Why is the exponent operator not giving the expected result?

Question

Here is the code

#include <stdio.h>

int main()
{
 int a;
 printf("%d\n",(3^6));
 printf("%d",(a^a));
 return 0;
}

I am getting the below output on executing the above code.

5
0

Shouldn't the output be 729? 3 to the power of 6 is 729. I could not understand the logic of why it is returning 5?

score 7 · Accepted Answer · edited May 23 '17 at 10:26

7

The ^ operator is bitwise xor. To understand why 3^6 is 5 let's look at binary representations of those values:

3 = 0011
6 = 0110

and so

3^6 = 0101 = 5

You did not initialise a, which means that a^a is undefined behaviour.

For exponentiation you need to use pow().

edited May 23 '17 at 10:26

Community

answered Jul 31 '14 at 10:58

David Heffernan

@David Heffernan: no UB? – Jul 31 '14 at 12:52
@dmcr_code you mean `a^a` with no init of `a`. Not sure. Interesting. – David Heffernan Jul 31 '14 at 18:38
@DavidHeffernan: as far as I know using uninitialized variables in C is UB anyway – Jul 31 '14 at 20:23
@dmcr http://stackoverflow.com/questions/25074180/is-aa-or-a-a-undefined-behaviour-if-a-is-not-initialized – David Heffernan Aug 01 '14 at 06:46
@DavidHeffernan: nice to see, I triggered that question:) – Aug 01 '14 at 07:12

score 4 · Answer 2 · answered Jul 31 '14 at 11:02

4

The ^ operator in C isn't for exponents : it is the bitwise xor.

00000011 (3)
xor
00000110 (6)
=
00000101 (5)

Use pow() for exponents.

And you didn't initialize a, so be careful with that.

answered Jul 31 '14 at 11:02

maxdefolsch

score 1 · Answer 3 · answered Jul 31 '14 at 10:59

1

The operator ^ in C is not power - C has no builtin operator for that, only pow function:

double x = pow(3,5);

The operator ^ is bitwise XOR.

answered Jul 31 '14 at 10:59

Wojtek Surowka

score 0 · Answer 4 · answered Jul 31 '14 at 11:01

0

is a operator used for XORing. include #include<math.h> header file and use pow(3,6) to evaluate power and u'll get 729

answered Jul 31 '14 at 11:01

jaimin

4 Answers4