问题
I want to convert *mut pointer to &mut reference.
// Both setting a value to ptr and getting a value from ptr succeeds.
let ptr: &mut usize = unsafe { &mut *(VIRTUAL_ADDRESS_TO_ACCESS_FREE_PAGE as *mut usize) };
This works. However, if &mut is outside of unsafe block, the code will not work partially. *ptr = foo will not store foo to the memory ptr points, but let foo = *ptr will assign the value of *ptr to foo.
// Setting a value to ptr fails, but getting a value from ptr succeeds.
let ptr: &mut usize = &mut unsafe { *(VIRTUAL_ADDRESS_TO_ACCESS_FREE_PAGE as *mut usize) };
What's the difference between unsafe { &mut } and &mut unsafe{ }?
回答1:
It doesn't have anything to do with unsafe per-se, rather with block boundaries.
&*ptr is a "reborrow", it's basically going to reinterpret the pointer in a new form. So you get different types of pointers (one raw and one reference) to the same object.
&{*ptr} is completely different, because {*ptr} will force a copy (more generally a move but here you're pointing to a usize which is Copy), then borrow the copy. This means you get two pointers to completely different objects.
The objects have the same value since one's pointee is a copy of the other's, and thus reading seems to work, but writing doesn't because... you're not writing where you think you are.
See this demonstration (not using mut pointers because there's no need to to demonstrate the issue)
来源：https://stackoverflow.com/questions/61605289/whats-the-difference-between-mut-unsafe-and-unsafe-mut