如何将泛型T转换为String？

Question

我正在尝试打印自定义类型：

struct Node<T> {
    prev: Option<Box<Node<T>>>,
    element: T,
    next: Option<Box<Node<T>>>,
}

现在，问题是：

print!(
    "{0} -> {1}",
    String::from(node.element),
    String::from(node.next)
);

error[E0277]: the trait bound `std::string::String: std::convert::From<T>` is not satisfied
  --> src/lib.rs:10:9
   |
10 |         String::from(node.element),
   |         ^^^^^^^^^^^^ the trait `std::convert::From<T>` is not implemented for `std::string::String`
   |
   = help: consider adding a `where std::string::String: std::convert::From<T>` bound
   = note: required by `std::convert::From::from`

error[E0277]: the trait bound `std::string::String: std::convert::From<std::option::Option<std::boxed::Box<Node<T>>>>` is not satisfied
  --> src/lib.rs:11:9
   |
11 |         String::from(node.next)
   |         ^^^^^^^^^^^^ the trait `std::convert::From<std::option::Option<std::boxed::Box<Node<T>>>>` is not implemented for `std::string::String`
   |
   = help: the following implementations were found:
             <std::string::String as std::convert::From<&'a str>>
             <std::string::String as std::convert::From<std::borrow::Cow<'a, str>>>
             <std::string::String as std::convert::From<std::boxed::Box<str>>>
   = note: required by `std::convert::From::from`

如何将qazxsw poi施放到qazxsw poi和qazxsw poi到qazxsw poi？

Answer 1

正如编译器告诉你的那样：

考虑添加node.element绑定

String

这对Option<Box<Node<T>>>不起作用，因为没有这样的实现。

---

如果要格式化结构，则需要实现String。为了显示它，没有理由将值转换为where std::string::String: std::convert::From<T>：

fn example<T>(node: Node<T>)
where
    String: From<T>,
{
    // ...
}

再次，这不适用于你的大案例，因为String: From<Option<Node<T>>>或Display都没有实施String。

也可以看看：

• fn example<T>(node: Node<T>)
  where
      T: std::fmt::Display
  {
      // ...
  }

• Node<T>

你可能想要这样的东西

Option<T>

要么

Display

甚至可以使用相同的代码自己实现Should I implement Display or ToString to render a type as a string? / The trait bound `T: std::fmt::Display` is not satisfied。

您还应该阅读

fn example<T>(mut node: Node<T>)
where
    T: std::fmt::Display,
{
    print!("{}", node.element);
    while let Some(n) = node.next {
        print!(" -> {}", n.element);
        node = *n;
    }
}

。您正在尝试创建一个双链表，这在安全的Rust中是不可能的。