Does garbage collection happen immediately after Hashmap.remove() is called?

问题

The Java code is as follows:

Random r = new Random(1234697890);
HashMap<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
List<Integer> list = new ArrayList<Integer>();

for(int i=0;i<100000;i++){
    for(int j=0;j<1000;j++){
        list.add(r.nextInt(100000));
    }
    map.put(i, list);
    map.remove(i);
}

when i reaches 37553 , java.lang.OutOfMemoryError: Java heap space happens.
It seems that garbage collection does not happen in the loop.
Now I wonder how to fix the problem.

回答1:

Try rewriting the code as follows and you should not get OOME's ...

Random r = new Random(1234697890);
HashMap<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();

for(int i=0;i<100000;i++){
    List<Integer> list = new ArrayList<Integer>();
    for(int j=0;j<1000;j++){
        list.add(r.nextInt(100000));
    }
    map.put(i, list);
    map.remove(i);
}

The problem with your original code is that:

• you only create one list,
• you keep adding more and more elements to it, and
• that list only becomes garbage when the code completes ... because it is "in scope" the whole time.

Moving the list declaration inside the loop means that a new ArrayList is created and filled in each loop iteration, and becomes garbage when you start the next iteration.

---

Someone suggested calling System.gc(). It won't help at all in your case because there is minimal¹ garbage to be collected. And in general it is a bad idea because:

• the GC is guaranteed to have run immediately before an OOME is thrown,
• the JVM can figure out better than you can when is the best (i.e. most efficient) time to run the GC,
• your call to System.gc() may be totally ignored anyway. The JVM can be configured so that calls to System.gc() are ignored.

_{1 - The pedant in me would like to point out that map.put(i, list); map.remove(i); is most likely generating an Integer object that most likely becomes garbage. However, this is "chicken feed" compared to your indefinitely growing ArrayList object.}

回答2:

You use the same List all the time, which contains 100000 * 1000 items when the loop exits. To enable GC to get rid of your list, you need to reduce its scope to within the for(i) loop.

In other words, both map and list are reachable at all time in that piece of code and are therefore not eligible for collection.

回答3:

In your case you keep filling the same list (even though you remove it from that HashMap, it still exists as a local variable).

The JVM promises to do a full garbage collect before throwing an OutOfMemoryError. So you can be sure there is nothing left to clean up.

回答4:

Your code

List<Integer> list = new ArrayList<Integer>();

for(int i=0;i<100000;i++){
    for(int j=0;j<1000;j++){
        list.add(r.nextInt(100000));
    }
    map.put(i, list);
    map.remove(i);
}

is the same as

List<Integer> list = new ArrayList<Integer>();

for(int i=0;i<100000 * 1000; i++) {
    list.add(r.nextInt(100000));
}

As you can see its the List, not the map which is retaining all the integers. BTW Try this instead and see what happens ;)

list.add(r.nextInt(128));

来源：https://stackoverflow.com/questions/13723983/does-garbage-collection-happen-immediately-after-hashmap-remove-is-called