For i = 0, why is (i += i++) equal to 0?

Question

Take the following code (usable as a Console Application):

static void Main(string[] args)
{
    int i = 0;
    i += i++;
    Console.WriteLine(i);
    Console.ReadLine();
}

The result of i is 0. I expected 2 (as some of my colleagues did). Probably the compiler creates some sort of structure that results in i being zero.

The reason I expected 2 is that, in my line of thought, the right hand statement would be evaluated first, incrementing i with 1. Than it is added to i. Since i is already 1, it is adding 1 to 1. So 1 + 1 = 2. Obviously this is not what's happening.

Can you explain what the compiler does or what happens at runtime? Why is the result zero?

Some-sort-of-disclaimer: I'm absolutely aware you won't (and probably shouldn't) use this code. I know I never will. Nevertheless, I find it is interesting to know why it acts in such a way and what is happening exactly.

Answer 1

This:

int i = 0;
i += i++

Can be seen as you doing (the following is a gross oversimplification):

int i = 0;
i = i + i; // i=0 because the ++ is a postfix operator and hasn't been executed
i + 1; // Note that you are discarding the calculation result

What actually happens is more involved than that - take a look at MSDN, 7.5.9 Postfix increment and decrement operators:

The run-time processing of a postfix increment or decrement operation of the form x++ or x-- consists of the following steps:

• If x is classified as a variable:

  • x is evaluated to produce the variable.
  • The value of x is saved.
  • The selected operator is invoked with the saved value of x as its argument.
  • The value returned by the operator is stored in the location given by the evaluation of x.
  • The saved value of x becomes the result of the operation.

Note that due to order of precedence, the postfix ++ occurs before +=, but the result ends up being unused (as the previous value of i is used).

---

A more thorough decomposition of i += i++ to the parts it is made of requires one to know that both += and ++ are not atomic (that is, neither one is a single operation), even if they look like they are. The way these are implemented involve temporary variables, copies of i before the operations take place - one for each operation. (I will use the names iAdd and iAssign for the temporary variables used for ++ and += respectively).

So, a closer approximation to what is happening would be:

int i = 0;
int iAdd = i; // Copy of the current value of i, for ++
int iAssign = i; // Copy of the current value of i, for +=

i = i + 1; // i++ - Happens before += due to order of precedence
i = iAdd + iAssign;

Answer 2

Disassembly of the running code:

int i = 0;
  xor         edx, edx
  mov         dword ptr i, edx         // set i = 0
i += i++;
  mov         eax, dword ptr i         // set eax = i (=0)
  mov         dword ptr tempVar1, eax  // set tempVar1 = eax (=0)
  mov         eax, dword ptr i         // set eax = 0 ( again... why??? =\ )
  mov         dword ptr tempVar2, eax  // set tempVar2 = eax (=0)
  inc         dword ptr i              // set i = i+1 (=1)
  mov         eax, dword ptr tempVar1  // set eax = tempVar1 (=0)
  add         eax, dword ptr tempVar2  // set eax = eax+tempVar2 (=0)
  mov         dword ptr i, eax         // set i = eax (=0)

Equivalent code

It compiles to the same code as the following code:

int i, tempVar1, tempVar2;
i = 0;
tempVar1 = i; // created due to postfix ++ operator
tempVar2 = i; // created due to += operator
++i;
i = tempVar1 + tempVar2;

Disassembly of the second code (just to prove they are the same)

int i, tempVar1, tempVar2;
i = 0;
    xor         edx, edx
    mov         dword ptr i, edx
tempVar1 = i; // created due to postfix ++ operator
    mov         eax, dword ptr i
    mov         dword ptr tempVar1, eax
tempVar2 = i; // created due to += operator
    mov         eax, dword ptr i
    mov         dword ptr tempVar2, eax
++i;
    inc         dword ptr i
i = tempVar1 + tempVar2;
    mov         eax, dword ptr tempVar1
    add         eax, dword ptr tempVar2
    mov         dword ptr i, eax

Opening disassembly window

Most people don't know, or even don't remember, that they can see the final in-memory assembly code, using Visual Studio Disassembly window. It shows the machine code that is being executed, it is not CIL.

Use this while debuging:

Debug (menu) -> Windows (submenu) -> Disassembly

So what is happening with postfix++?

The postfix++ tells that we'd like to increment the value of the operand after the evaluation... that everybody knows... what confuses a bit is the meaning of "after the evaluation".

So what does "after the evaluation" means:

• other usages of the operand, on the same line of code must be affected:
  • a = i++ + i the second i is affected by the increment
  • Func(i++, i) the second i is affected
• other usages on the same line respect short-circuit operator like || and &&:
  • (false && i++ != i) || i == 0 the third i is not affected by i++ because it is not evaluated

So what is the meaning of: i += i++;?

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

The order of evaluation is:

1. Store i + i (that is 0 + 0)
2. Increment i (i becomes 1)
3. Assign the value of step 1 to i (i becomes 0)

Not that the increment is being discarded.

What is the meaning of: i = i++ + i;?

This is not the same as the previous example. The 3rd i is affected by the increment.

The order of evaluation is:

1. Store i (that is 0)
2. Increment i (i becomes 1)
3. Store value of step 1 + i (that is 0 + 1)
4. Assign the value of step 3 to i (i becomes 1)

Answer 3

int i = 0;
i += i++;

is evaluated as follows:

Stack<int> stack = new Stack<int>();
int i;

// int i = 0;
stack.Push(0);                   // push 0
i = stack.Pop();                 // pop 0 --> i == 0

// i += i++;
stack.Push(i);                   // push 0
stack.Push(i);                   // push 0
stack.Push(i);                   // push 0
stack.Push(1);                   // push 1
i = stack.Pop() + stack.Pop();   // pop 0 and 1 --> i == 1
i = stack.Pop() + stack.Pop();   // pop 0 and 0 --> i == 0

i.e. i is changed twice: once by the i++ expression and once by the += statement.

But the operands of the += statement are

• the value i before the evaluation of i++ (left-hand side of +=) and
• the value i before the evaluation of i++ (right-hand side of +=).

Answer 4

First, i++ returns 0. Then i is incremented by 1. Lastly i is set to the initial value of i which is 0 plus the value i++ returned, which is zero too. 0 + 0 = 0.

Answer 5

This is simply left to right, bottom-up evaluation of the abstract syntax tree. Conceptually, the expression's tree is walked from top down, but the evaluation unfolds as the recursion pops back up the tree from the bottom.

// source code
i += i++;

// abstract syntax tree

     +=
    /  \
   i    ++ (post)
         \
         i

Evaluation begins by considering the root node +=. That is the major constituent of the expression. The left operand of += must be evaluated to determine the place where we store the variable, and to obtain the prior value which is zero. Next, the right side must be evaluated.

The right side is a post-incrementing ++ operator. It has one operand, i which is evaluated both as a source of a value, and as a place where a value is to be stored. The operator evaluates i, finding 0, and consequently stores a 1 into that location. It returns the prior value, 0, in accordance with its semantics of returning the prior value.

Now control is back to the += operator. It now has all the info to complete its operation. It knows the place where to store the result (the storage location of i) as well as the prior value, and it has the value to added to the prior value, namely 0. So, i ends up with zero.

Like Java, C# has sanitized a very asinine aspect of the C language by fixing the order of evaluation. Left-to-right, bottom-up: the most obvious order that is likely to be expected by coders.

Answer 6

Because i++ first returns the value, then increments it. But after i is set to 1, you set it back to 0.

Answer 7

The post-increment method looks something like this

int ++(ref int i)
{
    int c = i;
    i = i + 1;
    return c;
}

So basically when you call i++, i is increment but the original value is returned in your case it's 0 being returned.

Answer 8

Simple answer

int i = 0;
i += i++;
// Translates to:
i = i + 0; // because post increment returns the current value 0 of i
// Before the above operation is set, i will be incremented to 1
// Now i gets set after the increment,
// so the original returned value of i will be taken.
i = 0;

Answer 9

i++ means: return the value of i THEN increment it.

i += i++ means: Take the current value of i. Add the result of i++.

Now, let's add in i = 0 as a starting condition. i += i++ is now evaluated like this:

1. What's the current value of i? It is 0. Store it so we can add the result of i++ to it.
2. Evaluate i++ (evaluates to 0 because that's the current value of i)
3. Load the stored value and add the result of step 2 to it. (add 0 to 0)

Note: At the end of step 2, the value of i is actually 1. However, in step 3, you discard it by loading the value of i before it was incremented.

As opposed to i++, ++i returns the incremented value.

Therefore, i+= ++i would give you 1.

Answer 10

The post fix increment operator, ++, gives the variable a value in the expression and then do the increment you assigned returned zero (0) value to i again that overwrites the incremented one (1), so you are getting zero. You can read more about increment operator in ++ Operator (MSDN).

Answer 11

i += i++; will equal zero, because it does the ++ afterwards.

i += ++i; will do it before

Answer 12

The ++ postfix evaluates i before incrementing it, and += only evaluates i once.

Therefore, 0 + 0 = 0, as i is evaluated and used before it is incremented, as the postfix format of ++ is used. To get i incremented first, use the prefix form (++i).

(Also, just a note: you should only get 1, as 0 + (0 + 1) = 1)

References: http://msdn.microsoft.com/en-us/library/sa7629ew.aspx (+=)
http://msdn.microsoft.com/en-us/library/36x43w8w.aspx (++)

Answer 13

What C# is doing, and the "why" of the confusion

I also expected the value to be 1... but some exploration on that matter did clarify some points.

Cosider the following methods:

    static int SetSum(ref int a, int b) { return a += b; }

    static int Inc(ref int a) { return a++; }

I expected that i += i++ to be the same as SetSum(ref i, Inc(ref i)). The value of i after this statement is 1:

int i = 0;
SetSum(ref i, Inc(ref i));
Console.WriteLine(i); // i is 1

But then I came to another conclusion... i += i++ is actually the same as i = i + i++... so I have created another similar example, using these functions:

    static int Sum(int a, int b) { return a + b; }

    static int Set(ref int a, int b) { return a = b; }

After calling this Set(ref i, Sum(i, Inc(ref i))) the value of i is 0:

int i = 0;
Set(ref i, Sum(i, Inc(ref i)));
Console.WriteLine(i); // i is 0

This not only explains what C# is doing... but also why a lot of people got confused with it... including me.

Answer 14

A good mnemonic I always remember about this is the following:

If ++ stands after the expression, it returns the value it was before. So the following code

int a = 1;
int b = a++;

is 1, because a was 1 before it got increased by the ++ standing after a. People call this postfix notation. There is also a prefix notation, where things are exactly the opposite: if ++ stands before, the expression returns the value that it is after the operation:

int a = 1;
int b = ++a;

b is two in here.

So for your code, this means

int i = 0;
i += (i++);

i++ returns 0 (as described above), so 0 + 0 = 0.

i += (++i); // Here 'i' would become two

Scott Meyers describes the difference between those two notations in "Effective C++ programming". Internally, i++ (postfix) remembers the value i was, and calls the prefix-notation (++i) and returns the old value, i. This is why you should allways use ++i in for loops (although I think all modern compilers are translating i++ to ++i in for loops).

Answer 15

The only answer to your question which is correct is: Because it is undefined.

i+=i++; result in 0 is undefined.

a bug in the language evaluation mechanism if you will.. or even worse! a bug in design.

want a proof? of course you want!

int t=0; int i=0; t+=i++; //t=0; i=1

Now this... is intuitive result! because we first evaluated t assigned it with a value and only after evaluation and assignment we had the post operation happening - rational isn't it?

is it rational that: i=i++ and i=i yield the same result for i?

while t=i++ and t=i have different results for i.

The post operation is something that should happen after the statement evaluation.
Therefore:

int i=0;
i+=i++;

Should be the same if we wrote:

int i=0;
i = i + i ++;

and therefore the same as:

int i=0;
i= i + i;
i ++;

and therefore the same as:

int i=0;
i = i + i;
i = i + 1;

Any result which is not 1 indicate a bug in the complier or a bug in the language design if we go with rational thinking - however MSDN and many other sources tells us "hey - this is undefined!"

Now, before I continue, even this set of examples I gave is not supported or acknowledged by anyone.. However this is what according to intuitive and rational way should have been the result.

The coder should have no knowledge of how the assembly is being written or translated!

If it is written in a manner that will not respect the language definitions - it is a bug!

And to finish I copied this from Wikipedia, Increment and decrement operators :
Since the increment/decrement operator modifies its operand, use of such an operand more than once within the same expression can produce undefined results. For example, in expressions such as x − ++x, it is not clear in what sequence the subtraction and increment operators should be performed. Situations like this are made even worse when optimizations are applied by the compiler, which could result in the order of execution of the operations to be different than what the programmer intended.

And therefore.

The correct answer is that this SHOULD NOT BE USED! (as it is UNDEFINED!)

Yes.. - It has unpredictable results even if C# complier is trying to normalize it somehow.

I did not find any documentation of C# describing the behavior all of you documented as a normal or well defined behavior of the language. What I did find is the exact opposite!

[copied from MSDN documentation for Postfix Increment and Decrement Operators: ++ and --]

When a postfix operator is applied to a function argument, the value of the argument is not guaranteed to be incremented or decremented before it is passed to the function. See section 1.9.17 in the C++ standard for more information.

Notice those words not guaranteed...

Answer 16

The steps in calculation are:

1. int i=0 //Initialized to 0
2. i+=i++ //Equation
3. i=i+i++ //after simplifying the equation by compiler
4. i=0+i++ //i value substitution
5. i=0+0 //i++ is 0 as explained below
6. i=0 //Final result i=0

Here, initially the value of i is 0. WKT, i++ is nothing but: first use the i value and then increment the i value by 1. So it uses the i value, 0, while calculating i++ and then increments it by 1. So it results in a value of 0.

Answer 17

Be very careful: read the C FAQ: what you're trying to do (mixing assignement and ++ of the same variable) is not only unspecified, but it is also undefined (meaning that the compiler may do anything when evaluating!, not only giving "reasonnable" results).

Please read, section 3. The whole section is well worth a read! Especially 3.9, which explains the implication of unspecified. Section 3.3 gives you a quick summary of what you can, and cannot do, with "i++" and the like.

Depending on the compilers internals, you may get 0, or 2, or 1, or even anything else! And as it is undefined, it's OK for them to do so.

Answer 18

The ++ operator after the variable makes it a postfix increment. That means the expression i++ evaluates as the value of i prior to adding 1, which in this case is still the default 0. That means zero is what we are adding, so a result of i == 2 after this line executes is impossible.

However, it could, semantically speaking, still end up being 1; we are incrementing i, after all, it's not unreasonable to expect that to actually happen. Why doesn't it? Because i += i++ is syntax sugar for i = i + i++, and if you check the operator precedence rules for C#, assignment (=) happens dead last. That kinda makes sense if you think about it; if = had a higher precedence than, say +, x = a + 2 would result in x == a, because the assignment x = a would be evaluated prior to the math we're trying to do. So, basic assignment of the right-hand side to the left-hand side happens last, after everything else on the right-hand side is evaluated, including i++.

The result:

1. i is evaluated as 0 in all RValue usages (left to right).
2. Increment is evaluated; i is assigned the value of 1
3. Addition is evaluated; 0 + 0 == 0
4. Assignment occurs; i is assigned to the evaluated RValue 0

So, the incrementing is overwritten by the assignment, essentially being hidden by the combination of variable evaluation and assignment.

If the incrementing were done using the prefix version, ++i, then the answer would be 1, as the expression ++i will evaluate to the value after the operation. And, that incrementation would stack; i += ++i + ++i would leave i with a value of 3. However, when evaluating the "add and assign", what you're adding to is always the value of i before evaluation of this statement began.

The twist is that the combination of + and = into += changes the operator precedence of the addition; semantically it doesn't matter in the expression given in the question, however it can matter, especially when you use *= or /=. Consider the following:

var i = 5;
i *= 3 + 4;

The value of i after these lines execute is 35 (add 3 to 4, then multiply the result by 5 and assign to i), when if you break out the *= to the apparently-equivalent i = i * 3 + 4, by PEMDAS the multiplication occurs first and the answer would be 19 (multiply 5 by 3, then add 4 and assign to i). That's because the combined operator lowers the precedence of the multiplication to be the action performed just prior to assignment; i *= 3 + 4 is equivalent to i = i * (3 + 4), not i = i * 3 + 4 which would if written that way evaluate as i = (i * 3) + 4.

Answer 19

There are two options:

The first option: if the compiler read the statement as follows,

i++;
i+=i;

then the result is 2.

For

else if
i+=0;
i++;

the result is 1.

Answer 20

There's lot of excellent reasoning in above answers, I just did a small test and want to share with you

int i = 0;
i+ = i++;

Here result i is showing 0 result. Now consider below cases :

Case 1:

i = i++ + i; //Answer 1

earlier I thought above code resemble this so at first look answer is 1, and really answer of i for this one is 1.

Case 2:

i = i + i++; //Answer 0 this resembles the question code.

here increment operator doesn't come in execution path, unlike previous case where i++ has the chance to execute before addition.

I hope this helps a bit. Thanks

Answer 21

Hoping to answer this from a C programming 101 type of perspective.

Looks to me like it's happening in this order:

1. i is evaluated as 0, resulting in i = 0 + 0 with the increment operation i++ "queued", but the assignment of 0 to i hasn't happened yet either.
2. The increment i++ occurs
3. The assignment i = 0 from above happens, effectively overwriting anything that #2 (the post-increment) would've done.

Now, #2 may never actually happen (probably doesn't?) because the compiler likely realizes it will serve no purpose, but this could be compiler dependent. Either way, other, more knowledgeable answers have shown that the result is correct and conforms to the C# standard, but it's not defined what happens here for C/C++.

How and why is beyond my expertise, but the fact that the previously evaluated right-hand-side assignment happens after the post-increment is probably what's confusing here.

Further, you would not expect the result to be 2 regardless unless you did ++i instead of i++ I believe.

Answer 22

Simply put,

i++, will add 1 to "i" after the "+=" operator has completed.

What you want is ++i, so that it will add 1 to "i" before the "+=" operator is executed.

Answer 23

i=0

i+=i

i=i+1

i=0;

Then the 1 is added to i.

i+=i++

So before adding 1 to i, i took the value of 0. Only if we add 1 before, i get the value 0.

i+=++i

i=2

Answer 24

The answer is i will be 1.

Let's have a look how:

Initially i=0;.

Then while calculating i +=i++; according to value of we will have something like 0 +=0++;, so according to operator precedence 0+=0 will perform first and the result will be 0.

Then the increment operator will applied as 0++, as 0+1 and the value of i will be 1.