doesn it work like below ?
func2(8)= 8 * func1(7)
func1(7)= 7*6
so i=8*7*6???
On Sun, Aug 21, 2011 at 3:58 PM, Sanjay Rajpal <srn...@gmail.com> wrote:
> +1 to sagar
> Sanju
> :)
>
>
>
> On Sun, Aug 21, 2011 at 3:26 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> FUNC2(i) = i==0?i*(i-1*(i-1-1))
>>
>> 8==0?1:8*(8-1*(8-1-1))
>> 8==0?1:8*(8-6)
>> 8==0?1:8*2
>> 8==0?1:16
>>
>> hence it will return 16
>>
>>
>> On Sun, Aug 21, 2011 at 11:40 AM, SuDhir mIsHra <
>> sudhir08.mis...@gmail.com> wrote:
>>
>>> #include<stdio.h>
>>> #define FUNC1(i) (i*(i-1))
>>> #define FUNC2(i) (i==0?1:i*FUNC1(i-1))
>>> main()
>>> {
>>> int i=8;
>>>
>>>
>>> printf("\n%d",FUNC2(i));
>>> }
>>>
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>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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Regards,
Ayswarya S
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