Difference between 2^02 and (2^0)2?

Question

I have expected both expression's will give same answer:

System.out.println(2^0*2);
System.out.println((2^0)*2);

Output:

2
4

Is there a specific reason why 2^0*2 = 2 and (2^0)*2 = 4?

Please restrict yourself to one language only. A question that asks about two language is too broad. — Stephen C, Oct 01 '16 at 08:18
`*` has higher precedence than `^`. Most likely what you intended was `Math.pow(2, 0) * 2` — Peter Lawrey, Oct 01 '16 at 10:09

Nikolas Charalambidis · Accepted Answer · 2016-10-01T08:36:55.640

You have wrongly assumed that ^ operator behaves the same like the exponentiation in math.

At the first sight you can see that ^ is understood as + operator. Actually it means bitwise XOR operator.

System.out.println(2^0*2);   //  2 XOR 0  * 2 = 2
System.out.println((2^0)*2); // (2 XOR 0) * 2 = 4
System.out.println(2^4);     //  2 XOR 4      = 6

The XOR is exclusive disjunction that outputs true only when inputs differ. Here is the whole trick:

2^0 = 2 XOR 0 = (0010) XOR (0000) = (0010) = 2 
2^4 = 2 XOR 4 = (0010) XOR (0100) = (0110) = 6

score 2 · Answer 2 · answered Oct 01 '16 at 08:17

2

check this link

2^0*2=2

has higher priority thatn ^ so first you will evaluate 0*2 which is 0 and then xor it with 2 which will resutl 2

(2^0)*2

() has higher priority so you will first evaluate 2^0 then which is 2 then multiply it with 2

answered Oct 01 '16 at 08:17

Amer Qarabsa

and where i exactly mentioned that is power operator, i mentioned that its xor – Amer Qarabsa Oct 01 '16 at 08:31
NO check the link it specifies different meanings of '^' – Amer Qarabsa Oct 01 '16 at 08:32

score -3 · Answer 3 · answered Oct 01 '16 at 08:18

-3

In your first example you calculate: 0*2 = 2 ^ 0 = 2

In your second example you calculate: 2 ^ 0 = 2 * 2 = 4

answered Oct 01 '16 at 08:18

Tim L

Difference between 2^0*2 and (2^0)*2?