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froggy
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July 31, 2011, 04:46:50 PM
 #21
Quote from: drawoc on July 13, 2011, 04:08:58 AM
<snip>
And simply check hash32[1] against 0x051EB333 instead of 0x19998000.
<snip>
Should this be hash32[7] rather than hash32[1]   (because of the little endian reversal)?
 
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July 31, 2011, 05:01:22 PM
 #22
Quote from: froggy on July 31, 2011, 04:46:50 PM
Quote from: drawoc on July 13, 2011, 04:08:58 AM
<snip>
And simply check hash32[1] against 0x051EB333 instead of 0x19998000.
<snip>
Should this be hash32[7] rather than hash32[1]   (because of the little endian reversal)?

Thanks for catching that. It should be hash32[6] actually because we should be checking the second to last 32 bit part, and the last is 7 because we start counting at zero. I'll edit my post.
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