Mixing mutable and immutable references to a HashMap

Question

I want to create a function which gets the value associated to a key in a hashtable, and, if such value does not exist, inserts an arbitrary value (let us say 0).

use std::collections::HashMap;

fn get_or_insert(table: &mut HashMap<i32, i32>, key: i32) -> i32 {
    match table.get(&key) {
        None => table.insert(key, 0).unwrap(),
        Some(v) => *v,
    }
}

This code does not compile:

error[E0502]: cannot borrow `*table` as mutable because it is also borrowed as immutable
 --> src/main.rs:5:17
  |
4 |     match table.get(&key) {
  |           ----- immutable borrow occurs here
5 |         None => table.insert(key, 0).unwrap(),
  |                 ^^^^^ mutable borrow occurs here
6 |         Some(v) => *v,
7 |     }
  |     - immutable borrow ends here

Indeed, table is mutably borrowed in the method insert whereas it is immutably borrowed in the method get.

I can see no way to separate the mutable and immutable parts in this function.

Shepmaster · Accepted Answer · 2017-10-11T13:19:39.023

This is a great time to use the entry method:

use std::collections::HashMap;

fn get_or_insert(table: &mut HashMap<i32, i32>, key: i32) -> i32 {
    *table.entry(key).or_insert(0)
}

fn main() {}

But you are correct, otherwise you'd have to split the call:

fn get_or_insert(table: &mut HashMap<i32, i32>, key: i32) -> i32 {
    match table.get(&key) {
        None => {}
        Some(v) => return *v,
    }

    table.insert(key, 0).unwrap()
}

However, you have to calculate the hash a second time, which is part of the reason that entry was created.

As a side note, insert returns the previous value for the key. If I'm reading it correctly, your unwrap call will always fail.

thanks, I did not know there was a `return` keyword. This is useful ! — user19018, May 01 '15 at 19:23

Mixing mutable and immutable references to a HashMap

1 Answers1