type `RecursiveCall<Body + 'static>` in the current scope 
    --> src/main.rs:61:13 
    | 
61 |  let t = RecursiveCall::<B>::new(); 
    |    ^^^^^^^^^^^^^^^^^^^^^^^ 
    | 
    = note: the method `new` exists but the following trait bounds were not satisfied: `Body : std::marker::Sized`, `Body : Body` 
    = help: items from traits can only be used if the trait is implemented and in scope; the following traits define an item `new`, perhaps you need to implement one of them: 
    = help: candidate #1: `Body` 
    = help: candidate #2: `std::sys_common::thread_info::NewThread` 
    = help: candidate #3: `std::iter::ZipImpl` 

error[E0277]: the trait bound `Body + 'static: std::marker::Sized` is not satisfied 
    --> src/main.rs:61:13 
    | 
61 |  let t = RecursiveCall::<B>::new(); 
    |    ^^^^^^^^^^^^^^^^^^^^^^^ the trait `std::marker::Sized` is not implemented for `Body + 'static` 
    | 
    = note: `Body + 'static` does not have a constant size known at compile-time 
    = note: required by `RecursiveCall` 

error[E0277]: the trait bound `Body + 'static: Body` is not satisfied 
    --> src/main.rs:61:13 
    | 
61 |  let t = RecursiveCall::<B>::new(); 
    |    ^^^^^^^^^^^^^^^^^^^^^^^ the trait `Body` is not implemented for `Body + 'static` 
    | 
    = note: required by `RecursiveCall` 

error[E0038]: the trait `Body` cannot be made into an object 
    --> src/main.rs:61:13 
    | 
61 |  let t = RecursiveCall::<B>::new(); 
    |    ^^^^^^^^^^^^^^^^^^^^^^^ the trait `Body` cannot be made into an object 
    | 
    = note: method `new` has no receiver

来源

2017-02-21 k.nyao

答案的关键是从编译器这样一个字条：

注意：方法new存在，但不满足以下特征边界：Body : std::marker::Sized,Body : Body

您已经定义B作为一个类型别名Body，所以名称B和Body都意味着同样的事情在你的程序（至少在这个模块）。

定义特征时，编译器还定义了一个具有相同名称的类型。但是，该类型不能直接实例化（与C++/C＃/ Java /等中的类不同）。然而，这正是你想要做的！

不符合性状Body : std::marker::Sized，因为Body是编译器定义的对应于同名性状的类型，是未定义类型。未定义类型只能用于指针和引用（例如，&Body,Box<Body>等）。

特质绑定Body : Body不满意，因为你的特质不是object-safe。它不是对象安全的，因为方法new没有self参数（这是编译器在最后一个注意事项中的含义：method `new` has no receiver）。

通常情况下，您将定义一个结构或枚举并实现该类型的特征，然后在实例化RecursiveCall时使用该类型。尝试用以下代码替换type B = Body;：

struct B; 

impl Body for B { 
    fn new() -> B { 
     B 
    } 
}

来源

2017-02-22 01:44:26

我通过您的帮助解决了问题！谢谢你的好意。 –

如何处理仿制药？

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