pointers - 分配给 *mut T 和 &mut T 有什么区别？

Question

这段代码:

let mut a2 = 99;
let b: *mut i32 = &mut a2;
*b = 11; // does not compile , even after unsafe {*b}

产生错误:

error[E0133]: dereference of raw pointer requires unsafe function or block
 --> src/main.rs:4:5
  |
4 |     *b = 11;
  |     ^^^^^^^ dereference of raw pointer

但是这段代码有效:

let mut a2 = 99
let b = &mut a2;
*b = 11;

两者有什么区别？

Answer 1

What is the difference between the two?

一个是原始指针 (*mut _)，另一个是引用 (&mut _ ).正如书中所说:

the compiler guarantees that references will never be dangling

此外，引用永远不会是NULL。取消引用引用总是安全的。取消引用原始指针并不总是安全的，因为编译器不能保证其中任何一个。因此，您需要一个 unsafe block :

unsafe { *b = 11; }

另见:

• Understanding Pointer Types in Rust
• References and Borrowing
• Unsafe Rust