Can an FFI function modify a variable that wasn

问题:

fn main() {
    let val = 0;
    unsafe { foo(&val) }
}

extern "C" {
    pub fn foo(val: *const u32);
}

Implementation in C:

void foo(unsigned* val) { *val=1; }

Of course, I should pass val: *mut u32, but what happens in the case that I pass an immutable reference? What compiler rules apply? Does val remain unchanged even though I'm passing a pointer to the local variable?

回答1:

I'd say undefined behavior:

Mutating non-mutable data — that is, data reached through a shared reference or data owned by a let binding), unless that data is contained within an UnsafeCell<U>.

And this might include:

• if you use val after the FFI-call it might ignore the writes you did (e.g. cached the value in a register or due to constant propagation)
• segfault in FFI because the referenced memory might be read-only
• the write from FFI might show up in seemingly unrelated locations because the compiler reused the memory and assumed it had a well-defined value
• and worse :)