Why Rust prevents from multiple mutable references?

Question

Like in the topic, why Rust prevents from multiple mutable references? I have read chapter in rust-book, and I understand that when we have multi-threaded code we are secured from data races but let's look at this code:

fn main() {
    let mut x1 = String::from("hello");
    let r1 = &mut x1;
    let r2 = &mut x1;

    r1.insert(0, 'w');

}

This code is not running simultaneously so there is no possibility for data races. What is more when I am creating new thread and I want to use variable from parent thread in a new thread I have to move it so only new thread is an owner of the parent variable.

The only reason I can see is that programmer can lose himself in his code when it is growing up. We have multiple places in which one piece of data can be modified and even the code is not running parallel we can get some bugs.

You should only ask one question per post. I suggest you edit your post to remove the 2nd question and ask it in its own post. — mcarton
Ok, I did it. Here is link: stackoverflow.com/q/58367174/9620900 if u want to answer. — mikeProgrammer
What would be the advantage of having multiple mutable references to x1 in your code? Where would you want to use r2 where you couldn't also just use r1? — Bergi
Whole Java works in the way I described above. You can have million of objects containing reference to the same piece of data in the same scope or am I wrong? — mikeProgrammer

mcarton mcarton · Accepted Answer · 2019-10-13T15:28:46

The fact that Rust prevent two mutable references at the same time to prevent data races is a common misconception. This is only one of the reasons. Preventing two mutable references makes it possible to keep invariants on types easily and let the compiler enforce that the invariant are not violated.

Take this piece of C++ code for an example:

#include <vector>

int main() {
    std::vector<int> foo = { 1, 2, 3 };
    
    for (auto& e: foo) {
        if (e % 2 == 0) {
            foo.push_back(e+1);
        }
    }

    return 0;
}

This is unsafe because you cannot mutate a vector while you are iterating it. Mutating the vector might reallocate its internal buffer, which invalidates all references. In C++, this is a UB. In Python, Java or C# (and probably most other languages), you would get a runtime exception.

Rust however, prevents this kind of issues at compile time:

fn main() {
    let mut foo = vec![1, 2, 3];
    
    for e in foo {
        if e % 2 == 0 {
            foo.push(e+1);
        }
    }
}

gives an error:

error[E0382]: borrow of moved value: `foo`
 --> src/main.rs:6:13
  |
2 |     let mut foo = vec![1, 2, 3];
  |         ------- move occurs because `foo` has type `std::vec::Vec<i32>`, which does not implement the `Copy` trait
3 |     
4 |     for e in foo {
  |              ---
  |              |
  |              value moved here
  |              help: consider borrowing to avoid moving into the for loop: `&foo`
5 |         if e % 2 == 0 {
6 |             foo.push(e+1);
  |             ^^^ value borrowed here after move

Why Rust prevents from multiple mutable references?

1 Answers