Assignment and ++

[Bruno Arantes]

Given the following code fragment:

 public class Test
{
public static void main(String[] args)
{
int i = 0;
i = i++; // Line 6
i = i++; // Line 7
i = i++; // Line 8
System.out.println("i" + i);          
}
} 

When this code is run, it produces the output:

 i = 0 

I expected variable i to contain value 3 when it's value was printed out, but (for my surprise) value 0 has been showed. I thought this code was to work the following way:

In line 6, i receives 0 (original value)

then i is incremented by the unary operator ++.

In line 7, i receives the new value (1)

and then i is incremented by the unary operator ++.

and so on... until i holds the value 3, which does not happen in the real case, as we have seen.

The program does not behave as I expected. Could someone tell me the right behaviour of this code?
Thanks in advance.
Bruno.

[Dan Chisholm]

Here's a question from my mock exam that demonstrates how the postfix operator works in this situation.

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

What is the result of attempting to compile and run the above program? 
a.  Prints: 0,0 
b.  Prints: 1,0 
c.  Prints: 0,1 
d.  Prints: 1,1 
e.  Runtime error 
f.  Compiler error 
g.  None of the above 

The answer is b. The explanation is as follows.

The statement contains an addition operation. The left operand is a postfix operation and it is evaluated first. The result of the postfix operation is zero, but variable i is incremented as a side effect. The right hand operand of the addition operation is the result of method m. As a side effect, method m prints the current value of variable i which is one. Method m then returns the value zero. The result of the addition operation is zero and that is the value that is assigned to variable i.

[Francis Siu]

hi Bruno
It is a tricky question.

Within the assignment from i++ to i,the i++ become i and the effect cancel because it is the postfix opperator
The effect will take place after the syntax ';' in this case
so this action can not take effect

 
int i = 0;
i = i++; 
System.out.println("i" + i);
i = i++; 
System.out.println("i" + i);
i = i++;
System.out.println("i" + i); 

Hope this help

[Corey McGlone]

This gets discussed a lot here. Check out this thread.
Corey

[Bruno Arantes]

Thanks for all who helped! Such "mysteries" of java are now certainly less "mysterous" for me...