why rust won’t let you borrow twice

Berend De Schouwer

What Question Am I Answering?

A question came up in a coding class, about why in Rust there’s a borrow, and there’s a an error when borrowing.

You may have seen it like (copied from the rust book):

error[E0499]: cannot borrow `s` as mutable more than once at a time
 --> src/main.rs:5:14
  |
4 |     let r1 = &mut s;
  |              ------ first mutable borrow occurs here
5 |     let r2 = &mut s;
  |              ^^^^^^ second mutable borrow occurs here
6 |
7 |     println!("{}, {}", r1, r2);
  |                        -- first borrow later used here

For more information about this error, try `rustc --explain E0499`.
error: could not compile `ownership` due to previous error

The Rust book does explain what borrow is, and the theory behind why it’s allowed and not allowed.

I’m showing the risks practical example.

How Am I Answering It?

With as little computer theory as possible.

• No assembler
• No diagrams
• No multiple languages
• short, simple examples
• No theory until after the bug is shown

Caveat

The answer is in C. It’s in C because Rust won’t let me break borrow, not even in an unsafe {} block.

It’s all in plain C, though. No assembler, and no C++. As straightforward C as I can make it, specifically to allow for plain examples. As few as possible shortcuts are taken.

I’m not trying to write a C-programmer’s C code. I’m trying to write an example that shows the problem, for non-C programmers.

It should not be difficult to follow coming from Rust.

Inspiration

The inspiration comes from the C max() function, which is actually a macro.

That is great for explaining the differences between functions and macros, and it’s great for explaining pass-by-code instead of value or reference.

It’s also great for explaining surprises.

First Examples

First, we give example code for pass-by-value and pass-by-reference. Pass-by-reference is also called borrow in Rust. The examples are simple, and we do not yet discuss the difference.

For now, it’s simply about C syntax.

#include <stdio.h>

int double_by_value(int x) {
    x = x * 2;
    return x;
}

int double_by_reference(int *x) {
    *x = *x * 2;
    return *x;
}

int main() {
    int x;
    x = 5;
    printf("double_by_value(x) = %d\n", double_by_value(x));
    /* double_by_value(x) = 10 */
    x = 5;
    printf("double_by_reference(x) = %d\n", double_by_reference(&x));
    /* double_by_reference(x) = 10 */
    return 0;
}

The code is almost identical. You can see the syntax difference, but no functional difference yet.

The logic is the same, the input is the same, and the output is the same.

Extra variables

Let’s make it a tiny bit more challenging, and add a second variable.

#include <stdio.h>

int add_by_value(int x, int y) {
    x = x + y;
    return x;
}

int add_by_reference(int *x, int *y) {
    *x = *x + *y;
    return *x;
}

int main() {
    int x, y;

    x = y = 5;
    printf("add_by_value(x, y) = %d\n", add_by_value(x, y));
    /* add_by_value(x, y) = 10 */
    x = y = 5;
    printf("add_by_reference(x, y) = %d\n", add_by_reference(&x, &y));
    /* add_by_reference(x, y) = 10 */
    return 0;
}

The logic is the same, the input is the same, and the output is the same.

Put it together

Add the two and we get…

#include <stdio.h>

/* Previous example code here */

int add_and_double_by_value(int x, int y) {
    x = double_by_value(x);
    y = double_by_value(y);
    x = x + y;
    return x;
}

int add_and_double_by_reference(int *x, int *y) {
    *x = double_by_reference(x);
    *y = double_by_reference(y);
    *x = *x + *y;
    return *x;
}

int main() {
    int x, y;

    x = y = 5;
    printf("add_and_double_by_value(x, y) = %d\n", add_and_double_by_value(x, y));
    /* add_and_double_by_value(x, y) = 20 */
    x = y = 5;
    printf("add_and_double_by_reference(x, y) = %d\n", add_and_double_by_reference(&x, &y));
    /* add_and_double_by_reference(x, y) = 20 */
    return 0;
}

No surprised yet. Both examples print the same, correct answer.

The logic is the same, the input is the same, and the output is the same.

Surprise

Now let’s make it simpler…

#include <stdio.h>

/* Previous example code here */

int main() {
    int x;
    x = 5;
    printf("add_and_double_by_value(x, x) = %d\n", add_and_double_by_value(x, x));
    /* add_and_double_by_value(x, x) = 20 */
    x = 5;
    printf("add_and_double_by_reference(x, x) = %d\n", add_and_double_by_reference(&x, &x));
    /* add_and_double_by_reference(x, x) = 40 */
    return 0;
}

The logic is the same, the input is the same, but the output is different.

Why is it 40?

Double borrow

So what happened? Why did removing a variable, y, result in the “wrong” answer?

Pass by value copies the value, and passes the value, not the variable. The original is never modified.

Pass by reference (or borrow), passes a reference, not a copy. The original is modified.

Look at

int add_and_double_by_reference(int *x, int *y) {
    *x = double_by_reference(x);
    *y = double_by_reference(y);
    *x = *x + *y;
    return *x;
}

When x is doubled, at *x = double_by_reference(x);, and *x is also *y, y is also doubled. X is now 10, as expected, but y is also 10.

Then y is doubled. And since *y is _also *x, x is doubled again. Now both variables are 20.

Tada!

This is (one of the reasons) why Rust won’t let you borrow the same variable twice.

What do you do instead?

Borrow once, or .clone() /* copy */.

Bonus time

So why have pass by reference or borrow at all?

1. It’s faster, when the data is large.
2. It’s faster for streamed data.
3. It’s not possible in assembler pass complicated variables by value. Manual copies must be made, and the language you use might not implement implicit copies.

Extra Points

For extra points, do the same with threads.