该递归方法如何打印3次而不仅仅是退出？

Question

public static void foo(int n) {

    if(n > 0) {
        if(n % 2 == 0) {
            foo(n - 3); 
        } else {
            foo(n - 1); 
        }
    }

    System.out.println(n); 
}

I'm not understanding the recursive method here. I see how it would print -1, but I'm not understanding how it would print "-1 2 3". Any help would be appreciated.

The method was called by foo(3)

Answer 1

This is the process..

foo(3) --> foo(2)
//because  (3 % 2 != 0)

foo(2) --> foo(-1)
//because  (2 % 2 == 0)

foo(-1) --> print -1
//because (-1 < 0)  
//goes back & continue from last point where this method is called

print 2, then exit method
//goes back & continue from last point where this method is called

print 3, then exit method
//goes back & continue from last point where this method is called

back to main()

I did, I used pythontutor.com to visualize it, but I don't get why after it prints -1, it goes back to the "n % 2" line

It still continue to prints because after printing -1 , even though it exits the method which prints -1, but it returns and continue from where foo(-1) was called. It continues from there..

Answer 2

Recursive method are working continuesly by requesting same methods it self. In your Code when you request foo (3);

Time 1 : 3%2=1 so this will request ELSE part Then foo (3-1) = foo (2); will be called. Time 2 : 2%2=0 so this will request IF part Then foo (2-3) = foo (-1) will be called. Time 3 : -1 > 0 = False So Print Method will be called.

System will print -1 . Time 3 function will be ended and goes to Time 2 function

System will print 2 Time 2 function will be ended and goes to Time 1 function

System will print 3 .

So the output -1 2 3