Problem with post increment in c#?

Question

4.00/5 (1 vote)

See more:

C#

int x = 20;
x = x++;
Console.WriteLine(x);

In the above statement x=x++;

why x value is not increment.

I have a little bit confusion here. Please explain about this.

Posted 6-Mar-14 17:43pm

Srinubabu Ravilla

Updated 6-Mar-14 17:56pm

v2

Add a Solution

Comments

Sergey Alexandrovich Kryukov 7-Mar-14 0:08am

Could you explain why writing such thing at all? Just for fun?
—SA

Srinubabu Ravilla 7-Mar-14 0:26am

Dear Sergey thank you for your reply.
I am not writing it for fun.
I writing to know about whats happening here.
There is difference between
int x=10, y=10;
x=x++;
Console.WriteLine(x); // output: 10
y=x++;
Console.WriteLine(x);// output: 11

Can you explain why x value is increment first time but not second time.

Sergey Alexandrovich Kryukov 7-Mar-14 0:57am

Very good. So, let's see...
—SA

Aurus_Deepak 7-Mar-14 2:22am

x= x++, can be broken down to:

1) increment the value contained in x
now x contains 11

2) return the value that was contained in x before it was incremented
that is 10

3) assign that value to x
now, x contains 10

Sergey Alexandrovich Kryukov 7-Mar-14 8:51am

No, because increment should go after assignment. Look at the decompiled code to see.
—SA

2 solutions

Solution 3

U are using the post increment (x++) so only the value of x is not incremented at first time.
Use the pre increment (++x).

++x increments a before it is evaluated. x++ evaluates a and then increments it.

Posted 6-Mar-14 19:21pm

SViki

Updated 6-Mar-14 19:26pm

v3

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Sergey Alexandrovich Kryukov · Accepted Answer · 2014-03-06T18:55:00

This is pretty clear.

I would ask you: what result did you expected and why? Did you expect 21, as in case of y = x++? No way! Compiler is not supposed and should not make any assumption on the developer's intention, it should follow simple and strict principles. If you do weird thing which makes no practical sense, as in this case, the logical and apparent result may also appear weird.

The compiler needs to estimate expression on right two times, before ++ and after, because this is post-increment. The value "before" is needed for assignment; and value "after" is the result of increment. As x appears in both sides, the addition (temporary) slot on stack is reserved, to manipulate with the expression on right. The x estimated, this is 20. Then 20 is assigned to first slot representing variable x. It becomes 20. Then the memory location representing expression on right needs to be incremented by 1. It becomes 21. On exist from the stack frame, this value is discarded.

This is the only logical interpretation of post-increment with assignment to the same variable. You can easily see how it works if you disassemble this fragment of code.

You should understand that this is the pathological case which makes no sense at all in practice. Increment can be performed by writing ++x or x++. Combination of increment with assignment only makes sense if the result is assigned to a different variable.

I appreciate your interest to some pathological cases. It gives us the idea that some operations are not as trivial as they seem and that caution needs to be used when working with increments.

—SA