string - 为什么迭代HashMap <＆str，＆str>会产生&& str？

我迭代一个HashMap并通过一些局部变量中的模式匹配将值放入其中。
委托人

fn lyrics_no_bottles(song_template:&mut String){
    let mut template_partials = HashMap::new();

    template_partials.insert("start", "No more bottles");
    template_partials.insert("repeat", "no more bottles");
    template_partials.insert("remaining", "99 bottles");
    template_partials.insert("message", "Go to the store and buy some more");

    resolve_template(song_template, template_partials);
}

打电话

fn resolve_template(song_template:&mut String, partials: HashMap<&str, &str>){
    let start:&str;
    let repeat:&str;
    let remaining:&str;
    let message:&str;

    for key in partials.keys(){
        match key {
            "start" => start = partials.get(key),
            "repeat" => repeat = partials.get(key),
            "remaining" => remaining = partials.get(key),
            "message" => message = partials.get(key)
        }
    }

    *song_template = song_template.replace("%1", start);
    *song_template = song_template.replace("%2", repeat);
    *song_template = song_template.replace("%3", message);
    *song_template = song_template.replace("%4", remaining);
}

错误输出

lib.rs:51:5: 58:6 error: type mismatch resolving `<std::collections::hash::map::Keys<'_, &str, &str> as core::iter::Iterator>::Item == &str`:
 expected &-ptr,
    found str [E0271]
lib.rs:51     for key in partials.keys(){
lib.rs:52         match key {
lib.rs:53             "start" => start = partials.get(key),
lib.rs:54             "repeat" => repeat = partials.get(key),
lib.rs:55             "remaining" => remaining = partials.get(key),
lib.rs:56             "message" => message = partials.get(key)
          ...
lib.rs:51:5: 58:6 help: run `rustc --explain E0271` to see a detailed explanation
lib.rs:53:32: 53:49 error: mismatched types:
 expected `&str`,
    found `core::option::Option<&&str>`
(expected &-ptr,
    found enum `core::option::Option`) [E0308]
lib.rs:53             "start" => start = partials.get(key),

当我将参数声明为&&str时，我不明白为什么编译器认为存在HashMap<&str,&str>。

最佳答案

因为that particular method returns a reference wrapped by an Option。
基本上，如果你在那里存储一个String并使用get。。您将得到一个Option<&String>作为调用get的结果。存储&str将返回一个Option<&&str>作为调用get的结果。
值的示例以及在调用get时它们的显示方式：

String -> Option<&String>
&str -> Option<&&str>
u32 -> Option<&u32>

当您将其作为值插入时，&str的所有权将被转移到hashmap。因此，hashmap使用get返回对它们的引用。。。或mutable references using get_mut。
操场现在坏了（不能分享任何东西）。。。所以这里有一个工作。。代码的分解示例：

use std::collections::HashMap;

fn main() {
    let mut map: HashMap<&str, &str> = HashMap::new();
    map.insert("start", "value");

    let mut start: &str = "Unknown";
    let mut repeat: &str;
    let mut remaining: &str;
    let mut message: &str;

    for key in map.keys() {
        match &key[..] {
            "start" => start = map.get(key).unwrap(),
            "repeat" => repeat = map.get(key).unwrap(),
            "remaining" => remaining = map.get(key).unwrap(),
            "message" => message = map.get(key).unwrap(),
            _ => unreachable!()
        }
    }

    println!("Result: {}", start);
}