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 2000/12/21 [Uncategorized] UID:20140 Activity:nil 50%like:20889 12/20 Remove your own motd entries after they're posted over 3 days!
 2000/12/21 [Uncategorized] UID:20141 Activity:nil _ \__/ .\/ tweet tweet / /\ \____/ | | /\ /\
 2000/12/21 [Computer/SW/Languages] UID:20142 Activity:high 12/20 Can someone explain this: "Denormal numbers present a major stumbling block to implementing floating point multiplcation, because they require performing a variable shift in the multiplier, which wouldn't otherwise be needed. Thus, high- performance, floating point multipliers often do no handle denormalized numbers, but instead trap, letting software handle them." \_ Don't you always multiply the mantissas as if they are integers and then add the exponents, regardless of whether they are normalized or not? \_ normalizing a floating point number means shifting its mantissa to the left until the MSB of the mantissa is 1 (and compensating by changing the exponent). if you know a number is normalized, you don't need to store the 1, which means that assuming normalized gets you a free bit. however, being a certain number of shifts, when the exponent gets as small as it can get, you can't shift any more, and you might be forced to use a denormal float. this is only a problem with very small number. typically, hardware will throw and error when it can't normalize. good hardware doesn't and just says lets you know that something is a denorm. -ali. \_ That's easy. Just shift left by the leading zero count. Zero counting in hardware is pretty easy. For 32-bits that's just 5 levels of multiplexor and AND gates (AND gates are probaly has a higher delay). gates probably has a higher delay).