RE: [LIP] Doubt in ++ operator

int i=7; int j=i++ * i++; printf("%d",j);
The above set of instruction will give you 7 * 8 = 56.

i++ is an unary operator. (value will be used and the memory will
incremented i++).
(memory incremented and value will be take ++i).

Faruk
Singtel

-----Original Message-----
From: linux-india-programmers-admin-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org
[mailto:linux-india-programmers-admin-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org] On Behalf
Of Anand, Dheeraj KumarX
Sent: Monday, July 26, 2004 2:11 PM
To: linux-india-programmers-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org
Subject: RE: [LIP] Doubt in ++ operator


Next time , it should be 9 as you are calling ++ two times.

Dheeraj 

-----Original Message-----
From: linux-india-programmers-admin-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org
[mailto:linux-india-programmers-admin-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org] On Behalf
Of Vardhman Jain
Sent: Sunday, July 25, 2004 8:41 PM
To: linux-india-programmers-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org
Subject: [LIP] Doubt in ++ operator

Hi,
   I have a doubt whether the following code should really be compiler
dependent. int i=7; int j=i++ * i++; printf("%d",j);

I think the answer should always be 49, as the ++ operator comes to
action after the completion of the statement with the ';', But is there
some reason the i++ should increment i before the next i++ is read means
should the second i++ be read as 8 or 7?? Is this behaviour undefined in
c ??


Vardhman

-- 
*************************************
Vardhman Jain, Btech 4rd Year@IIIT Hyderabad
website: http://students.ac.in/~vardhman


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