为什么需要 num::One 来迭代一个范围?

本文介绍了为什么需要 num::One 来迭代一个范围?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

为什么以下带有字符范围的 for 循环无法编译?

Why does the following for loop with a char range fail to compile?

fn main() {
    for c in 'a'..'z' {
        println!("{}", c);
    }
}

错误...

main.rs:11:5: 14:2 error: the trait `core::num::One` is not implemented for the type `char` [E0277]
main.rs:11     for c in 'a'..'z' {
main.rs:12         println!("{}", c);
main.rs:13     }
main.rs:14 }
main.rs:11:5: 14:2 error: the trait `core::iter::Step` is not implemented for the type `char` [E0277]
main.rs:11     for c in 'a'..'z' {
main.rs:12         println!("{}", c);
main.rs:13     }
main.rs:14 }

为什么你甚至需要 core::num::One 来迭代一个范围?

Why do you even need core::num::One for a iterating over a range?

The x..y syntax is sugar for std::ops::Range { start: x, end: y }. This type (Range<A>) is iterable due to the implementation of Iterator for it, specifically, from that page:

impl<A> Iterator for Range<A>
   where A: One + Step,
         &'a A: Add<&'a A>,
         &'a A::Output == A {
    type Item = A;

这是说如果 A 类型实现了 OneRange 可以作为 A 上的迭代器/code> 和 Step，并且可以通过正确的方式添加.

This is saying that Range<A> can behave as an iterator over As if the type A implements One and Step, and can be added in the right way.

在这种情况下，char 不满足任何一个:char 具有 One 或可添加在语义上是无稽之谈，而且它也没有实现 Step.

In this case, char satisfies none of those: it is semantically nonsense for char to have One or be addable, and it doesn't implement Step either.

也就是说，因为 char 没有实现这些特征(因此 Range 不像迭代器那样通过 impl代码>)，应该可以有手动实现:

That said, since char doesn't implement those traits (and hence Range<char> doesn't behave like an iterator via that impl), it should be possible to have a manual impl:

impl Iterator for Range<char> {
    type Item = char;

这将允许 for x in 'a'..'z' 工作.

然而，这在语义上可能不是我们想要的:.. 范围不包括最后一个元素，这对于字符来说是令人惊讶的，必须编写 'a'..'{' 来获取字母 A 到 Z.已经提出了包含范围语法的建议，例如一个例子是 'a'...'z'(更多的点 == 更多的元素)，我想对于这种类型会有一个 Iterator 实现chars.

However, this probably isn't semantically what we want: the .. range doesn't include the last element, which would be suprising for characters, one would have to write 'a'..'{' to get the letters A through Z. There's been proposals for inclusive-range syntax, e.g. one example is 'a'...'z' (more dots == more elements), and I would imagine that there would be an Iterator implementation for this type with chars.

正如其他人所证明的，对于 ASCII 字符，可以使用字节文字，更一般地，可以将字符转换为 u32s:

As others have demonstrated, for ASCII characters one can use byte literals, and more generally, one can cast characters to u32s:

for i in ('à' as u32)..('æ' as u32) + 1 {
    let c = std::char::from_u32(i).unwrap();
    println!("{}", c);
}

给出:

à
á
â
ã
ä
å
æ

注意.这种方法并不完美，如果范围超过 surrogate 范围，0xD800，它会崩溃-0xDFFF.

NB. this approach isn't perfect, it will crash if the range crosses the surrogate range, 0xD800-0xDFFF.

我刚刚发布了一个 crate，char-iter，它正确地处理后者并且表现得像人们期望的那样.添加后(通过货物)，它可以像这样使用:

I just published a crate, char-iter, which handles the latter correctly and behaves like one would expect. Once added (via cargo), it can be used like:

extern crate char_iter;
// ...

for c in char_iter::new('a', 'z') {
    // ...
}
for c in char_iter::new('à', 'æ') {
    // ...
}

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为什么需要 num::One 来迭代一个范围?

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