Move semantics with references

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Where the compiler can choose between moving a borrowed reference and reborrowing, it will choose reborrowing. That's what's happening here: b reborrows the reference from a, and attempting to touch a itself or create any other references to the referrent of a while b is in scope will cause that error. Putting b in a more limited scope allows the rest of that program to compile:

fn main() {
    let a : &mut Box<usize> = &mut Box::new(0);
    
    {
        let b : &mut Box<usize> = a;
    } // b goes out of scope and releases its borrow here

    let c : &Box<usize> = a;
    
    let d : &Box<usize> = &a;
}

This thread goes into more detail about reborrowing, and cites some useful sources.

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