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.

well going from left to right won't solve it.. the compilers convert normal expressions to [reverse polish notations](https://en.wikipedia.org/wiki/Reverse_Polish_notation) and evaluates them and then constructs a **expression tree** and solves the expression. — Minato, Oct 14 '15 at 08:54
so simply going from left to right or insidest won't give you the result program will generate — Minato, Oct 14 '15 at 08:56
If you "know it is not recommended" why did you do it anyway? Why not do it the recommended way? — Thomas Weller, Oct 14 '15 at 09:01
@Thomas and what is the recommended way? That's the point right here. — Yassin Hajaj, Oct 14 '15 at 09:03
@Thomas This is simply curiosity indeed. I want to understand everything I'm learning. — Yassin Hajaj, Oct 14 '15 at 09:23
Can anyone confirm the first example actually returns `0`? Running the code on 1.6, I got result of `3`. — Jiri Tousek, Oct 14 '15 at 09:35
@Yassin I did. I got even more confusing results when I replaced one or both of the `i +=` with `i = i +` - replacing one of them (don't remember which) led to result `4`, replacing the other didn't change the result. — Jiri Tousek, Oct 15 '15 at 08:17
@JiriTousek I compiled with 1.4, then ran with following versions, and all produced `0`: 1.4.2_19, 1.5.0_22, 1.6.0_45, 1.7.0_79, 1.8.0_51 on Windows 7, Sun/Oracle JVMs. — Andreas, Oct 16 '15 at 03:53

score 22 · Accepted Answer · edited Jun 20 '20 at 09:12

Quoting Java Language Specification, 15.7 Evaluation Order:

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.

If the operator is a compound-assignment operator (§15.26.2), then evaluation of the left-hand operand includes both remembering the variable that the left-hand operand denotes and fetching and saving that variable's value for use in the implied binary operation.

So, essentially, i += ++i will remember the old value of i on the left side, before evaluating the right side.

Remember, evaluation order of operands and precedence of operators are two different things.

Showing evaluation order, step by step, with saved value in {braces}:

int i = 0;
i    = i    += (++i + (i    += 2 + --i) - ++i); // i = 0
i{0} = i    += (++i + (i    += 2 + --i) - ++i); // i = 0
i{0} = i{0} += (++i + (i    += 2 + --i) - ++i); // i = 0
i{0} = i{0} += (1   + (i    += 2 + --i) - ++i); // i = 1
i{0} = i{0} += (1   + (i{1} += 2 + --i) - ++i); // i = 1
i{0} = i{0} += (1   + (i{1} += 2 + 0  ) - ++i); // i = 0
i{0} = i{0} += (1   + (i{1} += 2      ) - ++i); // i = 0
i{0} = i{0} += (1   + 3                 - ++i); // i = 3
i{0} = i{0} += (4                       - ++i); // i = 3
i{0} = i{0} += (4                       - 4  ); // i = 4
i{0} = i{0} += 0                              ; // i = 4
i{0} = 0                                      ; // i = 0
0                                             ; // i = 0

Followup to edits to question

If we name the initial value I and the constant N:

int i = I;
i = i += (++i + (i += N + --i) - ++i);

Then we can see that the values are:

i{I} = i{I} += ((I+1) + (i{I+1} += N + I) - ((I+1+N+I)+1));
i{I} = i{I} += (I + 1 + (I + 1 + N + I) - (I + 1 + N + I + 1));
i{I} = i{I} += (I + 1 + I + 1 + N + I - I - 1 - N - I - 1);
i{I} = i{I} += I;
i{I} = I + I;
i = 2 * I;

Hi Andreas ! This seems to be an acceptable answer indeed but would you mind (if it is not too much asked) apply the same logic when i = something else and also when the hard coded value is bigger than 2 ? That would be awesome, I tried to do it but came with a wrong answer — Yassin Hajaj, Oct 14 '15 at 09:54

Tunaki · Answer 2 · 2015-10-14T10:15:23.973

This is the logic taking into account your first edit (with an unknown X):

public static void main(String[] args) {
    int i = 0;
    i = i+=(++i + (i+=X + --i) - ++i);
    // i = 0 += (++i + ((i += (X + --i)) - ++i));
    // i = 0 += (1 + ((i += (X + --i)) - ++i)); // i = 1
    // i = 0 += (1 + ((1 += (X + --i)) - ++i)); // i = 1 and i will then take the result of 1 += (X + --i)
    // i = 0 += (1 + ((1 += (X + 0)) - ++i)); // i = 0 and i will then take the result of 1 += (X + 0)
    // i = 0 += (1 + (X + 1 - ++i)); // i = X + 1
    // i = 0 += (1 + (X + 1 - X - 2)); // i = X + 2
    // i = 0 += (0); // i = X + 2
    // i = 0;
    System.out.println(i); // Prints 0
}

Tricks here:

+= is an assignement operator so it is right-associative: in the snippets, I added parenthesis to express this more clearly
The result of the assignment expression is the value of the variable after the assignment has occurred
The postfix increment operator ++ and postfix decrement operator -- add or subtract 1 from the value and the result is stored back into the variable.
The + additive operator first computes the left-hand operand then the right-hand operand.

For your second edit (with an unknown I added):

public static void main(String[] args) {
    int i = I;
    i = i+=(++i + (i+=X + --i) - ++i);
    // i = I += (++i + ((i += (X + --i)) - ++i));
    // i = I += (I+1 + ((i += (X + --i)) - ++i)); // i = I+1
    // i = I += (I+1 + ((I+1 += (X + --i)) - ++i)); // i = I+1 and i will then take the result of I+1 += (X + --i)
    // i = I += (I+1 + ((I+1 += (X + I)) - ++i)); // i = I and i will then take the result of I+1 += (X + I)
    // i = I += (I+1 + (X+2*I+1 - ++i)); // i = X + 2*I + 1
    // i = I += (I+1 + (X+2*I+1 - X-2*I-2)); // i = X + 2*I + 2
    // i = I += (I); // i = X + 2*I + 2
    // i = 2 * I;
    System.out.println(i); // Prints 2 * I
}

following that logic, could you eleborate why `i = i+= (++i + (i+=2)) == i = i+= (++i + (i+=2 + --i)); = 4` — SomeJavaGuy, Oct 14 '15 at 09:18
I'm editing the question because I came to something else really weird ! — Yassin Hajaj, Oct 14 '15 at 09:19
Hey @Tunaki, could you check the edit I made to the post. This logic does not apply anymore — Yassin Hajaj, Oct 14 '15 at 09:29

Thomas Weller · Answer 3 · 2015-10-14T09:38:07.010

I suggest the following: format the code differently, so that there's only 1 statement per line, e.g.

@Test
public void test() {
    int i = 0;
    i = 
    i+=
    (
    ++i 
    + 
    (
    i+=
    2 
    + 
    --i
    ) 
    -
    ++i
    );
    System.out.println(i); // Prints 0 instead of 5
}

Then run it under the debugger and press F5 ("Step into") always. This will help you to understand in which order items get evaluated:

int i=0;
i=: ... (needs to wait for result of calculation A)
i+= ... (needs to wait B)
++i: i=1
i+= ... (needs to wait C)
2+
--i: i=0
...: i=3 (result for wait C)
-
++i: i=4 and operand of - is also 4
...: i=0 (result for wait B)
...: i=0 (result for wait A)

Line 10 will always make the result of line 3 0, so the initial value of i will never be changed by the whole operation.

Cyrbil · Answer 4 · 2015-10-14T09:25:34.947

Ok, let's break down every thing:

int i = 0; // i = 0, no big deal

Then starting into the most inner parenthesis:

(i+=2 + --i)

it first decrement i and use the result (-1)
then add 2 (-1+2=1)
and add the result to i (which is 0 at the start of the operation) (0+1=1=i)

At the end, the first decrement is ignored by the reassignation.

Next parenthesis:

i+= (++i + previous_result - ++i)

it increase the i ( with ++i) at two points
then the operation becomes (i+1) + previous_result - (i+2) (notice how the increment are not simultaneous) that gives 2 + 1 - 3 = 0.
the result of the operation is added to the initial i (0)

Again the increment will get discard by the reassignation.

finally:

i = previous_result

Which gives 0 :)

I'm editing the question because I came to something else really weird ! — Yassin Hajaj, Oct 14 '15 at 09:19

Sanjay · Answer 5 · 2015-10-14T09:58:26.887

Because of the highest precedence (...) will be evaluated first then ++ & -- and then remaining operators. Your expression is like

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

//i = i = i + ( (++i) + (i+=2 + (--i)) - (++i) );
//i = i = 0 + ( (++i) + (i+=2 + (--i)) - (++i) );
//i = i = 0 + ( (1) + (i+=2 + (--i)) - (++i) );
//i = i = 0 + ( (1) + (i+=2 + (0)) - (++i) );
//i = i = 0 + ( (1) + (2 + (0)) - (++i) );
//i = i = 0 + ( (1) + (2) - (++i) );
//i = i = 0 + ( (1) + (2) - (3) );
//i = i = 0 + ( 3 - 3 );
//i = i = 0 + ( 0 );
//i = 0

Please fully parenthesize your expression. You will understand evaluation order of expression.

For your EDIT 1

i = i+=( (++i) + (i+=32500 + (--i) ) - (++i) ); 

// 0 + ( (++i) + (i+=32500 + (--i) ) - (++i) ); // i = 0
// 0 + ( (1) + (i+=32500 + (--i) ) - (++i) );  // i = 1
// 0 + ( (1) + (i+=32500 + (0) ) - (++i) ); // i = 0
// 0 + ( (1) + (32500 + (0) ) - (++i) );  // i = 32500
// 0 + ( (1) + (32500) - (++i) ); // i = 32500
// 0 + ( (1) + (32500) - (32501) ); // i = 32501
// 0 + ( 32501 - 32501 ); // i = 32501
// 0 // i = 0

Mr. M · Answer 6 · 2015-10-14T09:56:59.607

I traced the value of i and here are the results:

i = i+=(++i + (i+=2 + --i) - ++i);
initialization: i = 0;
++i: i = 1;(1st one) and store this value
(i+=2 + --i): In it
 --i: i = 0;(i was changed by the previous ++i)
 i += 2 + 0: i = 2;(value of the inner (i+=2 + --i), store it)
++i: i = 3;
1 + 2 -3: i = 0;
i += 0: i = 0;

The value of the 2nd i from the left is not zero, it is the value of i after all the operations to the right are finished.

score 0 · Answer 7 · edited Oct 14 '15 at 10:10

0

Paranthesis takes the precedence. order would be from inner most to outermost

(i+=2 + --i) =>i=i+=2==(2) --2 =0 
(++i - ++i)  =>1-1=0 
i=i+ =0+0 =0

Each expression evaluates to 0

edited Oct 14 '15 at 10:10

MichaelS

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answered Oct 14 '15 at 09:19

Muthu

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