Reputation: 136347

Pre- and postincrement in Java

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:

public class test {
    public static void main(String[] args) {
        int i = 1;
        i += ++i + i++ + ++i;

        System.out.println("i = " + i);
    }
}

It outputs 9.

My (at least partly) wrong explanation:

I'm not quite sure, but I think the term after i += gets evaluated like this:

enter image description here

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

is the same as

int i = 1;
i += ((++i) + (i++)) + (++i);

This gets evaluated from left to right (See Pre and postincrement java evaluation).

The first ++i increments i to 2 and returns 2. So you have:

i = 2;
i += (2 + (i++)) + (++i);

The i++ returns 2, as it is the new value of i, and increments i to 3:

i = 3;
i += (2 + 2) + ++i;

The second ++i increments i to 4 and returns 4:

i = 4;
i += (2 + 2) + 4;

So you end up with 12, not 9.

Where is the error in my explanation? What would be a correct explanation?

Upvotes: 11

Answers (6)

Denys Séguret

Reputation: 382130

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.

Upvotes: 3

Pshemo

Reputation: 124225

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

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

So i = 1 + 8 -> i=9

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

Upvotes: 0

Aaron Digulla

Reputation: 328594

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

Upvotes: 1

robi terebesi

Reputation: 65

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

Upvotes: 1