@umesh...thanks :) this explanation seems to be satisfactory but failed to
understand from where @anika and @himanshu were getting 7 6 8....i ran
it...output comes out to be 7 6 6  !!

On Sun, Jun 12, 2011 at 6:15 AM, UMESH KUMAR <[email protected]>wrote:

>
>>>> void call(int a,int b,int c)
>>>> {
>>>>          printf("%d %d %d",a,b,c);
>>>> }
>>>>
>>>> int main()
>>>> {
>>>> int a=5;
>>>> call(a++,a++,++a);
>>>> return 0;
>>>>
>>>> }
>>>>
>>>>
>>>> On Sat, Jun 11, 2011 at 8:21 PM, PRAMENDRA RATHi rathi <
>>>> [email protected]> wrote:
>>>>
>>>>> IN second program:
>>>>>  in function value are always push in the stack from right.
>>>>> so first value is --i that will make i=1 and value 1 will be passed to
>>>>> function
>>>>> and
>>>>> after that i++ that's means i will be passed.
>>>>> so 1 will be passed and after passing value. i will changed to 2.
>>>>>
>>>>> if u want to know why reverse order than can go through:
>>>>> http://cs.nyu.edu/courses/fall03/V22.0201-003/c_param.html<http://www.google.com/url?sa=D&q=http://cs.nyu.edu/courses/fall03/V22.0201-003/c_param.html>
>>>>> -----------------------------------------
>>>>> PRAMENDRA RATHI
>>>>> NIT ALLAHABAD
>>>>>
>>>>>
>>>>>
>>>>> When the function is called then formal values PUSH in a
>>>>>
>>>>
>              STACK from left to right and  POP from left to right finally
>             so we can obtained the output :7 6 6 ...............
>
>          Since function call ----------------------
>
> *         call(a++,a++,++a);*
> *         and *a=5
>
>        STACK PUSH AS:-
>         0::++a --------c
>         1::a++ --------b
>         2::a++ --------a
>
>       POP AS:
>
>        0::++a --------c=   6         ( *6)*
> *
> *
> *      *1::a++ --------b =  6          (*7*)
>
>       2::a++ --------a =  7          (*8*)
>
>  So. finally OUTPUT :-
> * a=7, b=6, c=6*
> *
> *
> *
> *
> **************************************
> *
> *
>
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