如何用函數引用創建一個lazy_static HashMap作爲值？

Question

我試圖創建一個功能HashMap的值：

#[macro_use] 
extern crate lazy_static; 

use std::collections::HashMap; 

lazy_static! { 
    static ref HASHES: HashMap<&'static str, &'static Fn([u8])> = { 
     let mut m = HashMap::new(); 
     m.insert("md5", &md5); 
     m 
    }; 
} 

fn md5(bytes: &[u8]) -> String { 
    String::default() 
} 

編譯器給我一個錯誤：

error[E0277]: the trait bound `std::ops::Fn([u8]) + 'static: std::marker::Sync` is not satisfied in `&'static std::ops::Fn([u8]) + 'static` 
    --> src/main.rs:6:1 
    | 
6 | lazy_static! { 
    | _^ starting here... 
7 | |  static ref HASHES: HashMap<&'static str, &'static Fn([u8])> = { 
8 | |   let mut m = HashMap::new(); 
9 | |   m.insert("md5", &md5); 
10 | |   m 
11 | |  }; 
12 | | } 
    | |_^ ...ending here: within `&'static std::ops::Fn([u8]) + 'static`, the trait `std::marker::Sync` is not implemented for `std::ops::Fn([u8]) + 'static` 
    | 
    = note: `std::ops::Fn([u8]) + 'static` cannot be shared between threads safely 
    = note: required because it appears within the type `&'static std::ops::Fn([u8]) + 'static` 
    = note: required because of the requirements on the impl of `std::marker::Sync` for `std::collections::hash::table::RawTable<&'static str, &'static std::ops::Fn([u8]) + 'static>` 
    = note: required because it appears within the type `std::collections::HashMap<&'static str, &'static std::ops::Fn([u8]) + 'static>` 
    = note: required by `lazy_static::lazy::Lazy` 
    = note: this error originates in a macro outside of the current crate 

我不明白，我應該做些什麼來解決這個錯誤，我不知道任何其他方式創建這樣的HashMap。

Answer 1

您的代碼有多個問題。由編譯器給出的錯誤是告訴你，你的代碼，可以讓內存不安全：

`std::ops::Fn([u8]) + 'static` cannot be shared between threads safely 

你是存儲在您的HashMap的類型沒有保證它可以共享。

您可以通過將您的值類型更改爲&'static (Fn([u8]) + Sync)來指定此類邊界來「修復」。這解鎖下一個錯誤，由於你的函數簽名不匹配：

expected type `std::collections::HashMap<&'static str, &'static std::ops::Fn([u8]) + std::marker::Sync + 'static>` 
    found type `std::collections::HashMap<&str, &fn(&[u8]) -> std::string::String {md5}>` 

「修復」與&'static (Fn(&[u8]) -> String + Sync)導致深奧，kinded更高的壽命錯誤：

expected type `std::collections::HashMap<&'static str, &'static for<'r> std::ops::Fn(&'r [u8]) -> std::string::String + std::marker::Sync + 'static>` 
    found type `std::collections::HashMap<&str, &fn(&[u8]) -> std::string::String {md5}>` 

哪可以「固定」鑄造用&md5 as &'static (Fn(&[u8]) -> String + Sync))的功能，從而導致

note: borrowed value must be valid for the static lifetime... 
note: consider using a `let` binding to increase its lifetime 

這見底，因爲the reference you've made is to a temporary value that doesn't live outside of the scope。

---

我把修復引入恐慌報價，因爲這不是真正的解決方案。正確的做法是隻用一個函數指針：

lazy_static! { 
    static ref HASHES: HashMap<&'static str, fn(&[u8]) -> String> = { 
     let mut m = HashMap::new(); 
     m.insert("md5", md5 as fn(&[u8]) -> std::string::String); 
     m 
    }; 
} 

老實說，我會說，一個HashMap可能是矯枉過正;我會使用一個數組。一小陣，可能比一個小HashMap更快：

type HashFn = fn(&[u8]) -> String; 

static HASHES: &'static [(&'static str, HashFn)] = &[ 
    ("md5", md5), 
]; 

您可以通過只通過列表迭代開始，也許是幻想和按字母順序排列，然後用binary_search當它變得有點大了。