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Published byJaylon Hollander Modified about 1 year ago

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COMP2130 Winter 2015 Storing signed numbers in memory

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Single Precision Floats IEEE Bit

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Floating Point (32 bit) Mantissa (0-22) Exponent (23-30) Signed bit (31) 0 = positive 1 = negative

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Floating Point (32 bit) 0 Convert Positive so sign bit 31 = 0

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Floating Point (32 bit) Convert Positive so bit 31 = 0 Need to convert to 1.xxxxxx number 0

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Floating Point (32 bit) Convert to 1.xxxxxx number / 2 = / 2 = / 2 = / 2 = / 2 = / 2 = / 2 = >= 1.0 use Log 2 (173.7), 173.7/2 x 0

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Floating Point (32 bit) = x2 7 Time to fill in the exponent Has a bias of = in unsigned 8 bit binary is

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Floating Point (32 bit) = x2 7 Time to fill in the exponent Has a bias of = in unsigned 8 bit binary is

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Floating Point (32 bit) x2 7 Now we will need to deal with the orange part of the number which will be placed into the mantissa

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) Numberx2Whole

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Floating Point (32 bit) =

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Floating Point (32 bit) What if was ?

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Floating Point (32 bit) What if was ? =

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Floating Point (32 bit) What if was ? = =

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