java.lang.StackOverflowError exception

This post shows how to create a StackOverflowError with a recursion. Let’s look at the error message first:
Exception in thread “main” java.lang.StackOverflowError
at ch.allenstudy.newway01.Recursion1.f(Recursion1.java:9)
at ch.allenstudy.newway01.Recursion1.g(Recursion1.java:14)
at ch.allenstudy.newway01.Recursion1.f(Recursion1.java:9)
at ch.allenstudy.newway01.Recursion1.g(Recursion1.java:14)
…

Here’s the code:
package ch.allenstudy.newway01;

public class Recursion1 {
public int i;

void f() {
g();
}

void g() {
f();
}

}

class RecursionTester {
public static void main(String[] args) {
Recursion1 n = new Recursion1();
n.g();
}
}

You can see that f() method calls g() method, then g() calls f(); it’s a recursion.

Why it gets the StackOverFlow?
That’s because in the program, the application executes the code one by one.
{
…
myFunct(int a); //Address in memory stack is A
next line //Address in memory stack is B
…
}

The program make the call to myFunct(int a), this method will be executed. While, after finish execution, the system need send a signal to myFunct(int a), tell it where you should come back and continue. The signal here is called “returned address”; but notice, it’s the return address of “next line”, so that it can continue to run.
And this “returned address” is pushing into Stack.
In up sample, you see, the f() and g() call each other in an infinite loop, every time they call each other, there will add a “returned address” into stack.
After a while, the stack is full. Then have the overflow.