What is the idiomatic Rust way to copy/clone a vec

问题:

I'm trying to write a parameterized function that takes an immutable vector, clones or copies it, does something to the new vector (such as shuffle it) and returns it as a new owned vector. How can this be done and what is the most idiomatic way to do it?

Attempt #1

pub fn shuffle<T>(vec: &mut [T]) {
    // ... contents removed: it shuffles the vector in place
    // ... so needs a mutable vector
}

pub fn shuffle_create_new<T: Clone>(vec: &[T]) -> Vec<T> {
    let mut newvec = vec.clone();
    shuffle(&mut newvec);
    return newvec.to_owned();
}

Fails with:

error[E0596]: cannot borrow immutable borrowed content as mutable
 --> src/main.rs:8:13
  |
8 |     shuffle(&mut newvec);
  |             ^^^^^^^^^^^ cannot borrow as mutable

even though I declared newvec as mutable. I don't understand why.

Attempt #2

pub fn shuffle_owned<T: Clone>(mut vec: Vec<T>) -> Vec<T> {
    shuffle(&mut vec);
    return vec;
}

While this compiles, it doesn't do what I want. The vector you pass into shuffle_owned gets moved into the function, shuffled and then has its ownership transferred back to the caller (via the return value). So the original vector is modified.

I want to know how to pass in a vector that will not be mutated, but have the values cloned into a new boxed vector and returned when finished - as you do in a functional programming language that has immutable data (such as Clojure).

回答1:

Your attempt #1 is almost correct, you just have to move to_owned() to the first line:

fn shuffle<T>(vec: &mut [T]) {
    // ...
}

fn shuffle_create_new<T: Clone>(vec: &[T]) -> Vec<T> {
    let mut newvec = vec.to_vec();
    shuffle(&mut newvec);
    newvec
}

This happens because calling clone() on a slice will return you a slice (i.e. &[T]), and you cannot go from &[T] to &mut [T] because you cannot choose mutability of a reference (borrowed pointer). When you call to_owned(), however, you are getting a fresh instance of Vec<T> and you can put it into a mutable variable to get mutable vector.

As of Rust 1.0, either slice::to_vec or to_owned() method from the ToOwned trait can be used to create Vec<T> from &[T].

There are also now several ways to obtain &mut [T] from Vec<T>: the slicing notation (&mut vec[..]), the deref conversion (&mut *vec) or the direct method call (vec.as_mut_slice(), though this one is deprecated):

回答2:

This is probably what you want:

pub fn copy_shuffle<T: Clone>(vec: &Vec<T>) -> Vec<T> {
    let mut vec = vec.clone();
    shuffle(&mut vec);
    vec
}

or

pub fn copy_shuffle<T: Clone>(vec: &[T]) -> Vec<T> {
    let mut vec = vec.to_vec();
    shuffle(&mut vec);
    vec
}