java LinkedList 方法 getFirst() 和 element()

Question

So I can see that from previous answers in stackoverflow that there is a difference between getFirst and peekFirst as peekFirst won't throw an exception if the LinkedList is empty.

But what is the difference between getFirst and element ?

getFirst() Returns the first element in this list.

element() Retrieves, but does not remove, the head (first element) of this list.

Both will throw out NoSuchElementException if empty.

Thanks!

Answer 1

They are equivalent as specified in the Deque docs.

Retrieves, but does not remove, the head of the queue represented by this deque (in other words, the first element of this deque). This method differs from peek only in that it throws an exception if this deque is empty.

This method is equivalent to getFirst().

https://docs.oracle.com/javase/7/docs/api/java/util/Deque.html#element()

Answer 2

element is in the Queue interface, while getFirst is not. But (as noted elsewhere) the behaviors are the same.

Answer 3

The source code for LinkedList says it all:

public E element() {
    return getFirst();
}

The element() method just calls getFirst() . So, in the case of the LinkedList implementation of List , both methods effectively do the same thing.

Note: The other implementation of the Deque interface, ArrayDeque , also has an element() implementation which just calls getFirst() .

Answer 4

There is no difference, the implementation of element is

public E element() {
    return getFirst();
}

The javadoc even states

This method is equivalent to {@link #getFirst()}