On Thu, Feb 16, 2006 at 04:44:02PM +0100, Michael Niedermayer wrote:
Hi
On Thu, Feb 16, 2006 at 01:47:32AM +0100, Alexander Strasser wrote: [...]
crc32 checksum The checksum is choosen so that c(x) is a multiple of x^32+x^26+x^23+x^22+x^16+x^12+x^11+x^10+x^8+x^7+x^5+x^4+x^2+x+1 over GF(2) c(x) is the polynom coresponding to the block from and including the forward ptr and upto and including the checksum c(x)= (1&(d[0]>>8))*x^(n-1) + (1&(d[0]>>7))*x^(n- 2) + ... + (1&(d[1]>>8))*x^(n-9) + (1&(d[1]>>7))*x^(n-10) + ... + ... + ... + (1&(d[n/8-1]>>1))*x^1 + (1&(d[n/8-1]>>0))*x^0 alternatively you can simply run crc=0; for(i=0; i<size; i++){ crc ^= buf[i]<<24; for(j=0; j<8; j++) crc= (crc<<1) ^ (0x04C11DB7 & (crc>>31)); } over the data
ok, while we are at it, should we use non zero initial value? if yes which?
personally i would simply include the forward ptr, so the all zero case would naturally be gone, if the others are against this then we must either use a
Is there an all-zero case? As far as I can tell, there's no NUT packet that's valid as all zeros, except possibly the very first syncpoint/header pair in the file. Rich