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将包含整数 ArrayList 转换成 原始整数数组

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    Jeffrey Wang
    Twitter

使用 java-8 的新特性，无需遍历即可轻松转换。

先上源码伺候。

Scanner scanner = new Scanner(System.in);
List<Integer> list = new ArrayList<>();
while (scanner.hasNext()) {
    list.add(scanner.nextInt());
}
list.sort((o1, o2) -> o2 - o1);
list.forEach(System.out::println);

// convert an ArrayList to primitive int[]
int[] arr = list.stream().mapToInt(i->i).toArray();

以下是来自 http://stackoverflow.com/questions/718554/how-to-convert-an-arraylist-containing-integers-to-primitive-int-array 的解释。

If you are using java-8 there's also another way to do this.

int[] arr = list.stream().mapToInt(i -> i).toArray();

What it does is:

  • getting a Stream from the list
  • obtaining an IntStream by mapping each element to itself (identity function), unboxing the int value hold by each Integer object (done automatically since Java 5)
  • getting the array of int by calling toArray

You could also explicitly call intValue via a method reference, i.e:

int[] arr = list.stream().mapToInt(Integer::intValue).toArray();

It's also worth mentioning that you could get a NullPointerException if you have any null reference in the list. This could be easily avoided by adding a filtering condition to the stream pipeline like this:

//.filter(Objects::nonNull) also works
int[] arr = list.stream().filter(i -> i != null).mapToInt(i -> i).toArray();

Example:

List<Integer> list = Arrays.asList(1, 2, 3, 4);
int[] arr = list.stream().mapToInt(i -> i).toArray(); //[1, 2, 3, 4]

list.set(1, null); //[1, null, 3, 4]
arr = list.stream().filter(i -> i != null).mapToInt(i -> i).toArray(); //[1, 3, 4]

最后再次感谢 Alexis C. 的帮助～

在 GitHub 查看