How can I get an item from Vec to Rust?

Question

What I need is a method:

fn take<T>(vec: Vec<T>, index: usize) -> Option<T>

However, I cannot find such a method. Am I missing something? Or there is a reason that it is really unsafe / cannot be done correctly.

Edit: is this another question from the Built-in * secure * way to navigate from Vec <T>? There was a method removethat did not panic outside of access (instead, it returned a result). Here I am looking for a method that uses Vec and returns one of the elements. None of the answers to this question apply to my question.

Edit 2: To clarify, I am looking for a method that consumes Vec and returns a single element without the overhead of restoring Vec invariants in the way removeand swap_removedo.

Answer 1

You can write your function as follows:

fn take<T>(mut vec: Vec<T>, index: usize) -> Option<T> {
    if vec.get(index).is_none() {
        None
    } else {
        Some(vec.swap_remove(index))
    }
}

This is guaranteed by O (1).

---

To mention another solution using iterators:

fn take<T>(vec: Vec<T>, index: usize) -> Option<T> {
    vec.into_iter().nth(index)
}

I was going to write this:

Although Iterator::nth()usually a linear time operation, an iterator over a vector overrides this method to make it an O (1) operation.

But then I noticed that this is only true for an iterator that iterates over slices. The iterator std::vec::IntoIterthat will be used in the above code does not cancel nth(). An attempt was made here , but it doesn't seem to be that simple.

So: O (n)!

Answer 2

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---

swap_remove:

fn take<T>(mut vec: Vec<T>, index: usize) -> Option<T> {
    if index < vec.len() {
        Some(vec.swap_remove(index))
    } else {
        None
    }
}

drain:

fn take<T>(mut vec: Vec<T>, index: usize) -> Option<T> {
    if index < vec.len() {
        vec.drain(index..index+1).next()
    } else {
        None
    }
}

, : O (1).

---

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