Simple but confusing C code

Question

0.00/5 (No votes)

See more:

XML

#include<stdio.h>
main()
{
      int i=5;
      printf("%d",++i + ++i + ++i);
}

Can anyone tell me how this prints 22..??

Posted 28-Nov-12 23:54pm

skrtbhtngr

Add a Solution

5 solutions

Solution 2

The GCC compiler with the -Wall switch, gives the following warnings:

j.c:5: warning: operation on ‘i’ may be undefined
j.c:5: warning: operation on ‘i’ may be undefined

Why the operation might be undefined is well explained at Stack Overflow: printf and ++ operator [closed][^].

Posted 29-Nov-12 0:10am

CPallini

Solution 6

I agree with Richard. Because in live project this kind of code will never used. This may give unexpected result. yes, this is very confusing. But compiler seeing this code section as it's own way. ie, it first evaluate
i = 5;
(++i) = 6. hope you got it.
but, next it take another two + operator as postfix ie, i++. hence now i have value 7. ie
1) ++i
2)i++
current i = 7.
then it take next + operator for addition. ie (i++) + i = 14. ok.
then compliler take i++. hence i have now value 15. next compliler take previous i result( it is stored in registry) and it to current result 15 + 7 = 22.
you can ensuare it using assembly debugging.

Note : Please avoid using of this kind of coding standard.

Posted 29-Nov-12 17:19pm

Sajeesh Payolam

Solution 4

Is it print 22 really?! Ok,have a look here..
At First,you declared i as int and assign 5 to it.
so at first time ++i will be 6(i+1)
at 2nd time ++i will be 7(i+1)
at 3rd time ++i will be 8(i+1)
So now sums up all those 3,(6+7+8)=21
For my case,it prints 21.

Posted 29-Nov-12 1:10am

ridoy

Comments

[no name] 30-Nov-12 0:12am

Yeah I also got 21...

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)

armagedescu · Accepted Answer · 2012-11-29T00:28:00

It first evaluates ++i + ++i, the unexpected result is 14, because the operation is i + i, but before doing this it is incremented. The result is saved to some place, for instance the registry, and after that i is incremented one more time and added to the saved result.

Even this may give 24, because of i + i + i. First is evaluated ++i, after that it is evaluated again and again. And after that is done operation result = i + i + i

Richard MacCutchan · Accepted Answer · 2012-11-29T00:03:00

Solution 1

It's all down to the order in which the compiler generates the code. In general this sort of construct should never be used in live code as it is prone to giving the wrong result. Step through the generated assembler code with your debugger to see what is happening.

Posted 29-Nov-12 0:03am

Richard MacCutchan