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.comwrote:
FUNC2(i) = i==0?i*(i-1*(i-1-1))
i guess the output is fff0
-1 = 0001
taking 1's complement
1110
taking 2's complement
-14
left shifting 4 bits
%x prints in hexa decimal format
so o/p- f f f 0
On Sat, Aug 20, 2011 at 11:39 AM,
can someone explain how this works-
*
*
*void* *main*(){
*int* a,b,d;
scanf(%d%d,a,b);
d=a+~b+1;
printf(%d,d);
getch();
}
say if a=5 b=6 then
output is -1
if ~ is one's complement operator
then
a=0101
~b=1001
d= 0101+1001+1. which is not -1
can someone explain how it works.
thank you dave and amit...
On Wed, Aug 10, 2011 at 8:36 PM, Dave dave_and_da...@juno.com wrote:
@Ayswarya: In twos-complement arithmetic, -x = ~x + 1, so
a + ~b + 1 = a +(~b +1) = a + (-b) = a - b.
Dave
On Aug 10, 9:23 am, Ayswarya Srinivasan krsayswa...@gmail.com wrote:
can someone
the written consisted of 45 questions for 90 minutes.. negative marking was
there.. the qs wer based on linux,networks, data structures and pattern
matching problems..
some of the qs were..
1. 25 horses, 5 race tracks, choose the top 3 . minimum number of races u
need ?
2. 2 clocks A and B are
