Question

try {
    for(;;) {
        s.add("Pradeep");
    }
} finally {
    System.out.println("In Finally");
}

In the try block the jvm runs out of memory,then how is jvm excuting finally block when it has no memory?

Output:

In Finally
Exception in thread "main" java.lang.OutOfMemoryError: Java heap space

Answer 1

Here is more simple code that demonstrated better what happens:

    try {
        int[] a = new int[Integer.MAX_VALUE];
    } catch( Exception e ) {
        e.printStackTrace();
    }

The allocation of the array fails but that doesn’t mean Java has no free memory anymore. If you add items to a list, the list grows in jumps. At one time, the list will need more than half of the memory of the VM (about 64MB by default). The next add will try to allocate an array that is too big.

But that means the VM still has about 30MB unused heap left for other tasks.

If you want to get the VM into trouble, use a LinkedList because it grows linearly. When the last allocation fails, there will be only very little memory to handle the exception:

    LinkedList<Integer> list = new LinkedList<Integer>();
    try {
        for(;;) {
            list.add(0); 
        }
    } catch( Exception e ) {
        e.printStackTrace();
    }

That program takes longer to terminate but it still terminates with an error. Maybe Java sets aside part of the heap for error handling or error handling doesn’t need to allocate memory (allocation happens before the code is executed).