Abd*_*lah 4 immutability dereference rust raw-pointer
我有一个必须以原始指针形式检索的结构。
pub struct BufferData {
/// Memory map for pixel data
pub map: Arc<Box<memmap::MmapMut>>,
pub otherdata: i32,
}
Run Code Online (Sandbox Code Playgroud)
我需要写入它的map字段,所以我将原始指针解引用到结构中,然后尝试写入它的数据字段。但是,我收到以下错误。
pub struct BufferData {
/// Memory map for pixel data
pub map: Arc<Box<memmap::MmapMut>>,
pub otherdata: i32,
}
Run Code Online (Sandbox Code Playgroud)
如何使map字段可变和可写?
使用以下代码可重现该错误:
extern crate memmap;
use std::fs::File;
use std::sync::Arc;
use std::boxed::Box;
use std::ops::Deref;
pub struct BufferData {
/// Memory map for pixel data
pub map: Arc<Box<memmap::MmapMut>>,
pub otherdata: i32,
}
fn main() -> () {
// Somewhere on other module
let mut mmap = Arc::new(Box::new(unsafe {
memmap::MmapMut::map_mut(&File::open("./shm").expect("file")).expect("MmapMut")
}));
let mut bfr = BufferData {
map: mmap,
otherdata: 0,
};
let ptr: *const _ = 𝔟
// Here, I must receive and process a pointer
let mut bdata: &BufferData = unsafe { &*(ptr as *const BufferData) };
let mut map = bdata.map.deref().deref();
// Problem lies here: need to write into it, so need to be mutable
map[0] = 9;
}
Run Code Online (Sandbox Code Playgroud)
箱: memmap = "0.6.2"
正如@Shepmaster 指出的那样,Arc在设计上是不可变的:
Rust 中的共享引用默认不允许突变,
Arc也不例外:您通常无法获得对Arc. 如果通过需要发生变异Arc,使用Mutex,RwLock或者一个Atomic类型。
我添加了一个Mutex来解决问题:
pub struct BufferData {
/// Memory map for pixel data
pub map: Arc<Mutex<Box<memmap::MmapMut>>>,
pub otherdata: i32,
}
Run Code Online (Sandbox Code Playgroud)
要访问该字段,我需要先锁定它。我添加了一个额外的块以Mutex在块完成时自动解锁:
{
let mut map = bdata.map.lock().unwrap();
map[0] = 9;
}
Run Code Online (Sandbox Code Playgroud)