Pre- and postincrement in Java

Question

I just wanted to create a little Java-Puzzle, but I puzzled myself. One part of the puzzle is:

What does the following piece of code do:

Answer 1

The output is 9 (try it)

int i = 1;
i += ++i + i++ + ++i;

becomes

i = 1 + 2 + 2 + 4

Answer 2

The code

int i = 1;
i += ++i + i++ + ++i

is equivalent to

int tmp1 = i // 1, +=

i ++; // 2
int tmp2 = i; // 2

int tmp3 = i; // 2
i ++; // 3

i ++; // 4
int tmp4 = i; // 4

i = tmp1 + tmp2 + tmp3 + tmp4; // 9

Answer 3

i += ++i + i++ + ++i; is the same as i = i + ++i + i++ + ++i;

The right-hand side is calculated from left-to-right, yielding i = 1 + 2 + 2 + 4; (which yields i = 9).

Answer 4

You're right regarding the right part evaluation, but you're missing a detail regarding the assignment.

Run this :

i = i++;

or this :

i += i++;

After both operations, i still has its original value.

That's because i is evaluated on the left before the right part of the assignment.

So in your case, you're adding 8 to 1, not to 4.

Answer 5

i += ++i + i++ + ++i;

1. i=1 at start
2. i += X -> i = i + X -> i = 1 + X (so lets count X)
3. ++i will be incremented to 2 and return 2
4. i++ will return 2 and then be incremented to 3
5. ++i will be incremented from 3 to 4 and return 4
6. X = 2 + 2 + 4 = 8

So i = 1 + 8 -> i=9

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You would get 12 if your code would be something like this

int i = 1;
int tmp = ++i + i++ + ++i;  
i += tmp;

because then your code would be i=1, and after calculating tmp i would be i=4, then i+=tmp -> i=4+8=12

Answer 6

it's very easy to understand how it works if you imagine it how java stores values in registers! he puts 1 in the first register, and than goes through = sign, and increments the i(++i), so now in i you have 2, and in the second register you have 2, but the first register is not updated, in the third register you'll have 2 and then i is incremented, and then i is incremented and in the last register you'll have 4. So you'll have something like this 1 = 2 + 2 + 4 == 9