Java 8-reduce(0,integer：：sum）与reduce(0,（a,b）->a+b)之间的差异

Question

List<Integer> numbers = Arrays.asList(1, 2, 1, 3, 3, 2, 4);

Integer sum1 = numbers.stream().reduce(0, (a, b) -> a+b);
System.out.println("SUM ="+sum1);

Integer product = numbers.stream().reduce(0, (a, b) -> a*b);
System.out.println("PRODUCT = "+product);

int sum2 = numbers.stream().reduce(0, Integer::sum);
System.out.println("SUM 2= "+sum2);

Optional<Integer> sum3 = numbers.stream().reduce((a, b) -> (a + b));
System.out.println("SUM3="+sum3);

// Updated as per  @Hadi J comment
int sum4 = numbers.stream().mapToInt(Integer::intValue).sum();
System.out.println("Sum ="+sum4);

Answer 1

使用性能的最佳方法是什么，如果使用大整数值，为什么？

主要思想是在处理原语及其对象表示时忽略自动装箱/取消装箱。

添加两个integer对象比添加两个int基元复杂得多。你可以在这个帖子中找到更多的细节。

更糟糕的（但不是最糟糕的）情况是将对象integer与原语int求和。它将导致第一个参数的取消装箱。这就是当您的初始数组或集合包含对象（例如integer[]、List等）时的工作方式而累加器是原始的。这将使集合的每个元素完全取消装箱一次，而累加器将保持不变。

最坏的情况是将对象集合汇总为integer累加器。

>

普通旧Java：

Integer[] array = {0,1,2,3,4,5,6,7,8,9};
Integer sum = 0;
// Each element of the array will be unboxed exactly once
// sum will be unboxed 10 times
// result of the sum will be boxed 10 times
for(int i : array) sum += i;

Integer[] array = {0,1,2,3,4,5,6,7,8,9};
int sum = 0;
// Each element of the array will be unboxed exactly once
// result of the sum will be boxed 10 times
for(int i : array) sum += i;

int[] array = {0,1,2,3,4,5,6,7,8,9};
int sum = 0;
// No boxing/unboxing at all
for(int i : array) sum += i;

流API

>

减少流 的和

Stream<Integer> stream = Stream.of(0,1,2,3,4,5,6,7,8,9);
// Each element will be unboxed exactly once
// The accumulator will be unboxed 10 times
// The result of the sum operation will be boxed 10 times
Integer sum = stream.reduce(0, Integer::sum);

减少流 映射到基元int的和

Stream<Integer> stream = Stream.of(0,1,2,3,4,5,6,7,8,9);
// Each element will be unboxed exactly once
// Neither accumulator not result will be unboxed/boxed
int sum = stream.mapToInt(Integer::intValue).reduce(0, Integer::sum);

IntStream stream = IntStream.of(0,1,2,3,4,5,6,7,8,9);
// No boxing/unboxing at all
int sum = stream.reduce(0, Integer::sum);
// same as
int sum2 = stream.sum();