Question about post-increment operator

Question

Hi guys,

Why does the following code

int i = 1; 
System.out.print(i += i++);
System.out.print(i);

output 2 two times instead of 3 for the 2nd print?

Could somebody please shed some light on it?

Thanks.

Answer 1

If you realise that a++ works as follows (pseudocode):

func ++(ref a)
{
    int b=a;
    a=a+1;
    return b;
}

then it all makes sense.

Answer 2

It may seems like i should be 3 in the end.

However, if you look into the statement more closely

i += (i++)

is equal to

i = ( i + (i++) ) 

which is, in this case, 1+1.

The side effect of i++ is i=i+1=1+1=2 as you may have expected, however, the value of i is override after the assignment.

Answer 3

1 + 1 == 2. 

Therefore:

i + i == 2  

and

i += i == 2

and then

i += i++ == 2

Pretty straight forward.

Answer 4

I don't know the Java bytecode syntax very well yet but according to me at bytecode level your code would look something like this :

int i = 1;  // iconst_1:    variables { }, stack {1}
            // istore_1:    variables {1}, stack { }
i += i++;   // iload_1:     variables {1}, stack {1}
            // iinc 1, 1:   variables {2}, stack {1}
            // iadd:        variables {2}, stack {2} ...Note that here 1 gets added to the value on stack
            // istore_1:    variables {2}, stack {2} ...Here the variable value will overwrite the stack value

I think this explains the output you are getting pretty well. :-)

Experts, please correct me if I am wrong...

Answer 5

I don't think this is an issue with not knowing how the postfix unary operator (expr++) works. It is the order in which the statements are evaluated that is creating the confusion.

int i = 1;
System.out.println(i += i++); // Output: 2

So the last statement is the same as the following two statements in this order:

i++; // i is now 2 for the rest of this statement and the program
i = 1 + 1; // i is assigned again  

So the postfix operator is evaluated first but then the whole line is evaluated but using the previous value of i.

So, to use another example that would make this more clear:

int i = 2;
System.out.println(i += i++); // Output: 4
System.out.println(i); // Output: 4  

And another example:

int i = 2;
System.out.println(i = i + i++ + i--); // Output: 7
System.out.println(i); // Output: 7  

The second line is assigning i. The first i is 2, the next i is also 2, but now the third i is 3 because i++ has changed the value of i. As the case from before, i-- will not have any affect on i because it will get rewritten with i = 2 + 2 + 3.

int i = 1;
System.out.println(i = i++ + i); // Output: 3
System.out.println(i); // Output: 3