Unexpected post-increment behavior in assignment

Question

Could you please help me understand why variable a is not incremented in the first case but it is in the second case?

Case 1:

int a = 10;            
a = a++;
Console.WriteLine(a); //prints 10

Case 2:

int a = 10;                        
int c = a++;
Console.WriteLine(a); //prints 11

I've gone through other similar questions but couldn't find any specifics.

UPDATE 1: How I think the program flows

Case 1:

1. 'a' is assigned 10
2. 'a' is assigned 10 before increment happens
3. 'a' is incremented by 1 (Why doesn't this step affect the final value of 'a'?)
4. 'a' is printed --> 10

Case 2:

1. 'a' is assigned 10
2. 'c' is assigned 10 before 'a' is incremented
3. 'a' is incremented by 1 (Why does the increment of 'a' work here?)
4. 'a' is printed --> 11

UPDATE 2: Thanks to all the answers, i think i've understood it, please correct me if i'm wrong.

Case 1:

1. `a` is assigned 10
2. Compiler evaluates `a++`, stores old value 10 and new value 11 as well. Since it's a post increment operation, assigns the old value to `a`. What i thought was, compiler would assign the old value 10 first and evaluate the `++` operation later. This is where i was wrong, compiler evaluates the RHS beforehand and assigns the value based on the operator.
4. 'a' is printed --> 10

Case 2:

1. `a` is assigned 10
2. Compiler evaluates `a++`, stores old value 10 and new value 11 as well. Since it's a post increment operation, assigns the old value to `c` but value of `a` is preserved with `11`.
4. 'a' is printed --> 11

Irwinirwinn

it's an Undefined behavior a=a++; – Cantilever

@Cantilever It compiles, if that's what you mean, or does Undefined behavior mean a different thing? – Irwinirwinn

Undefined Behavior is close to the term don't care state in digital, it means it compiles, it is correct, but can't expect the result of it – Cantilever

Thanks for the info, i'll read more on this. So, is this a bug with c#? – Irwinirwinn

@chouaib, it is NOT undefined behaviour. See my answer for details. – Ancheta

@Cantilever undefined behaviour means it's out of the specification for a language. As Eric Lippert writes, except for one edge case or unsafe parts, there's no such thing as "undefined behaviour" (not in managed, safe, compilable code) in c# – Yettie

Answer 1

For me the best way to understand some behaviour is to check IL generated. In your first case it's

IL_0001:  ldc.i4.s    0A // stack: 10
IL_0003:  stloc.0     // a = 10, stack: empty
IL_0004:  ldloc.0     // stack: 10
IL_0005:  dup         // stack: 10, 10
IL_0006:  ldc.i4.1    // stack: 10, 10, 1
IL_0007:  add         // stack: 10, 11
IL_0008:  stloc.0     // a = 11, stack: 10
IL_0009:  stloc.0     // a = 10, stack: empty
IL_000A:  ldloc.0     // stack: 10
IL_000B:  call        System.Console.WriteLine

You can see that there's still that original value hanging on the stack and so the created 11 gets overwritten in the end.

Let me try to explain it in plain words.

When you are assigning a value to a variable (a = a++) the whole right side of the assignment gets evaluated first in order to guarantee the correct value, that's how it is. So nothing like you get 10, application goes on and increments value when you execute the next line.

Now, imagine post-increment as someone, who first increments a value, but gave you his world that you will get back the original value from the expression. And now you should see why the 11 is overwritten. Increment goes first and in the end, you get the promise original value (as IL proves).

Shipwright

Janacel, thanks for this. However, without looking at the generated IL, would you have expected the output being generated? My point is that the result is unexpected. – Irwinirwinn

All the answers are extremely helpful but this one made me understand it properly, so marking it as the answer. – Irwinirwinn

Answer 2

The first case a = a++ is post-increment. Which says add 1 to a but return the previous value of a and then store the previous result back into a. This is basically a no-op.

If it was pre-increment, a = ++a, then a would be 11.

Ancheta

I'm trying to digest your answer :D – Irwinirwinn

@jSang you are outputting a in the case 2, so the variable c gets out of the question. When you do a++ the variable a is incremented, so that's what you get. If you output c, it'll be 10 – Yettie

@Yettie variable c is just a dummy variable to show how it affects the final value of a, it's there intentionally. – Irwinirwinn

@jSang well, to understand how the code works, since it's dummy, you can just remove it. Change the line int c = a++; to just a++; and see if you understand it better (will still compile and output 11) – Yettie

Answer 3

There is nothing undefined behavior here as someone describes in comment.

This is clearly defined behavior. To understand what is happening, you must first understand how the pre increment and post increment operator works.

Case1:

a++(post increment) will increment the value of a and stores it in a then returns the value before it was incremented.

So after executing a++;, value of a will be 11 but the operator will return 10.

Then a = a++; the assignment part becomes a = 10;.

Case2:

Sameway a++; will increment the value of a to 11 and returns the previous value (10). which will be assigned to c. c will be 10 but a will be 11 because you're not overwriting the value of a in this case.

Your Case1 is equal to:

int a = 10;
int temp = a;//10
a = a + 1;  //11
a = temp;   //10
Console.WriteLine(a);//10

and Case2 is equal to:

int a = 10;
int temp = a;//10
a = a + 1;   //11
int c = temp;//10
Console.WriteLine(a);//11

I hope now that should be clear why you see what you see.

Baileybailie

Answer 4

This is not undefined behavior nor is it a bug. From the MSDN Documentation:

The increment operator (++) increments its operand by 1. 
The increment operator can appear before or after its operand.
The first form is a prefix increment operation. The result of the operation is the value of the operand after it has been incremented.
The second form is a postfix increment operation. The result of the operation is the value of the operand before it has been incremented.

So literally, MSDN tells you that if you use this syntax (postfix):

a = a++;

Then the result of the operation will assign a to a then increment. However, since the assignment operation already took place, you are losing the result of the increment.

Using it like this (prefix):

a = ++a;

This will increment a first, then assign the incremented value to a.

EDIT

I will try to break this down so hopefully you understand better.

First of all, know that ++ always returns a value. If you use the prefix version (e.g. ++a), it returns the value of a+1. If you use the postfix version (e.g. a++), it returns the value of a, before the increment occurs.

When you execute this code:

int a = 10;
a = a++;

You are telling the compiler to assign a the value of a before the increment. Therefore a equals 10 after this executes. The incremented value of 11 gets lost in the "nether". Note that you are not assigning 11 to a. You are assigning the old value, before the increment, which is why you are getting an output of 10.

When you execute this code:

int a = 10;
int b = a++;

After the last line executes, it is incrementing a to equal 11 and assigning 10 to 'b'. Since you are assigning to a different variable, then a is not getting overwritten by its original value of 10 as in the first example.

To make this even more visual, look here:

a = a++;
    ^ a is increased to 11, but the postfix increment returns the old value (10)

This line effectively becomes:

a = 10;

int b = a++;
        ^ a is increased to 11, but b gets assigned a's old value

This line effectively becomes:

int b = 10;
a = a + 1;

Does this clear it up?

Landscape

I'm aware (or at least i think so :D) of how the pre/post increment works. My confusion is with step 3 where a is incremented by 1 but has different output based on the assigned variable. – Irwinirwinn

@jSang - Its because when C# compiles that code, it assigns the value first, then increments it. The assignment takes place first and doesn't occur a second time. The incremented value is lost. – Landscape

Sorry icemanind, "value of 11 gets lost" this is simply not true. Check my answer, examine the IL. 11 is not lost, it's, in fact, assigned to a before 10 is assigned there as well. – Shipwright

@OndrejJanacek - Internally, 11 is on the stack. However, at that point, it is lost in c#. There is no way in C# to access that value of 11, without embedding IL code or doing some write memory trick. Write me a C# program that can access that value of 11, after the line a=a++ and I will happily retract my answer. – Landscape

@icemanind You can't look at it this way. There would be so many "lost" values. Just because you cannot interact so closely with the stack from C# does not mean it's lost. You sure could write code directly in IL and access that value with no problem. – Shipwright

@OndrejJanacek - I'm looking at this through the eyes of the OP. In his eyes, his expected value of '11' does not exist anymore. He doesn't care if it exists on some stack or out in memory somewhere. He just wants to know why he can't access it through C# code. – Landscape

@icemanind All right, I guess I am just too pedantic. We can sure agree on value 11 being lost during the whole process. – Shipwright

@OndrejJanacek - Don't get me wrong. I agree that the value 11 doesn't just magically vanish. And if the OP wanted a deep technical explanation, I'd give it. But I dumbed down my answer so the OP could understand it easier. – Landscape

@jSang - Yes, now you are understanding it well! – Landscape

Answer 5

To make it clear, there is a different between ++a, and a++.

Both two of them produces the same result, that is to increment a by 1, but the process is a little bit different and it is only seen by assigning it to another variable.

This is difference between pre and post increment that you might looking for

Case 1

 int a,b;
 a = 10;
 b = a++; //Here b is assigned by a post-incremented a, therefore b value is still 10

Case 2

 int a,b;
 a = 10;
 b = ++a; //Here b is assigned by a pre-incremented a, b value is now 11

Taken from http://en.wikipedia.org/wiki/Increment_and_decrement_operators

UPDATE

In reply to this problem

int a = 10;
a = a++;

the process is as followed:

1. variable 'a' with data type integer is created

2. variable 'a' value is assigned with 10

3. check the availability and the data type correctness of the right side. This means all operation on the right side (increment 'a') is being executed. 'a' is now 11

4. assign the post-incremented value to 'a', which is 10, 'a' is now 10

Persist

Answer 6

I don't think that this is an undefined behavior , since in the first case :

int a = 10;            
a = a++;
Console.WriteLine(a); //prints 10

The value of a is the value before the increment action , meaning that if we simplify a = a++ , then :

a = a ; // `a` has `10`

a++; // `a` has `11`

and only then a has the value 11 .

The 2nd case :

int a = 10;                        
int c = a++;
Console.WriteLine(a); //prints 11

Now c has the value 10 since the assignment operation makes the value on the left side to get the value of the variable (on the right side) before the increment action , and only then the value on the right side goes up by 1 .

Foreleg

Answer 7

You should print out c in case 2. The post incement operator result is always the value before increment. So first give the value 10 back then increment it to 11 and after that assign the operator result (10) to the left operand.

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