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?

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

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

在这里，在for循环和闭包定义中有&code和&next. &表示这些含义是什么?为什么我们不能简单地使用code和next而没有&"号呢?它与模式匹配中的ref限定符有关吗?它与特征实现中self参数中的&有关吗?

在当前Rust参考手册或教程中.目前，我认为这是某种隐式取消引用(如果我在模式中省略了&，则会出现错误消息，这是隐式取消引用的原因)，但是我不确定.

解决方案

这是一个模式匹配，解构"了类型为&T的东西.也就是说，在

let &x = &1i; 

x的类型为int，值1.因此，它实际上与ref相反(@KerrekSB所说的话，ref x是通过引用而不是通过值捕获的.)

可以认为它类似于

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

&T的构造函数是&，而不是Some或None.

---

在这种特定情况下，我想seps是一个向量(您所指出的错误表明它可能是&[&str])，所以.iter()返回一个实现Iterator<& &str>的对象，即，它是一个对向量(&str)元素的引用进行迭代的迭代器，因此，您需要以某种方式取消引用next才能到达原始的&str.可以通过&进行模式匹配(如代码所示)来实现(如代码所示)，或者 与*next一起使用.

(请注意，&模式仅适用于隐式可复制类型，因为不能将所有权从引用/借用指针(即&T)移出.)

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| { ... }

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?

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.

解决方案

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

let &x = &1i; 

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).

One can regard it as similar to

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

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

---

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.

(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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