&mut T is copied in MIR?

1 
  let mut local = 0;
  let mut x = &mut local;
  f(x);
  *x = 101;

From the source codes, I know that f(x) will reborrow x. Thus, x is active later. However, when we transfer the above code into MIR. It seams that x is copied.

    let mut _0: ();
    let mut _1: i32;
    let mut _2: &mut i32;
    let  _3: ();
    debug local => _1;
    debug x => _2;
    bb0: {
        _1 = 0_i32;
        _2 = &mut _1;
        _3 = f(_2) -> [return: bb1, unwind continue];
    }
    bb1: {
        (*_2) = 101_i32;
        _1 = 10_i32;
        return;
    }

By the terminator of bb0, _2 is copied. Is it optimized?

Since &mut T does not implement Copy, is it ok to copy &mut T in MIR? Or, do I misunderstand the semantic of Copy in MIR(not trait)?

2

The reborrow semantics effectively happen as part of the Call terminator.

3

(I don't have the answer if you did, but) did you mean:

    bb0: {
        _1 = const 0_i32;
        _2 = &mut _1;
        _3 = f(copy _2) -> [return: bb1, unwind continue];
    }

(There is no copy in your OP MIR.)

4

You are right. The MIR from Rust Playground is right. I print the MIR on my computer. The copy is hidden.

5

I get it. But, why does MIR write copy _2?

6

There are multiple phases of MIR, and which one you're in matters. Before NLL runs, this is illegal. After NLL runs, this is entirely fine. ptr::read::<&mut T>(p) will give you a copy of a &mut in optimized MIR, for example.

3 Likes
7

The MIR is retrieved from rustc_internal API. I think that the NLL has ran.