Java: what is the difference between j += j++; and j = j + (j + 1);?

Question

for (i = 0, j = 0; i < 10; i++) j += j++;
System.out.println(j);

when i run this java prints the number 0 but when i type j = j + (j + 1); java prints 1023. Shouldn't I get the same outcome both times? And if not - why?

Anonym · Accepted Answer

Welcome to operations horror.

Let's look at "j += j++" (which really looks bad)
"j +=" is really a short way of writing "j = j +" so the calculation will actually be "j = j + j++"
Now this looks the same as "j = j + (j+1) but nope.

j++ is a postfix operator meaning the operation is done first, then j is incremented.
Note also the order of operations for java says postfix (++) precedes additive (+) which precedes assignment (=).

But postfix is done after the operation (hence the "post") so the first operation done is additive. In the first iteration, j is 0. j + j thus is 0. Now the operation is done and the postfix is evaluated next. j++ leads to j being 1 since it is currently zero. All seems good so far. Now comes assignment. But what is j assigned? Is it the result of the adding (0)? or postfix (1)? The java docs clearly state "the postfix version (result++) evaluates to the original value" so the assignment to j is given the value of j right before the postfix, that is, 0. That occurs 10 times. j at any number of iterations is therefore 0.

Afterthought: When j++ is evaluated, the value is assigned to j before the remaining assignment (the "j = ..."). If you were to print the value of j after the postfix but before the "j = ..." you would see j is always 1 at that instance.

Now "j = j + (j + 1)". You know brackets come first, so (j + 1) where j is originally 0 evaluates to 1. Added to 0 is again 1, and j is assigned 1. The next time, (j + 1) is 2, added to 1 is 3, etc. This line can be simplified to "j = (2j + 1)" Perform that ten times and you get 1023.

Change the postfix to a prefix (++j) and then they are the same.

Nitro · Answer

Hi,

j += j++; is not the same as j = j + (j +1);

j += j++;  I believe ++ will get processed first so if j was 0 it would be 1 and the 1 incremented (++) is 2 which is then added to 2 which is 4;

Where as j = j + (j + 1) would be again assuming j was 0, it would be 0 + 1 = 1 and 1 + 1 = 2 so j would be 2.

if your getting 1023 there is something else strange in the rest of the code.

Hope that helps

Java: what is the difference between j += j++; and j = j + (j + 1);?

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