Question

我尝试使用函数作为值创建HashMap：

#[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()
}

编译器给出了一个错误：

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

我不明白我该怎么做才能解决这个错误，而且我也不知道如何创建这样的HashMap。

Answer 1

您的代码有多个问题。编译器提供的错误告诉您，您的代码将使内存不安全：

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

您在HashMap中存储的类型无法保证可以共享。

你可以修复＆＃34;通过将值类型更改为&'static (Fn([u8]) + Sync)来指定此类限制。这会解锁下一个错误，因为您的功能签名不匹配：

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}>`

＆＃34;定影＆＃34;使用&'static (Fn(&[u8]) -> String + Sync)会导致更深奥的生命周期错误：

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}>`

可以＆＃34;修复＆＃34;通过使用&md5 as &'static (Fn(&[u8]) -> String + Sync))转换函数，这将导致

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

这是因为the reference you've made is to a temporary value that doesn't live outside of the scope。

我把修复引入恐慌报价，因为这不是真正正确的解决方案。正确的做法是使用函数指针：

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
    };
}

老实说，我说HashMap可能有点过分;我使用的是阵列。小数组可能比小HashMap：

更快

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

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

你可以从遍历列表开始，或者可能是花哨的并按字母顺序排列它，然后当它变得更大时使用binary_search。

如何使用函数引用创建一个lazy_static HashMap作为值？

1 个答案: