Question

Java object lifecycle:

Describe the lifecycle of the objects created within the method listed below. Discuss whether this program (assume that this method is embedded in a class) will fail with an out of memory error. And Explain when a stack overflow error can occur in Java, using an example program.

public static void main(String[] args) throwsException {

    java.util.Set set = new java.util.HashSet();

    while(set.size()>-1) {

        set = new java.util.HashSet();

        Thread.sleep(1000); // wait 1 s

        for (int i=0;i<1000000;i++) {

            set.add(newObject());

        }

    }

}

Answer 1

stack Line1 stack Line3 Heap i-4 public void Method10 cls1(ref) { y-2 cls1 int i-4 Object stack Line2 i-4 int y 2 y 2 new class10): class1 cls1 i-4 exiting method Heap stack cls1 Object

These errors occurs when program runs out of memory depending on the
available physical memory (RAM) for pc/laptop

Their are two parts of memory used by java programs.
1. stack memory
   It is used to store the variables and
   object referrences(which are also variables of class, Student s1's only address/reference not data).
  
2. heap memory
   It is used to store object's data into memory (student name, roll, grade
   or int i j k... variables in object for each object).
   if object holds more instance variables/fields then it needs more memory per object.
-------------------------------------------
Ans 1. This program will not fail with out of memory error because
   look the code in while loop,
  
   while(set.size()>-1) {
set = new java.util.HashSet();   // create new hashset everytime (discarding old data)
//this causes to loose 1000000 objects added previously after 2nd time loop runs
//so set holds 1000000 objects maximum, thus it can store 1000000 objects easily(normally) and no error
//but if at any time in the system other processes/applications are using more memory
//causing our program to have less memory then out of memory error can occur due to heap memory shortage
  
Thread.sleep(1000); // wait 1 s
for (int i=0;i<1000000;i++) {   // add 1000000 objects to hashset
set.add(newObject());
}
}
-------------------------------------------
but if we comment/delete below line in while loop
//set = new java.util.HashSet();

then it will cause while loop to add 1000000 objects each time loop runs
so set's size grows(and also heap size), and eventually heap runs out of memory and gives error

to test this change program like this and see output
comment this //set = new java.util.HashSet();
System.out.println("Set size:" + set.size()); //to see the set size
Thread.sleep(10); // change sleep time to 10 to get output faster
---------------------

-------------------------------------------
Ans 2.
   Stack overflow error can occur mostly due to bad recursive call to function
   (function calling same function repeatedly).
  
   Because each time the function calls to itself program needs to store
   some data into stack memory such as
       method parameters, its local parameters, and the return address of the method
       and this needs some memory
   this causes stack to grow(reducing available stack memory further)
   and at some point stack runs out of memory and gives error by JVM

see the following example that cause stack error
  
//StackOverflowErrorExample.java
public class StackOverflowErrorExample {

public static void recursivePrint(int num) {
System.out.println("Number: " + num);

if(num == 0)   // this never occurs
return;   // function never returs and runs infinitly
else
recursivePrint(++num); // recursive call, as each time we call same function by increasing num by 1
}

public static void main(String[] args) {
StackOverflowErrorExample.recursivePrint(1);   // start from 1
}
}
  
-------------------------------------------
thank you...

// MemLoadTest.java gives heap error :)
public class MemLoadTest {
public static void main(String[] args) throws Exception {
java.util.Set set = new java.util.HashSet();
while(set.size()>-1) {
//~ set = new java.util.HashSet();
System.out.println("Set size:" + set.size());
Thread.sleep(10); // wait 10 ms
for (int i=0;i<1000000;i++) {
set.add(new Object());
}
}
}
}

stack Line1 stack Line3 Heap i-4 public void Method10 cls1(ref) { y-2 cls1 int i-4 Object stack Line2 i-4 int y 2 y 2 new class10): class1 cls1 i-4 exiting method Heap stack cls1 Object