Construct an empty vector.

Get the length of a vector.

Time: O(1)

Examples

assert_eq!(5, vector![1, 2, 3, 4, 5].len());

Test whether a vector is empty.

Time: O(1)

Examples

let vec = vector!["Joe", "Mike", "Robert"];
assert_eq!(false, vec.is_empty());
assert_eq!(true, Vector::<i32>::new().is_empty());

Test whether a vector is currently inlined.

Vectors small enough that their contents could be stored entirely inside the space of std::mem::size_of::<Vector<A>>() bytes are stored inline on the stack instead of allocating any chunks. This method returns true if this vector is currently inlined, or false if it currently has chunks allocated on the heap.

This may be useful in conjunction with ptr_eq(), which checks if two vectors’ heap allocations are the same, and thus will never return true for inlined vectors.

Time: O(1)

Test whether two vectors refer to the same content in memory.

This uses the following rules to determine equality:

• If the two sides are references to the same vector, return true.
• If the two sides are single chunk vectors pointing to the same chunk, return true.
• If the two sides are full trees pointing to the same chunks, return true.

This would return true if you’re comparing a vector to itself, or if you’re comparing a vector to a fresh clone of itself. The exception to this is if you’ve cloned an inline array (ie. an array with so few elements they can fit inside the space a Vector allocates for its pointers, so there are no heap allocations to compare).

Time: O(1)

Get an iterator over a vector.

Time: O(1)

Get a mutable iterator over a vector.

Time: O(1)

Get an iterator over the leaf nodes of a vector.

This returns an iterator over the Chunks at the leaves of the RRB tree. These are useful for efficient parallelisation of work on the vector, but should not be used for basic iteration.

Time: O(1)

Get a mutable iterator over the leaf nodes of a vector. This returns an iterator over the Chunks at the leaves of the RRB tree. These are useful for efficient parallelisation of work on the vector, but should not be used for basic iteration.

Time: O(1)

Construct a Focus for a vector.

Time: O(1)

Construct a FocusMut for a vector.

Time: O(1)

Get a reference to the value at index index in a vector.

Returns None if the index is out of bounds.

Time: O(log n)

Examples

let vec = vector!["Joe", "Mike", "Robert"];
assert_eq!(Some(&"Robert"), vec.get(2));
assert_eq!(None, vec.get(5));

Get a mutable reference to the value at index index in a vector.

Returns None if the index is out of bounds.

Time: O(log n)

Examples

let mut vec = vector!["Joe", "Mike", "Robert"];
{
    let robert = vec.get_mut(2).unwrap();
    assert_eq!(&mut "Robert", robert);
    *robert = "Bjarne";
}
assert_eq!(vector!["Joe", "Mike", "Bjarne"], vec);

Get the first element of a vector.

If the vector is empty, None is returned.

Time: O(log n)

Get a mutable reference to the first element of a vector.

If the vector is empty, None is returned.

Time: O(log n)

Get the first element of a vector.

If the vector is empty, None is returned.

This is an alias for the front method.

Time: O(log n)

Get the last element of a vector.

If the vector is empty, None is returned.

Time: O(log n)

Get a mutable reference to the last element of a vector.

If the vector is empty, None is returned.

Time: O(log n)

Get the last element of a vector.

If the vector is empty, None is returned.

This is an alias for the back method.

Time: O(log n)

Get the index of a given element in the vector.

Searches the vector for the first occurrence of a given value, and returns the index of the value if it’s there. Otherwise, it returns None.

Time: O(n)

Examples

let mut vec = vector![1, 2, 3, 4, 5];
assert_eq!(Some(2), vec.index_of(&3));
assert_eq!(None, vec.index_of(&31337));

Test if a given element is in the vector.

Searches the vector for the first occurrence of a given value, and returns true if it's there. If it's nowhere to be found in the vector, it returns false`.

Time: O(n)

Examples

let mut vec = vector![1, 2, 3, 4, 5];
assert_eq!(true, vec.contains(&3));
assert_eq!(false, vec.contains(&31337));

Discard all elements from the vector.

This leaves you with an empty vector, and all elements that were previously inside it are dropped.

Time: O(n)

Binary search a sorted vector for a given element using a comparator function.

Assumes the vector has already been sorted using the same comparator function, eg. by using sort_by.

If the value is found, it returns Ok(index) where index is the index of the element. If the value isn’t found, it returns Err(index) where index is the index at which the element would need to be inserted to maintain sorted order.

Time: O(log n)

Binary search a sorted vector for a given element.

If the value is found, it returns Ok(index) where index is the index of the element. If the value isn’t found, it returns Err(index) where index is the index at which the element would need to be inserted to maintain sorted order.

Time: O(log n)

Binary search a sorted vector for a given element with a key extract function.

Assumes the vector has already been sorted using the same key extract function, eg. by using sort_by_key.

If the value is found, it returns Ok(index) where index is the index of the element. If the value isn’t found, it returns Err(index) where index is the index at which the element would need to be inserted to maintain sorted order.

Time: O(log n)

Construct a vector with a single value.

Examples

let vec = Vector::unit(1337);
assert_eq!(1, vec.len());
assert_eq!(
  vec.get(0),
  Some(&1337)
);

Create a new vector with the value at index index updated.

Panics if the index is out of bounds.

Time: O(log n)

Examples

let mut vec = vector![1, 2, 3];
assert_eq!(vector![1, 5, 3], vec.update(1, 5));

Update the value at index index in a vector.

Returns the previous value at the index.

Panics if the index is out of bounds.

Time: O(log n)

Swap the elements at indices i and j.

Time: O(log n)

Push a value to the front of a vector.

Time: O(1)*

Examples

let mut vec = vector![5, 6, 7];
vec.push_front(4);
assert_eq!(vector![4, 5, 6, 7], vec);

Push a value to the back of a vector.

Time: O(1)*

Examples

let mut vec = vector![1, 2, 3];
vec.push_back(4);
assert_eq!(vector![1, 2, 3, 4], vec);

Remove the first element from a vector and return it.

Time: O(1)*

Examples

let mut vec = vector![1, 2, 3];
assert_eq!(Some(1), vec.pop_front());
assert_eq!(vector![2, 3], vec);

Remove the last element from a vector and return it.

Time: O(1)*

Examples

let mut vec = vector![1, 2, 3];
assert_eq!(Some(3), vec.pop_back());
assert_eq!(vector![1, 2], vec);

Append the vector other to the end of the current vector.

Time: O(log n)

Examples

let mut vec = vector![1, 2, 3];
vec.append(vector![7, 8, 9]);
assert_eq!(vector![1, 2, 3, 7, 8, 9], vec);

Retain only the elements specified by the predicate.

Remove all elements for which the provided function f returns false from the vector.

Time: O(n)

Split a vector at a given index.

Split a vector at a given index, consuming the vector and returning a pair of the left hand side and the right hand side of the split.

Time: O(log n)

Examples

let mut vec = vector![1, 2, 3, 7, 8, 9];
let (left, right) = vec.split_at(3);
assert_eq!(vector![1, 2, 3], left);
assert_eq!(vector![7, 8, 9], right);

Split a vector at a given index.

Split a vector at a given index, leaving the left hand side in the current vector and returning a new vector containing the right hand side.

Time: O(log n)

Examples

let mut left = vector![1, 2, 3, 7, 8, 9];
let right = left.split_off(3);
assert_eq!(vector![1, 2, 3], left);
assert_eq!(vector![7, 8, 9], right);

Construct a vector with count elements removed from the start of the current vector.

Time: O(log n)

Construct a vector of the first count elements from the current vector.

Time: O(log n)

Truncate a vector to the given size.

Discards all elements in the vector beyond the given length.

Panics if the new length is greater than the current length.

Time: O(log n)

Extract a slice from a vector.

Remove the elements from start_index until end_index in the current vector and return the removed slice as a new vector.

Time: O(log n)

Insert an element into a vector.

Insert an element at position index, shifting all elements after it to the right.

Performance Note

While push_front and push_back are heavily optimised operations, insert in the middle of a vector requires a split, a push, and an append. Thus, if you want to insert many elements at the same location, instead of inserting them one by one, you should rather create a new vector containing the elements to insert, split the vector at the insertion point, and append the left hand, the new vector and the right hand in order.

Time: O(log n)

Remove an element from a vector.

Remove the element from position ‘index’, shifting all elements after it to the left, and return the removed element.

Performance Note

While pop_front and pop_back are heavily optimised operations, remove in the middle of a vector requires a split, a pop, and an append. Thus, if you want to remove many elements from the same location, instead of removeing them one by one, it is much better to use slice.

Time: O(log n)

Insert an element into a sorted vector.

Insert an element into a vector in sorted order, assuming the vector is already in sorted order.

Time: O(log n)

Examples

let mut vec = vector![1, 2, 3, 7, 8, 9];
vec.insert_ord(5);
assert_eq!(vector![1, 2, 3, 5, 7, 8, 9], vec);

Sort a vector.

Time: O(n log n)

Examples

let mut vec = vector![3, 2, 5, 4, 1];
vec.sort();
assert_eq!(vector![1, 2, 3, 4, 5], vec);

Sort a vector using a comparator function.

Time: O(n log n)

Examples

let mut vec = vector![3, 2, 5, 4, 1];
vec.sort_by(|left, right| left.cmp(right));
assert_eq!(vector![1, 2, 3, 4, 5], vec);

Verify the internal consistency of a vector.

This method walks the RRB tree making up the current Vector (if it has one) and verifies that all the invariants hold. If something is wrong, it will panic.

This method requires the debug feature flag.