rust - 如何安全地创建 UnsafeCell？

Question

UnsafeCell documentation说

The UnsafeCell<T> type is the only legal way to obtain aliasable data that is considered mutable.

唯一的构造方法是:

pub const fn new(value: T) -> UnsafeCell<T>

但是，不可能创建 c_void , 我们只能创建 *mut c_void或 *const c_void .

是否可以创建UnsafeCell<c_void>来自*mut c_void ？有了这个，我们可以让编译器知道指针可以指向可变的东西。

或者这是没有必要的？我们可以一直使用*mut c_void吗？即使我们知道某些 FFI 调用会改变它指向的数据，而且我们有多个对它的引用？

一个用例是:

struct FFIStruct { v: UnsafeCell<c_void>, other_fields: ... }
impl FFIStruct {
    // We don't want to require &mut self, as we 
    // are sure private call_ffi() will always be called 
    // sequentially, and we don't want to stop
    // status() being callable during the call
    fn call_ffi(&self){ ffi_function(self.v.get()) }
    pub fn status(&self) -> FFIStatus { ... }
}

现在我们如何创建FFIStruct ？或者只使用 *mut c_void会好吗？

创建&Cell<c_void>的示例代码

需要 #![feature(as_cell)] :

unsafe fn get_cell<'a>(p: *mut c_void) -> &'a Cell<c_void> {
    Cell::from_mut(&mut *p)
}

Answer 1

TL;DR:只需使用 *mut Foo。这里不需要任何类型的细胞。

---

免责声明:目前还没有正式的 Rust 内存模型。

你不能创建这个类型，period，因为你不能¹创建c_void的实例。

事实是，您不需要创建这样的类型。 Aliasing is not spatial but temporal .您可以有多个 *mut T 指向同一个地方，这并不重要，直到您尝试访问一个。这实质上将其转换为引用，并且在该引用存在时需要坚持别名要求。

raw pointers fall outside of Rust's safe memory model.

— The Rustonomicon

Different from references and smart pointers, raw pointers:

• Are allowed to ignore the borrowing rules by having both immutable and mutable pointers or multiple mutable pointers to the same location
• Aren’t guaranteed to point to valid memory
• Are allowed to be null
• Don’t implement any automatic cleanup

¸— The Rust Programming Language

另见:

• Why does modifying a mutable reference's value through a raw pointer not violate Rust's aliasing rules?
• What's the Rust idiom to define a field pointing to a C opaque pointer?
• Is it undefined behavior to do runtime borrow management with the help of raw pointers in Rust?
• Can an FFI function modify a variable that wasn't declared mutable?

¹ 从技术上讲，您可以，但这只是因为实现和向后兼容性限制。