我正在尝试实现一个用Rust编写的跨平台DLL/SharedObject。我在这个库中初始化了一个全局可变的哈希表。当我在这个哈希表中添加一个键和值时，键（&'static str）被释放，用\0字符重置，并重新分配给另一个用途。我不理解这种行为，因为我将变量声明为&'static str，所以Rust在进程完成之前不应该释放内存。你能给我解释一下你的行为吗？你知道解决这个问题的正确方法吗？
我在可执行文件中没有这种行为，只有在DLL中。
代码Rust DLL/SharedObject：

use std::{sync::Mutex, collections::HashMap};
use lazy_static::lazy_static;
use std::io::{stdout, Write};

#[derive(Clone)]
enum Color {
    Red,
}

impl Color {
    fn value(&self) -> i32 {
        match *self {
            Color::Red    => 1,
        }
    }
}

lazy_static! {
    static ref STATE_OK: State = {
        State {
            name: String::from("OK"),
            color: Color::Green,
            character: "+".to_string(),
        }
    };
    static ref STATES: Mutex<HashMap<&'static str, Box<dyn _State + Send>>> = {
        let _states: HashMap<&'static str, Box<dyn _State + Send>> = HashMap::from(
            [
                (
                    "OK",
                    Box::new(STATE_OK.clone()) as Box<dyn _State + Send>
                )
            ]
        );
        Mutex::new(_states)
    };
}

fn rust_from_c_string (string_c: *const u8) -> &'static str {
    let length = strlen(string_c);
    let slice = unsafe { std::slice::from_raw_parts(string_c, length) };
    std::str::from_utf8(slice).unwrap()
}

fn strlen(string_c: *const u8) -> usize {
    let mut length = 0;
    unsafe {
        while *string_c.add(length) != 0 {
            length += 1;
        }
    }
    length
}

#[no_mangle]
pub extern "C" fn add_state (key_: *const u8, character_: *const u8, color_: *const u8) {
    let key: &'static str = rust_from_c_string(key_);
    let color = rust_from_c_string(color_);
    let character = rust_from_c_string(character_);

    let mut _states = STATES.lock().unwrap();
    _states.insert(
        key,
        Box::new(State {
            name: String::from(key),
            color: match color {
                "red"    => Color::Red,
            },
            character: String::from(character),
        }) as Box<dyn _State + Send>
    );
}

#[no_mangle]
pub extern "C" fn messagef (text_: *const u8, state_: *const u8) {
    let text = rust_from_c_string(text_);
    let state = rust_from_c_string(state_);
    let to_print: String;
    let _states = STATES.lock().unwrap();
    let state = _states.get(&*state.unwrap_or("OK").to_string()).unwrap();
    print!("{}", to_print);
    let _ = stdout().flush();
}

#[no_mangle]
pub extern "C" fn print_all_state () {
    DEFAULT_STATE.lock().unwrap().print("not found");

    let _states = STATES.lock().unwrap();

    for (key, state) in _states.iter() {
        state.print(key);
    }
}

从Python中调用：

from ctypes import c_char_p, c_ubyte, c_ushort, pointer
from os.path import join, dirname, exists
from os import name, getcwd

if name == "nt":
    from ctypes import windll as sysdll

    filename = "TerminalMessages.dll"
else:
    from ctypes import cdll as sysdll

    filename = "libTerminalMessages.so"

filenames = (join(dirname(__file__), filename), join(getcwd(), filename))
for filename in filenames:
    if exists(filename):
        break
else:
    raise FileNotFoundError(f"Library {filename!r} is missing")

lib = sysdll.LoadLibrary(filename)

lib.print_all_state()

lib.add_state(
    c_char_p("TEST".encode()),
    c_char_p("T".encode()),
    c_char_p("red".lower().encode()),
)

lib.messagef(
    c_char_p("test".encode()),
    c_char_p("TEST".encode()),
)

lib.print_all_state()

完整的源代码是here。
我截图如下：
1.我添加了两个键和值，分别命名为TEST和TEST2

1. i列出了所有的键和值，TEST和TEST2都有正确的键值
   1.我使用我的库和哈希表与初始化的关键字，这是工作良好
   1.我列出了所有键和值，TEST2值与键\0\0\0\0\0（5个\0字符）一起出现，TEST值与键Ques（先前消息中使用的Question单词的一部分）一起出现。
   1.我使用我的库和哈希表与TEST，TEST2和初始化的关键字。未找到TEST，TEST2，但找到了初始化的密钥。

。