rust - 另一个关于 cannot borrow `x` as mutable in time more than once 错误的询问

Question

考虑以下代码:

struct Foo {
    x: i32,
}

struct Test {
    foos: Vec<Foo>,
    v: i32,
}

fn bar(test: &mut Test, foo: &mut Foo){
    // make something with test
    foo.x = 42;
}

fn main() {
    let test = &mut Test {
        foos: vec![
            Foo {x: 3}, 
            Foo {x: 4},
        ], 
        v: 55
    };

    // First time test got borrowed
    let iterator = test.foos.iter_mut();

    for elem in iterator {
        // !!Error second time test got borrowed as mutable
        bar(test, elem);
    }

}

假设我不能改变它的签名，有没有办法在循环中为 test.foos 的每个元素调用 bar 而不会违反唯一的 &mut 一次规则？

我能想到的唯一解决方案是在它第一次被借用为可变之前克隆测试，如:

let mut tutu = test.to_owned();
let iterator = test.foos.iter_mut();

for elem in iterator {
    bar(&mut tutu, elem);
}

但是根据结构的不同，克隆结构感觉可能是昂贵的操作。

Answer 1

您链接到的 GitHub Java 示例实际上并不要求父项和子项都是可变的。事实上，父级(在您的 Java 中，这是 safebox)只会调用 getUploadUrl()。

因此，我已将其稍微翻译成 Rust 语言，并使用与您的 Java 示例相似的类型名称。这个想法是方法本身只关心 URL ......所以直接传递它而不是采用可变引用:

struct Attachment {
    guid: String,
}

impl Attachment {
    fn new<S>(guid: S) -> Attachment
        where S: Into<String>
    {
        Attachment { guid: guid.into() }
    }

    fn set_guid<S>(&mut self, guid: S)
        where S: Into<String>
    {
        self.guid = guid.into();
    }
}

struct SafeBox {
    attachments: Vec<Attachment>,
    upload_url: &'static str,
}

impl SafeBox {
    fn get_upload_url(&self) -> &str {
        self.upload_url
    }
}

fn uploadAttachment(upload_url: &str, attachment: &mut Attachment) {
    attachment.set_guid("12345");
}

fn main() {
    let mut safebox = SafeBox {
        attachments: vec![Attachment::new("12345"), Attachment::new("67890")],
        upload_url: "http://upload.com/upload",
    };

    for mut attachment in &mut safebox.attachments {
        uploadAttachment(safebox.upload_url, &mut attachment);
    }
}

这是running in the playground .

作为曾尝试将简单的 C# 库移植到 Rust 的人，我可以根据经验告诉您，尝试逐行、逐类型地移植它是一个坏主意。 Rust Way (tm) 非常不同，C#(或 Java)根本无法直接移植。