E7_Su2010_Solutions3_9d+11

# E7_Su2010_Solutions3_9d+11 - E7 Laboratory Assignment 3...

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Unformatted text preview: E7 Laboratory Assignment 3 Partial Solutions Problem 9. (d) Single precision oating point numbers are represented by 4 bytes or 32 bits, which are divided into the sign, exponent, and mantissa as follows: u1D460 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright sign u1D452 1 u1D452 2 u1D452 3 . . .u1D452 8 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright exponent u1D45A 1 u1D45A 2 u1D45A 3 . . . u1D45A 23 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright mantissa . To simplify the notation, let u1D438 = ( u1D452 1 u1D452 2 u1D452 3 . . . u1D452 8 ) 2 and u1D440 = (0 .u1D45A 1 u1D45A 2 u1D45A 3 . . . u1D45A 23 ) 2 . Since u1D438 ranges from 0 to 255, the value of a single precision oating point number u1D453 is determined by the following rules: If 1 u1D438 254, then u1D453 is a normalized number and u1D453 = ( 1) u1D460 2 u1D438 127 (1 + u1D440 ) If u1D438 = 0, then u1D453 is a denormal number and u1D453 = ( 1) 2 2 126 u1D440. If u1D438 = 255 and u1D440 = 0, then u1D453 = Inf where the sign is determined by ( 1) u1D460 ....
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## This note was uploaded on 08/14/2010 for the course E 7 taught by Professor Patzek during the Summer '08 term at Berkeley.

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E7_Su2010_Solutions3_9d+11 - E7 Laboratory Assignment 3...

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