«

#include<stdio.h>
int main(void){
int i=5;
printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
return 0;
}


This gives me : 4 5 5 5 5

How ???

Can anyone please give me some hint as to why I am getting this ??

Thanks & regards,
Onkar

On Feb 16, 11:17 am, Richard Heathfield <r...@see.sig.invalid> wrote:
> onkar said:
>
> > #include<stdio.h>
> > int main(void){
> > int i=5;
> > printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
> > return 0;
> > }
>
> > This gives me : 4 5 5 5 5
>
> > How ???
>
> > Can anyone please give me some hint as to why I am getting this ??
>
> What answer were you expecting, and why?
>
> --
> Richard Heathfield <http://www.cpax.org.uk>
> Email: -http://www. +rjh@
> Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
> "Usenet is a strange place" - dmr 29 July 1999


I was expecting 4 5 5 4 5

onkar said:

> #include<stdio.h>
> int main(void){
> int i=5;
> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
> return 0;
> }
>
>
> This gives me : 4 5 5 5 5
>
> How ???
>
> Can anyone please give me some hint as to why I am getting this ??

What answer were you expecting, and why?


--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

onkar said:

> On Feb 16, 11:17 am, Richard Heathfield <r...@see.sig.invalid> wrote:
>> onkar said:
>>
>> > #include<stdio.h>
>> > int main(void){
>> > int i=5;
>> > printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
>> > return 0;
>> > }
>>
>> > This gives me : 4 5 5 5 5
>>
>> > How ???
>>
>> > Can anyone please give me some hint as to why I am getting this ??
>>
>> What answer were you expecting, and why?
>
> I was expecting 4 5 5 4 5

From your reply, I deduce that you were expecting the compiler to evaluate
the printf arguments from right to left. But in fact C doesn't guarantee
any particular evaluation order for function arguments. To get the output
you require, do this:

#include <stdio.h>

int main(void)
{
int i = 5;
int z = i;
int y = --i;
int x = ++i;
int w = i--;
int v = i++;
printf("%d %d %d %d %d\n", v, w, x, y, z);
return 0;
}

This guarantees the output 4 5 5 4 5

It is always a bad idea to make a line of C code do too much. Break complex
expressions down into stuff you can actually understand. (If you thought
you understood printf("%d %d %d %d %d\n",i++,i--,++i,--i,i), it should be
evident from your surprise at the output that you were mistaken to think
that.)

In general, if you're using ++ or -- on any particular object (or indeed
modifying the object's value in any way), that should be the only place
you use that object within that whole statement. Otherwise, the behaviour
of the code is not defined by the language spec.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

onkar <> writes:
> #include<stdio.h>
> int main(void){
> int i=5;
> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
> return 0;
> }
>
>
> This gives me : 4 5 5 5 5
[...]

The FAQ for this newsgroup is at <http://www.c-faq.com/>.

You desperately need to read section 3. (Then read the rest of it.)

--
Keith Thompson (The_Other_Keith) <kst->
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

onkar wrote:
> #include<stdio.h>
> int main(void){
> int i=5;
> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
> return 0;
> }
>
What's brought all these daft UB questions out of the woodwork this week?

--
Ian Collins.

Ian Collins said:

> onkar wrote:
>> #include<stdio.h>
>> int main(void){
>> int i=5;
>> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
>> return 0;
>> }
>>
> What's brought all these daft UB questions out of the woodwork this week?

What makes you think there's anything special about this week?

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

onkar wrote:
> #include<stdio.h>
> int main(void){
> int i=5;
> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
> return 0;
> }
>
>
> This gives me : 4 5 5 5 5
>
> How ???
>
> Can anyone please give me some hint as to why I am getting this ??

Have you not been paying attention? How many times do we have to answer
the same half-assed question here? Not only is this covered in three
separate questions in the FAQ, but

This post, with the same stupidity,
From: "" <>
Subject: Funny results with increment decrement operators
Date: Fri, 15 Feb 2008 16:00:33 -0800 (PST)

was posted (and answered) only 6 hours before yours

From: onkar <>
Subject: tricky output ???
Date: Fri, 15 Feb 2008 22:09:44 -0800 (PST)

"Martin Ambuhl" <> wrote in message
news:...
> onkar wrote:

>> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);

> This post, with the same stupidity,
> From: "" <>
> Subject: Funny results with increment decrement operators
> Date: Fri, 15 Feb 2008 16:00:33 -0800 (PST)
>
> was posted (and answered) only 6 hours before yours

And in comp.programming a week ago.

It can only be a favourite assignment of C/programming classes.

--
Bart

On Sat, 16 Feb 2008 01:04:31 -0600, Ian Collins wrote
(in article <>):

> onkar wrote:
>> #include<stdio.h>
>> int main(void){
>> int i=5;
>> printf("%d %d %d %d %d\n",i++,i--,++i,--i,i);
>> return 0;
>> }
>>
> What's brought all these daft UB questions out of the woodwork this week?

New semester started recently.



--
Randy Howard (2reply remove FOOBAR)
"The power of accurate observation is called cynicism by those
who have not got it." - George Bernard Shaw