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fp-example

# fp-example - Normalization 1. x 2 4 does not need to be...

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CSE 341 Floating Point Operations The Task Perform single precision floating point addition of 18.5625 + 0.25 Convert the Operands to Floating Point Notation 18.5625 10010.1001 2 Normalized 1.00101001 2 x 2 4 Exponent 127+4 = 131 = 10000011 Significand 00101001 Sign 0 Number 01000001100101001000000000000000 0.25 0.01 2 Normalized 1.0 2 x 2 -2 Exponent 127-2 = 125 = 01111101 Significand 0 Sign 0 Number 00111110100000000000000000000000 Floating Point Addition 1.00101001 2 x 2 4 + 1.0 2 x 2 -2 Step 1 Align decimal point of number with smaller exponent After doing so, the exponents are the same 1.0 x 2 -2 becomes 0.000001 x 2 4 Step 2 Add significands 100101001000000000000000 + 000000100000000000000000 = 100101101000000000000000

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Unformatted text preview: Normalization 1.00101101000000000000000 x 2 4 does not need to be modified Check for overflow/underflow Step 4 Round significand to correct number of digits It is possible that renormalization may be necessary if rounding increases the most significand digit from a 9 to a 10 1. 00101101000000000000000 x 2 4 does not require rounding Final Answer Exponent 127+4 = 131 = 10000011 Significand 00101101000000000000000 Sign Number 010000011 00101101000000000000000 References David A. Patterson and John L. Hennessy, Computer Organization and Design, The Hardware/Software Interface , 4 th Edition, Elsevier, 2009...
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