Use of moved values

Question

I'm trying to compute the 10,001st prime in Rust (Project Euler 7), and as a part of this, my method to check whether or not an integer is prime references a vector. However, when I try to reference the vector, I get the following error:

error: use of moved value: `primes`
note: `primes` was previously moved here because it has type `collections::vec::Vec<u64>`, which is non-copyable
error: use of moved value: `primes`
error: aborting due to 2 previous errors

My source code is below:

fn main(){

    let mut count: u32 = 1;
    let mut num: u64 = 1;
    let mut primes: Vec<u64> = Vec::new();
    primes.push(2);


    while count < 10001{
        num += 2;
        if vectorIsPrime(num, primes){
            count+=1;
            primes.push(num);
        }
    }

}

fn vectorIsPrime(num: u64, p: Vec<u64>) -> bool{

    for i in p{
        if num > i && num % i != 0{
            return false;
        }
    }

    true

}

What do I have to do to primes in order to be able to access it within the vectorIsPrime function?

Answer 1

Rust is, as I would say, a “value-oriented” language. This means that if you define primes like this

let primes: Vec<u64> = …

it is not a reference to a vector. It is practically a variable that stores a value of type Vec<u64> just like any u64 variable stores a u64 value. This means that if you pass it to a function defined like this

fn vec_is_prime(num: u64, vec: Vec<u64>) -> bool { … }

the function will get its own u64 value and its own Vec<u64> value.

The difference between u64 and Vec<u64> however is that a u64 value can be easily copied to another place while a Vec<u64> value can only move to another place easily. If you want to give the vec_is_prime function its own Vec<u64> value while keeping one for yourself in main, you have to duplicate it, somehow. That's what's clone() is for. The reason you have to be explicit here is because this operation is not cheap. That's one nice thing about Rust: It's not hard to spot expensive operations. So, you could call the function like this

if vec_is_prime(num, primes.clone()) { …

but that's not really what you want, actually. The function does not need its own a Vec<64> value. It just needs to borrow it for a short while. Borrowing is much more efficient and applicable in this case:

fn vec_is_prime(num: u64, vec: &Vec<u64>) -> bool { …

Invoking it now requires the “borrowing operator”:

if vec_is_prime(num, &primes) { …

Much better. But we can still improve it. If a function wants to borrow a Vec<T> just for the purpose of reading it, it's better to take a &[T] instead:

fn vec_is_prime(num: u64, vec: &[u64]) -> bool { …

It's just more general. Now, you can lend a certain portion of a Vec to the function or something else entirely (not necessarily a Vec, as long as this something stores its values consecutively in memory, like a static lookup table). What's also nice is that due to coersion rules you don't need to alter anything at the call site. You can still call this function with &primes as argument.

For String and &str the situation is the same. String is for storing string values in the sense that a variable of this type owns that value. &str is for borrowing them.

Answer 2

You move value of primes to the function vectorIsPrime (BTW Rust use snake_case by convention). You have other options, but the best one is to borrow vector instead of moving it:

fn vector_is_prime(num: u64, p: &Vec<u64>) -> bool { … }

And then passing reference to it:

vector_is_prime(num, &primes)

Answer 3

With the current definition of your function vectorIsPrime(), the function specifies that it requires ownership of the parameter because you pass it by value.

When a function requires a parameter by value, the compiler will check if the value can be copied by checking if it implements the trait Copy.

• If it does, the value is copied (with a memcpy) and given to the function, and you can still continue to use your original value.
• If it doesn't, then the value is moved to the given function, and the caller cannot use it afterwards

That is the meaning of the error message you have.

However, most functions do not require ownership of the parameters: they can work on "borrowed references", which means they do not actually own the value (and cannot for example put it in a container or destroy it).

fn main(){

    let mut count: u32 = 1;
    let mut num: u64 = 1;
    let mut primes: Vec<u64> = Vec::new();
    primes.push(2);

    while count < 10001{
        num += 2;
        if vector_is_prime(num, &primes){
            count+=1;
            primes.push(num);
        }
    }
}

fn vector_is_prime(num: u64, p: &[u64]) -> bool{

    for &i in p {
        if num > i && num % i != 0{
            return false;
        }
    }
    true
}

The function vector_is_prime() now specifies that it only needs a slice, i.e. a borrowed pointer to an array (including its size) that you can obtain from a vector using the borrow operator & (this is recent in Rust, you may still find examples where a method .as_slice() is called explicitly)

For more information about ownership, I invite you to read the part of the book dealing with ownership.