震惊 Guava 竟然有坑

点击上方蓝色“方志朋”，选择“设为星标”

回复“666”获取独家整理的学习资料！

转自：Java布道

最近，团队里边一个兄弟突然叫我：快来看，有个奇怪的事情，无法解释…跑过去一看，是这么一段代码：

 private static class Person {private int age;private String name;public Person(int age, String name) {this.age = age;this.name = name;}public void setAge(int age) {this.age = age;}@Overridepublic String toString() {return name + ":" + age;}}@Testpublic void test_collections2_filter() {Person lxy = new Person(35, "lxy");Person xhf = new Person(34, "xhf");Person nws = new Person(31, "nws");List<Person> names = Lists.newArrayList(lxy, xhf, nws);Collection<Person> personAgeOver30 = Collections2.filter(names, p -> p.age > 30);System.out.println(personAgeOver30);//[lxy:35, xhf:34, nws:31]nws.setAge(25);System.out.println(personAgeOver30);//[lxy:35, xhf:34]}

确实是比较奇怪，personAgeGt30中的元素怎么会少了一个呢？本着任何表现奇怪的程序都有其背后原因的指导思想，打开了Guava Collections2类的源代码。其实，源代码的注释已经解释得非常清楚了：returned collection is a live view of {@code unfiltered};changes to one affect the other.

/*** Returns the elements of {@code unfiltered} that satisfy a predicate. The returned collection is* a live view of {@code unfiltered}; changes to one affect the other.** <p>The resulting collection's iterator does not support {@code remove()}, but all other* collection methods are supported. When given an element that doesn't satisfy the predicate, the* collection's {@code add()} and {@code addAll()} methods throw an {@link* IllegalArgumentException}. When methods such as {@code removeAll()} and {@code clear()} are* called on the filtered collection, only elements that satisfy the filter will be removed from* the underlying collection.** <p>The returned collection isn't threadsafe or serializable, even if {@code unfiltered} is.** <p>Many of the filtered collection's methods, such as {@code size()}, iterate across every* element in the underlying collection and determine which elements satisfy the filter. When a* live view is <i>not</i> needed, it may be faster to copy {@code Iterables.filter(unfiltered,* predicate)} and use the copy.** <p><b>Warning:</b> {@code predicate} must be <i>consistent with equals</i>, as documented at* {@link Predicate#apply}. Do not provide a predicate such as {@code* Predicates.instanceOf(ArrayList.class)}, which is inconsistent with equals. (See {@link* Iterables#filter(Iterable, Class)} for related functionality.)** <p><b>{@code Stream} equivalent:</b> {@link java.util.stream.Stream#filter Stream.filter}.*/

Collections2.filter方法返回的只是原有列表的一个视图。所以：改变被过滤列表会影响过滤后列表，反之亦然。并且我们从这段文字中还能get到下面几个注意事项：

过滤后列表的迭代器不支持remove()操作
add不符合过滤条件的元素，会抛出IllegalArgumentException
当过滤后列表中的元素不再满足过滤条件时，会影响到已经过滤出来的列表出于程序员的本能，决定还是看下源码心里更踏实。核心源码如下：

1. add方法：public boolean add(E element) {//不符合过滤条件抛IllegalArgumentException的原因checkArgument(predicate.apply(element));return unfiltered.add(element);}不支持通过迭代器删除的原因：public abstract class UnmodifiableIterator<E> implements Iterator<E> {/** Constructor for use by subclasses. */protected UnmodifiableIterator() {}/*** Guaranteed to throw an exception and leave the underlying data unmodified.*
* @throws UnsupportedOperationException always
* @deprecated Unsupported operation.*/@Deprecated@Overridepublic final void remove() {throw new UnsupportedOperationException();}打印结果变化的原因：public String toString() {Iterator<E> it = iterator();if (! it.hasNext())return "[]";StringBuilder sb = new StringBuilder();sb.append('[');for (;;) {E e = it.next();sb.append(e == this ? "(this Collection)" : e);if (! it.hasNext())return sb.append(']').toString();sb.append(',').append(' ');}}@Overridepublic Iterator<E> iterator() {return Iterators.filter(unfiltered.iterator(), predicate);}

问题已经弄清楚了，怎么修改这个问题呢？

方案一（可行）：

干脆不用guava，

List<Person> names = Lists.newArrayList(lxy, xhf, nws);ArrayList<Person> persons = new ArrayList<>();for (Person p : names) {if(p.age > 30) {persons.add(p);}
}

方案二(可行)：

用个容器再包装一下：

Collection<Person> personAgeGt30 = new ArrayList<(Collections2.filter(names, p -> p.age > 30));

方案三（可行，改用Java8的过滤器）：

List<Person> personAgeGt30 = names.stream().filter((Predicate<Person>) p -> p.age >30).collect(Collectors.toList());

方案四（可行，改用Guava的连贯接口，IDEA编辑器会提示你替换成Java8 API）

ImmutableList<Person> personAgeGt30 = FluentIterable.from(names).filter(p -> p.age > 30).toList();

上述方案中，支持java8的生产环境推荐方案三，不支持java8的生产环境推荐方案二。

总结

其实，Java语言中类似的坑还有很多，比如：

1.Arrays.asList()生成的列表是不可变的。
2.subList生成的子列表，对原列表元素的修改，会导致子列表的遍历、增加、删除抛出ConcurrentModificationException,
3.subList对子列表的修改会影响到原列表数据
4.修改Map的keySet()方法和values()生成的容器，都会影响Map本身。

总之，使用任何API之前多看看源码，至少看看源码的注释。

热门内容：

Java身份证号码识别系统
看看人家那后端API接口写得，那叫一个优雅！
给IDEA换个酷炫的主题，这个有点哇塞啊！

最近面试BAT，整理一份面试资料《Java面试BAT通关手册》，覆盖了Java核心技术、JVM、Java并发、SSM、微服务、数据库、数据结构等等。获取方式：点“在看”，关注公众号并回复 666 领取，更多内容陆续奉上。
明天见(｡･ω･｡)ﾉ♡