近日使用ArrayListMultimap
的putAll时犯了一个错误,先看一下Code,Pos1、2、3、4输出什么值?
//test1
Set<Long> sets = new HashSet<>();
sets.add(1L);
sets.add(2L);
Multimap<String, Long> maps = ArrayListMultimap.create();
maps.putAll("A", sets); // ①
System.out.println("Pos1:" + maps.get("A").size());
maps.putAll("A", sets); // ②
System.out.println("Pos2:" + maps.get("A").size());
// test2
Multimap<String, Long> maps2 = ArrayListMultimap.create();
maps2.put("A", 1L);
maps2.put("A", 2L);
System.out.println("Pos3:" + maps2.get("A").size());
maps2.put("A", 1L);
maps2.put("A", 2L);
System.out.println("Pos4:" + maps2.get("A").size());
结果如下:
Pos1:2
Pos2:4
Pos3:2
Pos4:4
这里的putAll,不是覆盖相同key的value,而是追加;putAll或put,对同一key下相同的value并不会过滤。这里和map.put(k,v)
是不一样的。所以,如果重复调用ArrayListMultimap的putAll或put方法,value就会造成OOM的异常,这里要特别注意。
下面,我们从源码进行一次分析:
public boolean putAll(@Nullable K key, Iterable<? extends V> values) {
checkNotNull(values);
// make sure we only call values.iterator() once
// and we only call get(key) if values is nonempty
if (values instanceof Collection) {
Collection<? extends V> valueCollection = (Collection<? extends V>) values;
return !valueCollection.isEmpty() && get(key).addAll(valueCollection); // ①
} else {
Iterator<? extends V> valueItr = values.iterator();
return valueItr.hasNext() && Iterators.addAll(get(key), valueItr);
}
}
下面看下addAll方法①:
public boolean addAll(Collection<? extends V> collection) {
if (collection.isEmpty()) {
return false;
}
int oldSize = size(); // calls refreshIfEmpty
boolean changed = delegate.addAll(collection); // ②
if (changed) {
int newSize = delegate.size();
totalSize += (newSize - oldSize);
if (oldSize == 0) {
addToMap();
}
}
return changed;
}
其中addAll方法②调用了java.util.ArrayList
的addAll:
public boolean addAll(Collection<? extends E> c) {
Object[] a = c.toArray();
int numNew = a.length;
ensureCapacityInternal(size + numNew); // Increments modCount
System.arraycopy(a, 0, elementData, size, numNew);
size += numNew;
return numNew != 0;
}
至此,ArrayListMultimap
的putAll是调用了java.util.ArrayList
的addAll。我们还记得list.add(v)
方法:
List<Long> lists = new ArrayList<>();
lists.add(1L);
lists.add(1L);
lists.add(1L);
System.out.println("Pos1:" + lists.size());
List<Long> list2 = new ArrayList<>();
list2.addAll(lists);
list2.addAll(lists);
System.out.println("Pos2:" + list2.size());
结果如下:
Pos1:3
Pos2:6