1 - Sign bit 7 - Exponent in Excess-64 form 8 - Mantissa
$(0.148)_{10} = (0.00100101;111…)_2$ We shift it 3 bits to left to make it normalized $(1.00101;111)_2 * 2^{11}$. Exponent = $11+64 = (75)_{10} = (1001011)_2$ and Mantissa = $(01001;111)_2$. So floating point representation is $(0;1001011;00101111)_2 = (4B2F)_{16}$ Representation A But if we store the denormalized mantissa into 8 bit register, then it won’t have stored the last three $1$s and then the mantissa would have normalized from $(0.00100101)_2$ to $(1.00101;000)_2$ by inserting 3 $0$s instead of $1$s. The representation would have been $(0;1001011;00101000)_2 = (4B28)_{16}$ Representation B So while normalizing, does the processor takes into account the denormalized mantissa bits beyond 8 bits too? Or just rounds it off? Which one is correct: A or B? Does it store the mantissa in fixed point representation? How does it all work?
Asked By : Shashwat
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Question Source : http://cs.stackexchange.com/questions/7828
