HashMap.clear() reduce memory?

Question

System.gc();
System.out.println((Runtime.getRuntime().totalMemory()-Runtime.getRuntime().freeMemory())/1024/1024);// print 1M

Map map=new HashMap();
for(int i=0;i<100000;i++){
    map.put("key"+i,"i");
}
System.gc();
System.out.println((Runtime.getRuntime().totalMemory()-Runtime.getRuntime().freeMemory())/1024/1024); //print 10M

map.clear();
System.gc();
System.out.println((Runtime.getRuntime().totalMemory()-Runtime.getRuntime().freeMemory()));//print less than 1M

It seems the memory is reduced when call the clear method. However, from looking at other answers it seems the clear method never shrinks the HashMap. So why is the memory reduced?

Answer 1

If you're referring to this question's answers, they're telling you that the entries array (table) in the HashMap is never shrunk. Instead, its entries are all set to null.

But clearing the map makes the 100,000 strings you created ("key0", "key1", ...) and their associated Map.Entry objects eligible for garbage collection, despite table not getting smaller.

Answer 2

It's an implementation detail, so the exact answer may change depending on the exact version of Java.

Here's the Java 8 implementation of HashMap::clear:

public void clear() {
    Node<K,V>[] tab;
    modCount++;
    if ((tab = table) != null && size > 0) {
        size = 0;
        for (int i = 0; i < tab.length; ++i)
            tab[i] = null;
    }
}

The table of buckets is completely emptied, but the table itself, and so the non-default capacity, is retained.

Regardless of the exact implementation, you would expect to free up a significant chunk of memory, because all of those non-interned strings created by "key"+i will be eligible for collection.

If you really care about reducing the capacity back to default then just reassign the HashMap with a new instance.