Question

I try very simple test:

public static void main(String[] args) {
     test(2);
}

public static void test(int i) {
    i--;
    System.out.println("i="+i);
    if (i < 0)
        return;

    System.out.println("test1");
    test(i);
    System.out.println("test2");
    test(i);
}

Output:

i=1
test1
i=0
test1
i=-1
test2
i=-1
test2
i=0
test1
i=-1
test2
i=-1

I cannot understand why variable i in second call (test2) has value 0 after that already has 0 ?
Thanks.

Answer 1

I find indenting the output will help explain these things, each level here corresponds to the depth of the call stack (how many recursions deep you are), and the sequence corresponds to when these things are executed:

i=1
test1
  i=0 (invoked on the test1 path)
  test1
    i=-1 (invoked on the test1 path)
  test2
    i=-1 (invoked on the test2 path)
test2
  i=0 (invoked on the test2 path)
  test1
    i=-1 (invoked on the test 1 path)
  test2
    i=-1 (invoked on the test 2 path)

Notice that under each indentation level there are are invocations under the headings “test1” and “test2”, this is because you recursively call test under each of these headings, and so at each execution of test you recurse twice.

Let’s backup a bit and take a simpler case, if you were to execute test(1), you would expect to see:

i=0
  test1
    i=-1
  test2
    i=-1

Because you call test under the headings “test1” and “test2” you’re causing two paths to be navigated, one path under the “test1” heading, and a second under the “test2” heading.

When you call test(2), your code roughly does this: (recursions omitted)

(i = 2)
Decrement i  (i = 1)
Print i
Print "test1"
Call test(i) (test(1))
Print "test2"
Call test(i) (test(1))

…and remember, each time you call test(1), your code roughly does this: (recursions omitted)

(i = 1)
Decrement i  (i = 0)
Print i
Print "test 1"
Call test(i) (test(0))
Print "test 2"
Call test(i) (test(0))

If you replace each call in the first block with the output of the test(1) block you’ll see it exactly generates your output.

Basically, you get two i=0 prints because you recurse twice in each function call.