参数/模式中“&variable"的含义 [英] Meaning of '&variable' in arguments/patterns

问题描述

&variable 在模式或闭包参数中使用时是什么意思?

What does &variable mean when it is used in patterns or closure arguments?

for &code in self.exit_code.iter() { ... }

let mut new_seps = do seps.iter().fold(~[]) |result, &next| { ... }

这里我们有 &code 和 &next 在 for 循环和闭包定义中.& 登录这些是什么意思?为什么我们不能简单地使用 code 和 next 而没有 & 符号?它与模式匹配中的 ref 限定符有关吗?它是否与 trait 实现中 self 参数中的 & 相关?

Here we have &code and &next in for loop and closure definition. What do & sign in these mean? Why cannot we simply use code and next, without the ampersand? Is it related to ref qualifier in pattern matching? Is it related to & in self argument in trait implementations?

我在 current 中都找不到有关此语法的任何内容Rust 参考手册 和教程.目前，我认为这是某种隐式取消引用(这是从错误消息中得出的，如果我在模式中省略 &，则会出现该错误消息)，但我不确定.

I couldn't find anything on this syntax neither in current Rust reference manual nor in tutorial. Currently I think this is some kind of implicit dereferencing (this follows from error message which appears if I omit & in the pattern), but I'm not sure.

推荐答案

这是一个模式匹配，解构"&T 类型的东西.也就是说，在

It's a pattern match, "destructuring" something of type &T. That is, in

let &x = &1i; 

x 的类型为 int，值为 1.所以它实际上与 ref 相反(它做了@KerrekSB 所说的，ref x 通过引用而不是通过值捕获).

x has type int, and value 1. So it's actually the opposite to ref (which does what @KerrekSB is saying, ref x captures by reference rather than by value).

可以认为它类似于

match returns_an_option() {
    Some(a) => { ... }
    None => { ... }
}

除了&T的构造函数是&，不是Some或None.

except the constructor of &T is &, not Some or None.

在这个特定的例子中，我猜 seps 是一个向量(你陈述的错误表明它可能是一个 &[&str])，所以 .iter() 返回一个实现 Iterator<& 的对象&str>，也就是说，它是对向量元素(&str)的引用的迭代器，因此，您需要取消引用next 以某种方式获得原始 &str.这可以通过 & 在模式匹配中完成(如代码所示)或与 *next 一起使用.

In this specific instance, I guess seps is a vector (the error you state indicates it's probably a &[&str]), so the .iter() returns an object that implements Iterator<& &str>, that is, it is an iterator over references to the elements of the vector (&str), thus, you need to dereference next somehow to get to the original &str. This can be done by & in a pattern match (as the code demonstrates) or with *next when it is used.

(请注意，& 模式仅适用于隐式可复制类型，因为不能从引用/借用 指针(即 &T).)

(Note that the & patterns only work with implicitly copyable types, as one cannot move ownership out from a reference/borrowed pointer (i.e. &T).)

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