为什么编译器声称泛型不实现`Display`,即使它应该？

Question

我正在构建一个实现字符串连接的库; 也就是说,打印由分隔符分隔的容器的所有元素.我的基本设计如下:

use std::fmt;

#[derive(Debug, Clone, PartialEq, Eq)]
pub struct Join<Container, Sep> {
    container: Container,
    sep: Sep,
}

impl<Container, Sep> fmt::Display for Join<Container, Sep>
where
    for<'a> &'a Container: IntoIterator,
    for<'a> <&'a Container as IntoIterator>::Item: fmt::Display,
    Sep: fmt::Display,
{
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        let mut iter = self.container.into_iter();

        match iter.next() {
            None => Ok(()),
            Some(first) => {
                first.fmt(f)?;

                iter.try_for_each(move |element| {
                    self.sep.fmt(f)?;
                    element.fmt(f)
                })
            }
        }
    }
}

这种特质实施无需投诉即可编译.注意绑定&'a C: IntoIterator.许多容器实现IntoIterator对自身的引用,以允许迭代对包含项的引用(例如,在此处Vec实现它).

但是,当我实际尝试使用我的Join结构时,我得到一个不满意的特征限制:

fn main() {
    let data = vec!["Hello", "World"];
    let join = Join {
        container: data,
        sep: ", ",
    };
    println!("{}", join);
}

此代码生成编译错误:

error[E0277]: `<&'a std::vec::Vec<&str> as std::iter::IntoIterator>::Item` doesn't implement `std::fmt::Display`
  --> src/main.rs:38:20
   |
38 |     println!("{}", join);
   |                    ^^^^ `<&'a std::vec::Vec<&str> as std::iter::IntoIterator>::Item` cannot be formatted with the default formatter
   |
   = help: the trait `for<'a> std::fmt::Display` is not implemented for `<&'a std::vec::Vec<&str> as std::iter::IntoIterator>::Item`
   = note: in format strings you may be able to use `{:?}` (or {:#?} for pretty-print) instead
   = note: required because of the requirements on the impl of `std::fmt::Display` for `Join<std::vec::Vec<&str>, &str>`
   = note: required by `std::fmt::Display::fmt`

关键线似乎是这样的:

the trait `for<'a> std::fmt::Display` is not implemented for `<&'a std::vec::Vec<&str> as std::iter::IntoIterator>::Item`

不幸的是,编译器实际上并没有告诉我这Item是什么类型,但根据我对文档的阅读,它似乎是&T,在这种情况下意味着&&str.

为什么编译器不认为&&str实现Display？我已尝试过许多其他类型的东西,比如usize和String,并且它们都不起作用; 他们都失败了同样的错误.我知道这些引用类型没有直接实现Display,但实现应该通过deref强制自动获取,对吗？

Answer 1

看起来像编译器限制.您现在可以通过编写表示"使用生命周期显示"的私有帮助器特征的impl绑定来解决它.这使编译器能够看到这for<'a> private::Display<'a>意味着fmt::Display.

use std::fmt;

pub struct Join<Container, Sep> {
    container: Container,
    sep: Sep,
}

mod private {
    use std::fmt;
    pub trait Display<'a>: fmt::Display {}
    impl<'a, T> Display<'a> for T where T: fmt::Display {}
}

impl<Container, Sep> fmt::Display for Join<Container, Sep>
where
    for<'a> &'a Container: IntoIterator,
    for<'a> <&'a Container as IntoIterator>::Item: private::Display<'a>,
    Sep: fmt::Display,
{
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        let mut iter = self.container.into_iter();

        match iter.next() {
            None => Ok(()),
            Some(first) => {
                first.fmt(f)?;

                iter.try_for_each(move |element| {
                    self.sep.fmt(f)?;
                    element.fmt(f)
                })
            }
        }
    }
}

fn main() {
    println!("{}", Join {
        container: vec!["Hello", "World"],
        sep: ", ",
    });
}