“方法存在但特征边界不满足"是什么意思? [英] What does it mean when "method exists but trait bounds not satisfied"?

问题描述

我是 Rust 的新手，观察到一些我无法反驳的东西.

当我写

fn main() {('a'..'z').all(|_| true);}

编译器报错:

error[E0599]: 在当前范围内没有找到类型为 `std::ops::Range<char>` 的名为 `all` 的方法-->src/main.rs:2:16|2 |('a'..'z').all(|_| 真)|^^^|= 注意:方法 `all` 存在，但不满足以下特征边界:`std::ops::Range: std::iter::Iterator`

当我把它改成

fn main() {(b'a'..b'z').all(|_| 真);}

它编译.<​​/p>

这里发生了什么?Rust 说方法......存在但不满足以下特征边界是什么意思?

解决方案

all() 方法是 Iterator trait 的一个方法，所以你只能调用它在实现该特征的类型上.类型 Range 没有实现 Iterator 特性，因为在一般情况下，Unicode 字符的范围没有明确定义.一组有效的 Unicode 代码点存在差距，构建一系列代码点通常不被认为是有用的.另一个类型 Range 确实实现了 Iterator，因为在一个字节范围内迭代具有明确定义的含义.

更一般地，错误消息告诉您 Rust 找到了一个具有正确名称的方法，但该方法不适用于您调用它的类型.

I'm new to Rust and observed something that I couldn't reason against.

When I write

fn main() {
    ('a'..'z').all(|_| true);
}

The compiler reports an error:

error[E0599]: no method named `all` found for type `std::ops::Range<char>` in the current scope
 --> src/main.rs:2:16
  |
2 |     ('a'..'z').all(|_| true)
  |                ^^^
  |
  = note: the method `all` exists but the following trait bounds were not satisfied:
          `std::ops::Range<char> : std::iter::Iterator`

When I change it to

fn main() {
    (b'a'..b'z').all(|_| true);
}

it compiles.

What's happening here? What does Rust mean when it says the method ... exists but the following trait bounds were not satisfied?

解决方案

The method all() is a method of the Iterator trait, so you can only call it on types that implement that trait. The type Range<char> does not implement the Iterator trait, since a range of Unicode characters is not well-defined in the general case. The set of valid Unicode code points has gaps, and building a range of code points is not in general considered useful. The type Range<u8> on the other does implement Iterator, since iterating over a range of bytes has a well-defined meaning.

More generally, the error message tells you that Rust has found a method with the correct name, but that method does not apply to the type you call it for.

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