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Unexpected post-increment behavior in assignment

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

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).

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.

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.

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?

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 .

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

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.

C# in a nutshell example misunderstanding ++ unary operator increment [duplicate]

This question already has answers here:
What is the difference between i++ and ++i in C#?
(7 answers)
Closed 9 years ago.
I am reading C# 5.0 in a nutshell and I have the following statements:
int x = 0;
while (true)
{
if (x++ > 5)
break ; // break from the loop
}
x.Dump();
Executing the statements on LINQPad 4 the output is 7.
I still don't understand WHY?. Why it is not 6 being the condition: x++>5

The ++ operator increments the value and returns the original value.
So when x was 5, x was incremented and the condition 5 > 5 is evaluated.
Then, when x was 6, x was incremented and the condition 6 > 5 is evaluated, which results in the break. Because x was still incremented, the value of x in the end is 7.
People often say that when the ++ operator is use as a postfix, the increment is executed after the comparison, but this is not technically true, which is shown by decompilation in LinqPad:
int x = 5;
if (x++ > 5)
Console.WriteLine(x);
IL:
IL_0001: ldc.i4.5
IL_0002: stloc.0 // x
IL_0003: ldloc.0 // x
IL_0004: dup
IL_0005: ldc.i4.1
IL_0006: add // <---- Increment
IL_0007: stloc.0 // x <-- Result of increment is popped from the stack and stored in a variable
IL_0008: ldc.i4.5
IL_0009: cgt // <---- Comparison
IL_000B: ldc.i4.0
IL_000C: ceq
IL_000E: stloc.1 // CS$4$0000
IL_000F: ldloc.1 // CS$4$0000
IL_0010: brtrue.s IL_0019
IL_0012: ldloc.0 // x
IL_0013: call System.Console.WriteLine

x++ > 5 means check the value of x against 5 then increment it. Which means your break is not reached until x == 7.
The result would be 6 if you used ++x, i.e. increment x and then test it against 5.

The following are the iterations of the loop:
x == 0 -> comparison returns false -> x increased to 1
x == 1 -> comparison returns false -> x increased to 2
x == 2 -> comparison returns false -> x increased to 3
x == 3 -> comparison returns false -> x increased to 4
x == 4 -> comparison returns false -> x increased to 5
x == 5 -> comparison returns false -> x increased to 6
x == 6 -> comparison returns true -> x increased to 7 -> if-block is entered and the loop is left
The last value of x, namely 7, will be output.
Note that the x++ operation modifies the value of x, after the comparison x > 5 is evaluated, but regardless of whether or not the comparison returns true and the if-block is entered.

On C#, the ++ unary operator, when wrote after its operand (it can be either ++x or x++) executes itself after evaluations in the expression. That means:
When x is 5, the expression is gonna be evaluated as false (5 is not > than 5) and then increment it to 6.
Then, when x is 6, the expression is gonna be evaluated as true yes (6 is > than 5), but then it is gonna increment again, breaking when x is 7.

The reason is that x++ will evaluate after the check. So when it finally gets to x == 6 the condition will be met (6 > 5) and then afterwards it will be incremented (to 7)
It is easier to explain with some alternate code, which will produce the same result:
while (true)
{
if(x > 5)
{
//x = 6
x++;
//x = 7
break;
}
x++;//x increments from 0 - 6, over 6 loops
}

You're evaluating x before you increment it, so your condition of 6 will cause the break and then it goes up 1 to 7. I think you're expecting this:
int x = 0;
while (true)
{
if (++x > 5)
break ; // break from the loop
}
x.Dump();
This will increment x before you evaluate its condition.

It is because of the ++
The ++ operator counts x one up everytime it hits the line.
And as you have a while(true) loop it will count x one up in the if statement everytime it reaches that if-statement.

x++ increments x after the value has been read. This means that when it increments once after the comparison fails.
++x on the other hand increments before the value is read.

Sure, it's 7 because even when it does the check for 6 > 5, it increments by one. Because of the unary operator x++.

Post-increment x by n (n != 1)

So, if I execute the following code...
int x = 0;
Debug.WriteLine(x++);
Debug.WriteLine(x += 4);
Debug.WriteLine(x);
... I get 0, 5, and 5, respectively. What I'd like to get, however is 0, 1, and 5. Is there any way to do a post-increment by n in C#? Or do I have to write out the += as its own statement?
Just for context, what I'm actually doing is a bunch of BitConverter operations on a buffer, and it'd be really nice to have each one as a self-sufficient statement where the offset is incremented by the size of the data type being converted to. That way, if the buffer format is later changed, I can just add or remove the one line without having to worry about any of the surrounding code.

You should be able to abuse Interlocked.Exchange to get the old value of a variable while at the same time replacing its value:
Debug.WriteLine(Interlocked.Exchange(ref x, x+4));
in other words, replace the value of variable x with x + 4 but return the previous value of x.
Edit:
Disassembly shows this "increment by 4 and swap" is done in 4 instructions - no call needed at all, so performance should be good:
Interlocked.Exchange(ref x, x + 4);
0000005e mov eax,dword ptr [rbp+2Ch]
00000061 add eax,4
00000064 xchg eax,dword ptr [rbp+2Ch]
00000067 mov dword ptr [rbp+28h],eax
The non-intuitive nature of this (and other) solutions probably boils down to the violation of the CQS principle - we are mutating a variable and returning a value all at once, i.e. not something we should be doing in the mainstream.

As far as I know it's not possible. You could write your wrapper method like that:
static void ExecuteWithPost(ref int value, Action<int> operation, Func<int, int> postOperation)
{
operation(value);
value = postOperation(value);
}
and use it:
int i = 0;
ExecuteWithPost(ref i, x => Debug.WriteLine(x), x => x + 1);
ExecuteWithPost(ref i, x => Debug.WriteLine(x), x => x + 4);
ExecuteWithPost(ref i, x => Debug.WriteLine(x), x => x);
Prints what you want.
The wrapper method can be generic to make it work with types other than int:
static void ExecuteWithPost<T>(ref T value, Action<T> operation, Func<T, T> postOperation)
{
operation(value);
value = postOperation(value);
}

Operator ++ is short form of x = x + 1; so using += is not a bad idea:
If I understand you correctly:
int x = 0;
Debug.WriteLine(x++); //return 0
Debug.WriteLine(x); //return 1
Debug.WriteLine(x += 4); //return 5
I would suggest you to use operator += because any other way operator overloading or something else; will just be a overhead.

Aggregate lambda expressions

int sum0 = 0;
for (int i = 0; i < 10; i++)
{
sum0 += i;
}
int sum1 = Enumerable.Range(0, 10).Sum();
int sum2 = Enumerable.Range(0, 10).Aggregate((x, y) => x + y);
int sum3 = Enumerable.Range(0, 10).Aggregate(0, (x, y) => x + y);
All of the above 4 expressions are doing the same thing: find sum from 0 to 10. I understand the calculation of sum0 and sum1. But what are sum2 and sum3? Why the lambda uses two parameters (x, y) here?

Expanding on bdukes' answer, the lambda takes
( x = [value of last lambda expression], y = [next value] ) => x+y
and sum3 allows you to set the initial x value.

The Enumerable.Aggregate method expects a function that takes the current value of the aggregation and a value from the enumeration. The overload for sum3 also provides a starting value for the aggregation.

X is holding the current total, Y is being added to it for each element.
Foreach(y)
X = X + Y;

The Aggregate extension methods take a function (Func<T1,T2,TResult>) that calculates an aggregate..
The function specified for sum2 is one that adds x to y, for every supplied x and y (that is it sums all the items in the enumeration).
The additional parameter for sum3 is an accumulator - a value that is to be added for each operation - as this is 0, it is essentially summing up all the items in the enumeration without any additional values.

sum2 uses a custom function x + y to aggregate each element of the list. The aggregation starts with the default value for an integer 0 and adds the first element to it. It then takes that value and adds the next element, and so on, until it runs out of elements. It then returns the final figure.
sum3 does exactly the same as sum2 but it also explicitly starts the aggregation with a specific value of 0.
Semantically all three are the same - as presented here - but by varying the aggregation function and the initial starting value you can generate all sorts of custom aggregations.
Another way of looking at it is that .Sum() is simply short-hand for .Aggregate(0, (x, y) => x + y);.

x is the variable that "accumulates" the values so its value is
step 1) 1
step 2) 3
step 3) 6
step 4) 10
and so on...
the 0 in the sum3 is the starting value :) (which is redundant since 0 is the default value for int)

The parameter x holds the aggregation, and y is the next enumeration item.
string sentence = "the quick brown fox jumps over the lazy dog";
// Split the string into individual words.
string[] words = sentence.Split(' ');
// Prepend each word to the beginning of the
// new sentence to reverse the word order.
string reversed = words.Aggregate((workingSentence, next) =>
next + " " + workingSentence);
Console.WriteLine(reversed);
// This code produces the following output:
//
// dog lazy the over jumps fox brown quick the
from http://msdn.microsoft.com/en-us/library/bb548651.aspx

int[] arr={0}; int value = arr[arr[0]++]; Value = 1?

Today I came a cross an article by Eric Lippert where he was trying to clear the myth between the operators precedence and the order of evaluation. At the end there were two code snippets that got me confused, here is the first snippet:
int[] arr = {0};
int value = arr[arr[0]++];
Now when I think about the value of the variable value, I simply calculate it to be one. Here's how I thought it's working.
First declare arr as an array of int
with one item inside of it; this
item's value is 0.
Second get the value of arr[0] --0 in
this case.
Third get the value of arr[the value
of step 2] (which is still 0) --gets
arr[0] again --still 0.
Fourth assign the value of step 3
(0) to the variable value. --value =
0 now
Add to the value of step 2 1 --Now
arr[0] = 1.
Apparently this is wrong. I tried to search the c# specs for some explicit statement about when the increment is actually happening, but didn't find any.
The second snippet is from a comment of Eric's blog post on the topic:
int[] data = { 11, 22, 33 };
int i = 1;
data[i++] = data[i] + 5;
Now here's how I think this program will execute --after declaring the array and assigning 1 to i. [plz bear with me]
Get data[i] --1
Add to the value of step 1 the value
5 --6
Assign to data[i] (which is still 1)
the value of step 2 --data[i] = 6
Increment i -- i = 2
According to my understanding, this array now should contain the values {11, 27, 33}. However, when I looped to print the array values I got: {11, 38, 33}. This means that the post increment happened before dereferencing the array!
How come? Isn't this post increment supposed to be post? i.e. happen after everything else. What am I missing guys?

The postincrement operation occurs as part of evaluating the overall expression. It's a side effect which occurs after the value is evaluated but before any other expressions are evaluated.
In other words, for any expression E, E++ (if legal) represents something like (pseudo-code):
T tmp = E;
E += 1;
return tmp;
That's all part of evaluating E++, before anything else is evaluated.
See section 7.5.9 of the C# 3.0 spec for more details.
Additionally, for assignment operations where the LHS is classified as a variable (as in this case), the LHS is evaluated before the RHS is evaluated.
So in your example:
int[] data = { 11, 22, 33 };
int i = 1;
data[i++] = data[i] + 5;
is equivalent to:
int[] data = { 11, 22, 33 };
int i = 1;
// Work out what the LHS is going to mean...
int index = i;
i++;
// We're going to assign to data[index], i.e. data[1]. Now i=2.
// Now evaluate the RHS
int rhs = data[i] + 5; // rhs = data[2] + 5 == 38
// Now assign:
data[index] = rhs;
The relevant bit of the specification for this is section 7.16.1 (C# 3.0 spec).

For the first snippet, the sequence is:
Declare arr as you described:
Retrieve the value of arr[0], which is 0
Increment the value of arr[0] to 1.
Retrieve the value of arr[(result of #2)] which is arr[0], which (per #3) is 1.
Store that result in value.
value = 1
For the second snippet, the evaluation is still left-to-right.
Where are we storing the result? In data[i++], which is data[1], but now i = 2
What are we adding? data[i] + 5, which is now data[2] + 5, which is 38.
The missing piece is that "post" doesn't mean "after EVERYTHING else." It just means "immediately after I retrieve the current value of that variable." A post increment happening "in the middle of" a line of code is completely normal.

data[i++] // => data[1], then i is incremented to 2
data[1] = data[2] + 5 // => 33 + 5

I would expect the post-increment operator to increment the variable after its value is used.
In this case, the variable is incremented before the second reference to the variable.
If it would not be so, you could write
data[i++] = data[i++] + data[i++] + data[i++] + 5
If it would be like you say, then you could remove the increment operator because it doesn't do actually anything, in the instruction I reported.

You have to think of assignments in three steps:
Evaluate left hand side (=get address where the value should be stored)
Evaluate right hand side
Assign the value from step 2 to the memory location from step 1.
If you have something like
A().B = C()
Then A() will run first, then C() will run, and then the property setter B will run.
Essentially, you have to think of your statement as
StoreInArray(data, i++, data[i] + 5);

The cause might be that some compilers optimize i++ to be ++i. Most of the time, the end result is the same, but it seems to me to be one of those rare occasions when the compiler is wrong.
I have no access to Visual Studio right now to confirm this, but try disabling code optimization and see if the results will stay the same.