Incrementor logic?

Question

I'm trying to get deeper with post and pre incrementors but am a bit stuck with the following expression : public static void main(String[] args) { int i = 0; i = i+=(++i + (i+=2 + --i) - ++i); // i = 0 + (++i + (i+=2 + --i) - ++i); // i = 0 + (1 + (3 + 2) - 1); // i = 0 + (6 - 1); System.out.println(i); // Prints 0 instead of 5 } I know I'm missing the logic somewhere but where? What I've tried : - Going from left to right (though I know it is not recommended) - Going from the insidest bracket and starting from there. Thanks for the help PS : The comments are the details of my calculus **EDIT 1** I tried to change de hard coded value from the expression from `2` to something else and the result always gives `0` Look at this example : int i = 0; i = i+=(++i + (i+=32500 + --i) - ++i); System.out.println(i); // Prints 0 This expression should logically be nowhere near `0` but somehow it does print it. The same happens when I use a negative : int i = 0; i = i+=(++i + (i+=(-32650) + --i) - ++i); System.out.println(i); // Prints 0 -------------- **EDIT 2** Now, I changed the value of `i` to begin with : int i = 1; i = i+=(++i + (i+=2 + --i) - ++i); System.out.println(i); // Prints 2 i = 2; i = i+=(++i + (i+=10000 + --i) - ++i); System.out.println(i); // Prints 4 i = 3; i = i+=(++i + (i+=(-32650) + --i) - ++i); System.out.println(i); // Prints 6 It gives the double of `i` each time, whatever the hard coded value is.

Answer 1

The first thing I recommend is not to have equal signs all written in the same expression like what you have now. Instead, break the whole thing down into the format: someNumber = someNumber + anotherNumber; OR someNumber += anotherNumber; where you only have one assignment per executable statement. This will make debugging a lot easier and also clearer on the intention of the statement.

Next, if you are trying to understand pre- and post- increment, then here is the explanation on increment:

* The number that you want to increment will always be one value higher at the end of the executable statement.

* The difference between the pre- and post- increment is what "value" is being assigned to the variable on the left of the assignment (= sign). In other words:

- For pre-increment, the value of the right variable is incremented BEFORE assigned to the left variable.

- For post-increment, the value of the right variable is assigned to the left THEN its value gets incremented.

- Note that the order of execution by the system is always from left to right.

Let's do some examples to make things clearer:

int i, j; // j will be used as the left variable (left side of the assignment), and i is the right variable.

i = 5;

j = ++i; // order of execution is from left to right. Thus, i is incremented to 6 first. Then, its value (6) is assigned to j. Thus, j now has the value 6, and i is also 6.

i = 5; // reset i back to 5 for the next example

j = i++; // order of execution is from left to right. Thus, value of i is assigned to j first. Then, i is incremented. Thus, j now has the value of 5, while i has the value 6.

i = 5; // reset i back to 5 for the next example

j = 1; // set j to 1 for the next example

j += i; // this statement is equivalent to: j = j + i. Thus, j has the final value of 6 => (1 + 5) where 1=orig value of j and 5=value of i. The variable i remains at 5.

Hope this helps.