code snippet ? functionality ??

Question

Hii static void Main() { int i j = 1 k; for (i = 0; i { k =j++ + ++j; Console.WriteLine(k+ ); } Console.ReadKey(); } In he above code .. i am getting out put as 4 8 12 16 20 ... well its compiler generated output it cant be wrong but the thing is i am confused .. with the block of code K= j++ + ++J ; Here acc to pre & post increment operation mechanism say initially j=1 next step j++ it remains 1 only after that ++j it gets 2 so 1+2 should be 3 but y am i getting 4 .. Hope u got my point TY

Vulpes · Answer

The steps are: 1) j is stored as the first operand of the + operator. 2) j is incremented because of the postfix ++ operator. 3) j is incremented again because of the prefix ++ operator. 4) j is stored as the second operand of the + operator. 5) The addition takes place of the j values in steps 1) and 4) and is assigned to k So following those steps through: 1) j is 1 (first operand) 2) j is 2 3) j is 3 4) j is 3 (second operand) 5) k = 4 (printed to console) and on the next iteration: 1) j is 3 (first operand) 2) j is 4 3) j is 5 4) j is 5 (second operand) 5) k = 8 (printed to console) and so on.

Vulpes · Answer

The debugger can only show you the values of variables before a line is executed and after it has been executed. There doesn't appear to be any way that it can display variable values during the execution of a line (not even in the Watch window) and so the nuances of the postfix and prefix ++ and -- operators are difficult to observe. You therefore need to understand how these operators work but if you remember that all operators produce a value and that: 1. in the case of j++ the value is the value of j before it's incremented (incrementation follows immediately afterwards); 2. in the case of ++j the value is the value of j after it's incremented; and that 3. these values are used when j++ is itself an operand (or part of an operand) of some other operator then you shouldn't have any problems reading code which uses these operators.

Vulpes · Answer

When the debugger steps into this line: k = j++ + ++j; and you hover your mouse over the variables you see the value of those variables after the line has been executed i.e. j = 2 m = 2 and k = 3 on the first iteration. The point here is that the expression j++ has a value which is equal to the value of j before the incrementation takes place and it's this value which is stored as the first operand of the + operator. The incrementation then takes place before the second operand is evaluated. So in other words it's the value of j on entering the for statement (i.e. 1) which is used in the addition.

SUNIL GUTTA · Answer

:D so true and nice sense of humor & firstly entrant for a C obfuscation contest! oooh man this is too good :D :D :D sorry sorry sorry dam dam sometime i suck .. just thinking only in single point of view .. got my answer in simple the post operator make the value incremented i.e can say virtually inside the loop itself and make available .. now mud is very much clear hahah haaa haa

Vulpes · Answer

Maha The behavior of the C# compiler is in fact completely predictable. In your example: 1) j is 1 (first operand) 2) j is then incremented to 2 (postfix ++) 3) m gets incremented to 2 (prefix ++) 4) m is 2 (second operand) 5) The addition of 1) and 4) is 3 (printed to console). On the next iteration: 1) j is 2 (first operand) 2) j is then incremented to 3 (postfix ++) 3) m gets incremented to 3 (prefix ++) 4) m is 3 (second operand) 5) The addition of 1) and 4) is 5 (printed to console). and so on.

Vulpes · Answer

The order of execution is left to right. However the postfix and prefix ++ operator has higher priority than the + operator and so the ++ operations are performed first. Now the value of j++ is the value of j before the operation and the value of ++j is the value of j after the operation. So on the first iteration of the loop the value of j++ is 1 and this is then stored as the first operand of the + operator. The incrementation then takes place giving j a value of 2. The value of ++j is then 3 and this gets stored as the second operand of the + operator. Adding 1 and 3 gives k = 4. Clear as mud eh :) If it isn't very clear I wouldn't worry too much as nobody uses stuff like this in practice unless they're an idiot or an entrant for a C obfuscation contest!

Maha · Answer

I had an experience of struggling with post and pre increment operators after that I came to the conclusion that we couldn't predict the compiler behavioural pattern for this particular operation. For example see this: Let us say m=1 and change the 2nd j into m. Output is 3 5 7 9 11 using System; namespace ConsoleApplication6 { class Program { static void Main(string[] args) { int i j = 1 k m=1 ; for (i = 0; i { k = j++ + ++ m ; Console.WriteLine(k +  ); } Console.ReadKey(); } } }

Maha · Answer

There are many problems Vulpes. It will continue in a different thread.

Maha · Answer

Hi Sunil Gutta as I am an amateur not an adept in a C# language I always look for simple explanation for a problem. Anyhow thank you for taking trouble to give me an explanation for the problem. I will go through it and let you know.

SUNIL GUTTA · Answer

Hey maha please do check my last comment .. it explains more than clear to you .. hope it helps ..

Maha · Answer

Vulpes Thank you for your explanation. If you use the Step Into(F11) key in the following program in first iteration J=3 could you please explain the reason. using System; namespace ConsoleApplication6 { class Program { static void Main(string[] args) { int i j = 1 k; for (i = 0; i { k = j++ + ++j; Console.Write(k +  ); } Console.ReadKey(); } } } /* 4 8 12 16 20 */

SUNIL GUTTA · Answer

hi maha & ravi ..!! what i understood is it maY help u to clarify a bit The thing is in k =j++ + ++m ...!! if m & j=1 initially right 1) j++ is post increment i.e it will set j=1 initially and right away it increments after setting j=1 & stores j=2 in for the next instance of j lets say virtually :) As in your case there is only one J later to it u have ++m .. so the thing is the virtual values i.e j=2 assigned will be applied in reality to next j instance  Thank you ..!! Here in j++ + ++j ... after j is assigned to j=1 its internally asuming loop is done so next step is just to increment the value i.e j=2  which is made available to next j instance ie as our next one is ++j it will be +1 of virtual value what j hold .. so 2+1 =3 .. so finally we get 3+1 =4 ..... continues Hope you got it

Ravi Shekhar · Answer

I agree with maha . i am also confused . pls explain the reason

Maha · Answer

Vulpes if you use the Step Into(F11) key when j=2 and m=2 k is giving the result 3. Could you please explain the reason.

SUNIL GUTTA · Answer

well just distinct & awesome explanation but i have small doubt 2) j is incremented because of the postfix ++ operator. as u said but does it mean iteration done ?? loop over .. i'm confused .. just only this point . i didn't find any loop to make j from 1 to 2 my silly view : when j++ is there after loop only the j should increment right so it will be k=1 + 2 /* 1 beocz it cant be 2 as it is post increment j++  j remains same ?? */ /* 2 beocz of pre-increment of j */ after looping j becomes 3 becoz post increment that we have to apply after circular notion /// vulpes what the order of excecution i.e from right to left or left to right haaa in above thing