Borrowing rules

Learn the rules for multiple references at the same time.

Borrowing rules

At any given time, only one of these can be true for a value:

any number of &T (readers), or
a single &mut T (a writer).

let mut v = vec![1, 2, 3];
let ilk = &v[0];        // a reader
// v.push(4);           // error E0502: can't write while a reader exists
println!("{ilk}");
v.push(4);              // ilk is no longer used, fine

This rule prevents a reference from pointing at an old address after the vector grows and moves to new memory.

Fixing the error

The fix for these errors is usually one of: finishing the use of the reference before the change, copying the value (let ilk = v[0]; is Copy for i32), or separating the operations:

let mut v = vec![1, 2, 3];
let ilk = v[0];          // a copy, not a reference
v.push(ilk * 10);
println!("{:?}", v);     // [1, 2, 3, 10]

Exercise Duplicate the first element

Declare let mut v = vec![1, 2, 3];. Take the first element by copying (let ilk = v[0];), then do v.push(ilk * 10); and print with {:?}.

Show example solution

fn main() {
    let mut v = vec![1, 2, 3];
    let ilk = v[0];
    v.push(ilk * 10);
    println!("{:?}", v);
}

Test yourself

In the example above, why would <code>v.push(4)</code> fail on the first line?
- push might move to new memory and ilk would point at an invalid address
- v is immutable
- 4 is the wrong type
- It wouldn't fail
Show answer

push might move to new memory and ilk would point at an invalid address — When capacity runs out, the vector allocates a new array and moves the elements; the borrow checker catches this use-after-free bug at compile time.

Fix with a copy

let mut v = vec![5, 6];
let ilk = ____;   // referans değil
v.push(ilk);

Show answer

let mut v = vec![5, 6];
let ilk = v[0];   // referans değil
v.push(ilk);

Case and result

&v + &v

Allowed
&mut v + &mut v

E0499
&v + v.push()

E0502
let x = v[0]; v.push()

Allowed (copy)

Next lesson: Dangling reference

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