question from Marcus Green's Mock Exam 3

[Ray Chang]

Question 54)
What will happen when you attempt to compile and run the following code?
public class Inc{
public static void main(String argv[]){
Inc inc = new Inc();
int i =0;
inc.fermin(i);
i = i++;
System.out.println(i);
}
void fermin(int i){
i++;
}
}
1) Compile time error
2) Output of 2
3) Output of 1
4) Output of 0
the given answer is 4. I'm confused. I know that if the statement is System.out.println(i++); then the output will be the current i, say 0. But is there any difference between the statement System.out.println(i++); &
i = i++;
System.out.println(i); ?
Can anybody give me a clear expanation?Thank you

[Alan Phillips]

This is an excellent question. There are two things being tested here, both pass by value and operators. The value passed to fermin goes nowhere and disappears. The value is passed to fermin which increments its own i variable local to the method and does nothing to the i in Inc. The second thing is that i = i++; is really
i = 0(++);
Postfix operators happen after the expression i++(1) is never added to i. If you change the code to ++i, then you get a result of 1 because the prefix operator is added first!

[Valentin Crettaz]

For an excellent explanation on how the + operator works, please check out:
https://coderanch.com/t/190825/java-programmer-SCJP/certification/Array

[Ray Chang]

Thank you. I know the code below will give an output 0.
int i=0;
System.out.println(i++);
But I guess it's different when the code is like below:
int i=0;
i=i++;
System.out.println(i);
In my understanding, immediately after the statement i=i++; is excuted, i is 0. But then i is incremented after that and is 1. then the System.out.println(i); excuted and the output is 1. If I was wrong, when will i be incremented at this kind of statements?Thanks again

[Anthony Villanueva]

Originally posted by Ray Chang:
Thank you. I know the code below will give an output 0.
int i=0;
System.out.println(i++);
But I guess it's different when the code is like below:
int i=0;
i=i++;
System.out.println(i);
In my understanding, immediately after the statement i=i++; is excuted, i is 0. But then i is incremented after that and is 1. then the System.out.println(i); excuted and the output is 1. If I was wrong, when will i be incremented at this kind of statements?Thanks again

Actually, there is no difference:

public class Test58
{
public static void main (String[] args)
{
int i=0;
System.out.println("i = " + i++);
int j=0;
j=j++;
System.out.println("j = " + j);
}
}

What happens in
int j=0;
j = j++;
is that the assignment goes last. The return value of j++ is 0, but it is not assigned yet. j gets incremented to 1, then the assignment kicks in, assigning the return value 0 to j.

[Shishio San]

Originally posted by Anthony Villanueva:
then the assignment kicks in, assigning the return value 0 to j.

Hi Anthony,
I do understand that i gets 0 knowing that i = 0(1) but i was just wondering about the lasted value of i which is 1. I compiled the program myself and the value 1 of i seems to be flashed.
i really appreciated more insights on this.

[david eberhardt]

I think this section from the java language specification may help reveal what happens:

PostIncrementExpression:
PostfixExpression ++
A postfix expression followed by a ++ operator is a postfix increment expression.
The result of the postfix increment expression is not a variable, but a value.
At run time, if evaluation of the operand expression completes abruptly, then the postfix increment expression completes
abruptly for the same reason and no incrementation occurs.
The value of the postfix increment expression is the value of the variable before the new value is stored.

public class Inc{
  public static void main(String argv[]){
    Inc inc = new Inc();

    int i =0;
    i = i++; // increment exec doesn't complete
    
    System.out.println("what is i after i=i++;");
    System.out.println(i); // you should see 0 


    System.out.println("now execute this ->i++;");
    
    i++; // execution of increment completes 
    
    System.out.println(i);// you should see 1 
}
}
  

therefore, I believe that since i++ doesn't complete until after the assigment expression ( "i = i++; " completes, you get an abnormal termination to the increment operation.
just a guess ....