int[] ret = new int[list.size()];       
Iterator<Integer> iter = list.iterator();
for (int i=0; iter.hasNext(); i++) {       
    ret[i] = iter.next();                
}                                        
return ret;                              

Java 8为我们提供了一种通过流来实现这一点的简单方法……

使用collections stream()函数，然后映射到int，你将得到一个IntStream。对于IntStream，我们可以调用toArray()，得到int []

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

int []

to IntStream

Use:

int[] toIntArray(List<Integer> list)  {
    int[] ret = new int[list.size()];
    int i = 0;
    for (Integer e : list)
        ret[i++] = e;
    return ret;
}

对代码的这一轻微更改是为了避免昂贵的列表索引(因为list不一定是ArrayList，但它可以是链表，对于链表，随机访问是昂贵的)。

除了Commons Lang，你还可以使用Guava的方法Ints。toArray(收集<整数>收集):

List<Integer> list = ...
int[] ints = Ints.toArray(list);

这使您不必自己进行Commons Lang对等物要求的中间数组转换。

在Java 8中添加流后，我们可以编写如下代码:

int[] example1 = list.stream().mapToInt(i->i).toArray();
// OR
int[] example2 = list.stream().mapToInt(Integer::intValue).toArray();

思维过程:

The simple Stream#toArray returns an Object[] array, so it is not what we want. Also, Stream#toArray(IntFunction<A[]> generator) doesn't do what we want, because the generic type A can't represent the primitive type int So it would be nice to have some stream which could handle the primitive type int instead of the wrapper Integer, because its toArray method will most likely also return an int[] array (returning something else like Object[] or even boxed Integer[] would be unnatural here). And fortunately Java 8 has such a stream which is IntStream So now the only thing we need to figure out is how to convert our Stream<Integer> (which will be returned from list.stream()) to that shiny IntStream. Quick searching in documentation of Stream while looking for methods which return IntStream points us to our solution which is mapToInt(ToIntFunction<? super T> mapper) method. All we need to do is provide a mapping from Integer to int. Since ToIntFunction is functional interface we can provide its instance via lambda or method reference. Anyway to convert Integer to int we can use Integer#intValue so inside mapToInt we can write: mapToInt( (Integer i) -> i.intValue() ) (or some may prefer: mapToInt(Integer::intValue).) But similar code can be generated using unboxing, since the compiler knows that the result of this lambda must be of type int (the lambda used in mapToInt is an implementation of the ToIntFunction interface which expects as body a method of type: int applyAsInt(T value) which is expected to return an int). So we can simply write: mapToInt((Integer i)->i) Also, since the Integer type in (Integer i) can be inferred by the compiler because List<Integer>#stream() returns a Stream<Integer>, we can also skip it which leaves us with mapToInt(i -> i)

也可以试试Dollar(检查这个修订版):

import static com.humaorie.dollar.Dollar.*
...

List<Integer> source = ...;
int[] ints = $(source).convert().toIntArray();