Java9中对集合类扩展的of方法解析 Java9中对集合类扩展的of方法解

Java9 集合类扩展of方法
Java9集合类中重载多个of方法原因

有如下描述

Java9 集合类扩展of方法

package com.jd.collections; import org.junit.Test; import java.util.List; import java.util.Map; import java.util.Optional; import java.util.Set; import java.util.stream.IntStream; import java.util.stream.Stream; public class StreamTest {@Testpublic void testSet() {Set integerSet = Set.of(1, 2, 3, 4, 5, 6, 7, 8); System.out.println(integerSet); }@Testpublic void testList() {List integerSet = List.of(1, 2, 3, 4, 5, 6, 7, 8); System.out.println(integerSet); }@Testpublic void testMap() {Map stringMap = Map.of("k1", "v1", "k2", "v2", "k3", "v3"); System.out.println(stringMap); Map.Entry entry1 = Map.entry("k1", "v1"); Map.Entry entry2 = Map.entry("k11", "v11"); Map.Entry entry3 = Map.entry("k12", "v12"); Map mapOfEntries = Map.ofEntries(entry1, entry2, entry3); System.out.println(mapOfEntries); }@Testpublic void testStream1() {Optional integerOptional = Stream.ofNullable(Integer.valueOf("1232")).findAny(); System.out.println(integerOptional.get()); }@Testpublic void testStream2() {Stream.of(1, 2, 3, 4, 5, 6).dropWhile(x -> x == 6)/*.takeWhile(x -> x == 2)*/.forEach(System.out::println); }@Testpublic void testStream3() {IntStream.of(1, 2, 3, 4, 5, 6).forEach(System.out::println); }@Testpublic void testStream4() {IntStream.iterate(1, i -> i < 10, i -> i + 2).forEach(System.out::println); }//@Test//public void testFlow() {//Flow.Processor//}}

Java9集合类中重载多个of方法原因在java9 api的集合类中，有很多看似一样的重载of方法：

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那这里有个问题是为什么有了VarArgs（可变长参数）方法，还需要定义那么多重载的方法呢？查看官方的更新日志中可以发现
【Java9中对集合类扩展的of方法解析】
有如下描述
http://openjdk.java.net/jeps/269

These will include varargs overloads, so that there is no fixed limit on the collection size. However, the collection instances so created may be tuned for smaller sizes. Special-case APIs (fixed-argument overloads) for up to ten of elements will be provided. While this introduces some clutter in the API, it avoids array allocation, initialization, and garbage collection overhead that is incurred by varargs calls. Significantly, the source code of the call site is the same regardless of whether a fixed-arg or varargs overload is called.

大致得意思是，虽然重载了这么多of方法会造成api的混乱，但它避免了varargs调用引起的数组分配,初始化和垃圾收集开销。因为固定参数的重载方法，返回的是一个immutable list（不可变集合）。
以上为个人经验，希望能给大家一个参考，也希望大家多多支持脚本之家。