Why am i Getting correct in bitwise shifting by a negative value?

Asked Jul 19 '12 at 17:21

Active Jul 19 '12 at 17:21

Viewed 65 times

Possible Duplicate:
Right shifting negative numbers in C

This question i have found on SO, but everywhere i found the result as undefined.

int main()
{
   printf("%d\n",32<<-2); <-------------1
   printf("%d\n",32>>-2); <-------------2
   return 0;
}

I am getting the output as: http://ideone.com/fIaXo

8
128

It seems that in the statement numbered 1, 32 is shifted 2 times right while in the second, 32 is being shifted 2 times left.
Is it really the case that when a value is shifted left by -p, the compiler treats it as shifting p times right?
Why am i getting the correct output here?

edited May 23 '17 at 11:48

Community

asked Jul 19 '12 at 17:21

Green goblin

9,746
13
68
100

Is there a difference when using `unsigned` values? And FWIW, I get a warning saying `shift count is negative`. – Richard J. Ross III Jul 19 '12 at 17:23
Undefined behavior means that *anything* can happen. The compiler might give you a "correct" result, an error, a crash, or almost anything. What compiler are you on? – Linuxios Jul 19 '12 at 17:23
@Linuxios: Ideone says they use gcc-4.3.4 – Robert Harvey Jul 19 '12 at 17:23

Why am i Getting correct in bitwise shifting by a negative value?

0 Answers0