3 Apr
2011
3 Apr
'11
9:10 p.m.
Thank you very much Sven, your feedback was very helpful. i ended up
with the code below which i am still testing.
int weighted_bit_packet_count(TYPE_OF_WORD *seq_bits)
{
int i,check, count = 0;
TYPE_OF_WORD word;
for (i = 0; i < NUM_WORDS; i++) {
word = seq_bits[i];
int j = WORD_BIT_SIZE, k = 1;
while (j > 0 && k <= 32){
check = (word & (1 << j)) >> j;
if (check == 1)
count += k;
j--;
k++;
}
}
return count;
}
Kind Regards
Hlabishi
On 3/22/11, Sven Eckelmann <sven(a)narfation.org> wrote:
On Monday 21 March 2011 23:40:57 hlabishi kobo wrote:
[...]
a single "word"/unsigned long. seq_bits consist of NUM_WORDS words (words
are
defined in context of seq_bits as unsigned long - that's why we removed
TYPE_OF_WORD completely and just use unsigned long).
Regards,
Sven
--
Kind Regards
Hlabishi I. Kobo, Mr
B S.c Computer Science (Msc)
University of the Western Cape
Bellville, Cape Town, South Africa
+27 72 840 7719
+27 21 959 2461
I actually printed the seq_bits[i] and that is
basically why i used 32
but that was just for debugging purpose. But actually if i use
NUM_WORDS, my weighted sum become very small whereas if i use 32 it
becomes very big, trying to understand what might be causing it to be
that way.
NUM_WORDS == amount of unsigned long/words
WORD_BIT_SIZE == amount of bits in a unsigned long
TQ_LOCAL_WINDOW_SIZE == sum of bits in seq_bits
Nothing really fancy here.
you only
calculate the weighted sum of a single "word" (which you static
casted to int....?) and not all words.
i actually dont understand this very well, how many words does seq_bits[i]
have?