Question

我想使用泛型计算阶乘，但我在main中收到错误。

我的完整代码：

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();
}

main()：

发生此错误

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

Answer 1

答案的关键在于编译器的这一说明：

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

您已将B定义为Body的类型别名，因此名称B和Body在您的程序中都表示相同的内容（至少在此模块）。

定义特征时，编译器还会定义具有相同名称的类型。但是，该类型无法直接实例化（与C ++ / C＃/ Java /等中的类不同）。然而，这正是你想要做的！

不满足特征绑定Body : std::marker::Sized，因为Body是由与同名特征相对应的编译器定义的类型，是未确定的类型。 Unsized类型只能用于指针和引用（例如&Body，Box<Body>等。）

由于您的特质不是object-safe，因此不满足特征限制Body : Body。它不是对象安全的，因为方法new没有self参数（这是编译器在上一个注释中的含义：method `new` has no receiver）。

通常，您将定义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。尝试使用以下内容替换type B = Body;：

struct B;

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

我该如何对待仿制药？

1 个答案: