the nth gray code

Question

the formula for calculating nth gray code is :

(n-1) XOR (floor((n-1)/2))  
(Source: wikipedia)

I encoded it as:

int gray(         


        

Answer 1

Incrementing a number, when you look at it bitwise, flips all trailing ones to zeros and the last zero to one. That's a whole lot of bits flipped, and the purpose of Gray code is to make it exactly one. This transformation makes both numbers (before and after increment) equal on all the bits being flipped, except the highest one.

Before:

011...11
     + 1 
---------
100...00

After:

010...00
     + 1
---------
110...00
^<--------This is the only bit that differs 
          (might be flipped in both numbers by carry over from higher position)

n ^ (n >> 1) is easier to compute but it seems that only changing the trailing 011..1 to 010..0 (i.e. zeroing the whole trailing block of 1's except the highest 1) and 10..0 to 11..0 (i.e flipping the highest 0 in the trailing 0's) is enough to obtain a Gray code.