Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
97 views
in Technique[技术] by (71.8m points)

Convert list to array in Java


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Either:

Foo[] array = list.toArray(new Foo[0]);

or:

Foo[] array = new Foo[list.size()];
list.toArray(array); // fill the array

Note that this works only for arrays of reference types. For arrays of primitive types, use the traditional way:

List<Integer> list = ...;
int[] array = new int[list.size()];
for(int i = 0; i < list.size(); i++) array[i] = list.get(i);

Update:

It is recommended now to use list.toArray(new Foo[0]);, not list.toArray(new Foo[list.size()]);.

From JetBrains Intellij Idea inspection:

There are two styles to convert a collection to an array: either using a pre-sized array (like c.toArray(new String[c.size()])) or using an empty array (like c.toArray(new String[0]).

In older Java versions using pre-sized array was recommended, as the reflection call which is necessary to create an array of proper size was quite slow. However since late updates of OpenJDK 6 this call was intrinsified, making the performance of the empty array version the same and sometimes even better, compared to the pre-sized version. Also passing pre-sized array is dangerous for a concurrent or synchronized collection as a data race is possible between the size and toArray call which may result in extra nulls at the end of the array, if the collection was concurrently shrunk during the operation.

This inspection allows to follow the uniform style: either using an empty array (which is recommended in modern Java) or using a pre-sized array (which might be faster in older Java versions or non-HotSpot based JVMs).


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...