Question

Here is an example that outputs 6:

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

static void m(int i) {
    i = ++i + i++; 
    System.out.println(i);
}

We get 6 because in the m(int i) method at first 3 and 3 are summarized, then i becomes 4 (due to the i++), but after that i from the left part takes the summarized 6 value.

But if the method is changed to the following, we get 7:

static void m(int i) {
    i = ++i + ++i; 
    System.out.println(i);
}

But I expected to see 7 in both cases (I’ve been guided by the fact that unary operations, in this case incrementing, have a higher priority than binary operations). Could someone please provide an explanation (or a reference to an explanation) of the ignored i++ in the first example?

Answer 1

++i increments i and returns the new value of i.

i++ increments i and returns the old value of i.

Expressions are evaluated from left to right, taking operator precedence into account.

So in ++i + i++, when you start with i == 2, you get: ++i which increments i to 3 and returns 3; then i++ which increments i to 4 and returns 3. Then finally you have i = 3 + 3 so i becomes 6.

Note that these are funny tricks that are not really relevant to real-world programming.