如何在Rust中的通用类型枚举上实现fmt :: Display？

Question

我使用这个递归枚举实现了我的链表,但现在我想为它实现一个自定义显示格式

use std::fmt;

#[derive(Debug)]
enum List<A> {
    Empty,
    Cons(A, Box<List<A>>),
}

impl<T> fmt::Display for List<T> {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        match *self {
            List::Empty => write!(f, "()"),
            List::Cons(x, ref xs) => write!(f, "({} {})", x, xs),  
        }
    }
}

错误

error[E0277]: the trait bound `T: std::fmt::Display` is not satisfied
  --> src/main.rs:13:59
   |
13 |             List::Cons(x, ref xs) => write!(f, "({} {})", x, xs),  
   |                                                           ^ the trait `std::fmt::Display` is not implemented for `T`
   |
   = help: consider adding a `where T: std::fmt::Display` bound
   = note: required by `std::fmt::Display::fmt`

如果重要的话,这是我的其余代码

fn cons<A>(x: A, xs: List<A>) -> List<A> {
    return List::Cons(x, Box::new(xs));
}

fn len<A>(xs: &List<A>) -> i32 {
    match *xs {
        List::Empty => 0,
        List::Cons(_, ref xs) => 1 + len(xs),
    }
}

fn map<A, B>(f: &Fn(&A) -> B, xs: &List<A>) -> List<B> {
    match *xs {
        List::Empty => List::Empty,
        List::Cons(ref x, ref xs) => cons(f(x), map(f, xs)),
    }
}

fn main() {
    let xs = cons(1, cons(2, cons(3, List::Empty)));
    println!("{}", xs);
    println!("{:?}", len(&xs));

    let f = |x: &i32| (*x) * (*x);
    let ys = map(&f, &xs);
    println!("{}", ys);
    println!("{}", List::Empty);
}

预期产出

(1 (2 (3 ())))
3
(1 (4 (9 ())))
()

---

真的,我想看到这个,但我完全不知道如何使用这种输出fmt::Result

(1 2 3)
3
(1 4 9) 
()

Answer 1

解决编译器错误

你错过了一个特质限制.也就是说,你需要告诉Rust T可以显示:

impl<T: fmt::Display> fmt::Display for List<T> {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        match *self {
            List::Empty => write!(f, "()"),
            List::Cons(ref x, ref xs) => write!(f, "({} {})", x, xs),  
        }
    }
}

注意特征限制T: fmt::Display.这基本上意味着:if Timplements fmt::Display,然后List<T>实现fmt::Display.

蛋糕上的糖衣

我不确定你是否可以使用递归定义获得良好的格式.此外,Rust不保证尾调用优化,因此总会有堆栈溢出的可能性.

另一种定义可能是:

fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
    write!(f, "(")?;
    let mut temp = self;
    while let List::Cons(ref x, ref xs) = *temp {
        write!(f, "{}", x)?;

        // Print trailing whitespace if there are more elements
        if let List::Cons(_, _) = **xs {
            write!(f, " ")?;
        }

        temp = xs;
    }

    write!(f, ")")
}

请注意?最多的write!宏调用.它基本上意味着:如果这write!导致错误,现在返回错误.否则,继续执行该功能.