关于程序输出的困惑

Question

public class Main {
    public static void main(String[] args) {        
        pop1(1234);     
    }

    public static void pop1(int x){
        System.out.print(x % 10);   
        if ((x / 10) != 0){
            pop1(x/10);
        }   
        System.out.print(x % 10);
    }   
}

Output : 43211234

I dont understand the output. First past '4321' is ok; but how last part '1234' is generated?

Answer 1

The last System.out.print(x % 10); --> This line causes the values 1,2,3,4 to be printed. This is the last line of each method call.

While entering X:4->3->2->1 (now printing starts here) 1 is printed, then stack unwinds (method returns), then 2 prints and so on..

Answer 2

The second print is called when the pop1 function returns from the recursion - that is the reason the other four numbers are printed in reverse order. Remove it and you'll be fine.

Answer 3

The first call to:

System.out.print(x % 10); 

generates the reverse number 4321 while the second call generates the direct order 1234 . So remove on of the lines if you need only direct/reverse order.

So you get a direct order 1234 from the second print() call because it's executed after the recursion call ie innermost recursion call prints 1 at first place while the outermost call prints 4 at the last place.

Answer 4

To visualize what happen you might amend the code like this

public class Main {
    public static void main(String[] args) {        
        pop1(1234);     
    }

    public static void pop1(int x){
        System.out.println("enter: " + x % 10);   
        if ((x / 10) != 0){
            pop1(x/10);
        }   
        System.out.println("leave: " + x % 10);
    }   
}

produced output

enter: 5
enter: 4
enter: 3
enter: 2
enter: 1
leave: 1
leave: 2
leave: 3
leave: 4
leave: 5

combine this output with the answer from @TheLostMind and you will understand how the code is working.