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C code 1ULL and bit calculation

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# 1  
Old 06-13-2013
C code 1ULL and bit calculation
I found this block of C code to create combinations of A, T, C & G characters (DNA bases). Can anybody explain this code for me, especially with 1ULL<< and [y & 3]?
Code:
16  for (x = 0; x < 1ULL << (2 * k); ++x)
17    {
18      for (i = 0, y = x; i < k; ++i, y >>= 2)
19      putchar ("ACGT"[y & 3]);
20    }

Thanks a lot!
# 2  
Old 06-13-2013
1ULL is 'unsigned long long'. This is probably intended to be an unsigned, 64-bit number. They make sure it's 64-bit to avoid accidentally overflowing the number of (1<<(2*k)) for high values of k.

"y & 3" masks off all but the three two lowest bits of y, making it effectively "y % 8 4", which would produce a modulous between 0 and 7 3.

Doy, thanks Alister.
Last edited by Corona688; 06-13-2013 at 03:53 PM..
# 3  
Old 06-13-2013
Quote:
Originally Posted by Corona688
"y & 3" masks off all but the three lowest bits of y
I'm sure its just a momentary lapse, but 3 (11 base 2) only passes the 2 least significant bits.

Regards,
Alister
This User Gave Thanks to alister For This Post:
Corona688
# 4  
Old 06-13-2013
Thanks Corona & Alister!

k is the arugment from stdin. Say k =7, can you walk me thru the two loops
Code:
1ULL << (2 * k) 
 y >>= 2

and the line
Code:
putchar("ACGT[y & 3]")?

I am trying to catch C at this moment, barely have ideas about bitwise operations.
Thanks again!
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