How the increment work

[Ruwan Scott]

Can some one explain me the out put of following program

class Class1{
static int f1(int i) {
System.out.print( i + "," );
return 0;
}
public static void main (String[] args) {
int i = 0;
System.out.print( i + "," );
i = i++ + f1(i);
System.out.print(i);
}
}

When I run this I get 0,1,0 as the output. I dont understand how we get 1 and 0 as last 2 digits. It is confusing.

[Jesper de Jong]

Hello "Ruwan Ruwan", welcome to JavaRanch. Please check your private messages (see the link at the top right of the page) for an administrative JavaRanch matter.

Have a look at PostIncrementOperatorAndAssignment in our FAQ.

[Ruwan Scott]

Hi Jasper,
I already read it. I couldnt relate that logic to this piece of code. problem is what happens inside method f1. we pass 0 to that method , but inside the method it is printed as 1. that is what confuse me.
Thanks.

[Ankit Garg]

Ruwan please Use Code Tags when you post a source code.

i = i++ + f1(i); 

In this statement, when i++ is executed, the original value of i i.e. 0 is taken in the expression and then i is incremented to 1. So the statement would look as

i = 0 + f1(i); 

But as I said, the value of i is incremented after the original value is used in the expression, so when f1 is called, the value of i is 1. Thus 1 is displayed as the second last digit in the output. The method f1 returns 0 so the expression becomes

i = 0 + 0; 

And thus the last digit in the output is 0 as the value of i becomes 0 after this statement is executed...

[Ruwan Scott]

Hi Ankit,
thanks for explanation. Now i know how f1 method prints 1.
I have a small quesiton here . when compiler evaluates i++ + f1(i) part, Why doesn't it copy 0 to i++ and f1(i) at the same time.
Is it evaluates from Left to Right ? Can you explain this?

Thanks
Ruwan.

[Ankit Garg]

If you see the expression, there is a binary operator in it i.e. the + operator. The operands of the + operator are evaluated from left to right that's why 0 is not put in the call to f1(). It will first completely execute the left operand and then execute the right operand. Try this to clear your doubt

static int method1() {
    System.out.println("1");
    return 1;
}
static int method2() {
    System.out.println("2");
    return 2;
}
public static void main(String[] args) {
    int i = method1() + method2();  //output 12
}

[Mo Jay]

Hi,

Ankit's old explanation was good but the last one with the code example is not accurate as the last output will be 3 instead of 12, the reason why the previous output is printing 12 is because you are printing the string literals "1" and "2" respectively.

Now back to the original question, i++ is a post increment which means that i=0 is saved first then i++=1 is used to call the f() method which returns 0 ofcourse. Now the old saved i=0 is used + 0 from the returned function which yield to 0.

Try i = ++i + f1(i); and you will see the difference because i will be incremented first and saved as 1 then used in the method call, the final i result should be 1.

Cheers!!!

[Ankit Garg]

Mo Jay wrote:Ankit's old explanation was good but the last one with the code example is not accurate as the last output will be 3 instead of 12,

What do you mean by the last output. I never displayed the value of i so 3 will not be in the output...

[Mo Jay]

 # public static void main(String[] args) {  
          #     int i = method1() + method2();  //output 12  
          #     System.out.print(i);   

The value of i will print 3 and not 12 as you commented in your code above.

[Ankit Garg]

Mo Jay wrote:

 # public static void main(String[] args) {  
          #     int i = method1() + method2();  //output 12  
          #     System.out.print(i);   

The value of i will print 3 and not 12 as you commented in your code above.

I think you misinterpreted it or I didn't explain it properly. When I said output 12, I didn't mean that 12 will be assigned to i. I meant to say that the output of execution of that expression will be 12...

[Ruwan Scott]

Hi Ankit , Mo
Thanks for your explanations. That clears all my doubts on this issue.

/Ruwan