Why does this method print 4?

Question

I was wondering what happens when you try to catch an StackOverflowError and came up with the following method:

class RandomNumberGenerator {

    static int         


        

Answer 1

After digging around for a while, I can't say that I find the answer, but I think it's quite close now.

First, we need to know when a StackOverflowError will be thrown. In fact, the stack for a java thread stores frames, which containing all the data needed for invoking a method and resume. According to Java Language Specifications for JAVA 6, when invoking a method,

If there is not sufficient memory available to create such an activation frame, an StackOverflowError is thrown.

Second, we should make it clear what is "there is not sufficient memory available to create such an activation frame". According to Java Virtual Machine Specifications for JAVA 6,

frames may be heap allocated.

So, when a frame is created, there should be enough heap space to create a stack frame and enough stack space to store the new reference which point to the new stack frame if the frame is heap allocated.

Now let's go back to the question. From the above, we can know that when a method is execute, it may just costs the same amount of stack space. And invoking System.out.println (may) needs 5 level of method invocation, so 5 frames need to be created. Then when StackOverflowError is thrown out, it has to go back 5 times to get enough stack space to store 5 frames' references. Hence 4 is print out. Why not 5? Because you use cnt++. Change it to ++cnt, and then you will get 5.

And you will notice that when the size of stack go to a high level, you will get 50 sometimes. That is because the amount of available heap space need to be taken into consideration then. When the stack's size is too large, maybe heap space will run out before stack. And (maybe) the actual size of stack frames of System.out.println is about 51 times of main, therefore it goes back 51 times and print 50.