为什么＆v [1] +＆v [2]与Rust中的v [1] + v [2]具有相同的结果？

Question

I'm learning about ownership and borrowing.

The difference between borrow1 and borrow2 is the usage of & while printing in borrow2 :

fn borrow1(v: &Vec<i32>) {
    println!("{}", &v[10] + &v[12]);
}

fn borrow2(v: &Vec<i32>) {
    println!("{}", v[10] + v[12]);
}

fn main() {
    let mut v = Vec::new();

    for i in 1..1000 {
        v.push(i);
    }

    borrow1(&v);
    println!("still own v {} , {}", v[0], v[1]);

    borrow2(&v);
    println!("still own v {} , {}", v[0], v[1]);
}

Why do they give the same output, even though borrow1 doesn't have & ?

Answer 1

The index operator ( [] ) for a Vec<T> returns a T . In this case, that's an i32 . Thus v[0] returns an i32 and &v[0] returns an &i32 :

let a: i32 = v[0];
let b: &i32 = &v[0];

v[0] only works because i32 implements Copy .

i32 has implemented Add for both the (left-hand side, right-hand-side) pairs of ( i32 , i32 ) and ( &i32 , &i32 ) . The two implementations add values in the same way, so you get the same result.

See also:

• What is the return type of the indexing operation on a slice?
• Understanding (automatic?) Deref/coercion when adding references and values of numbers