>
>
>>> 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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