Unformatted text preview: to 3 2;23. We get
( 1:1000000000000000000000 )2 21
+ ( 0:0000000000000000000001j1 )2 21
= ( 1:1000000000000000000001j1 )2 21 :
This time, the result is not an IEEE single precision oating point number,
since its signi cand has 24 bits after the binary point: the 24th is shown
beyond the vertical bar. Therefore, the result must be correctly rounded.
Rounding down gives the result (1:10000000000000000000001)2 21, while
rounding up gives (1:10000000000000000000010)2 21. In the case of rounding to nearest, there is a tie, so the latter result, with the even nal bit, is
For another example, consider the numbers 1, 2;15 and 215, which are
all oating point numbers. The result of adding the rst to the second is
(1:000000000000001)2, which is a oating point number the result of adding
the rst to the third is (1:000000000000001)2 215, which is also a oating
point number. However, the result of adding the second number to the third
which is not a oating point number in the IEEE single precision format,
since the fraction eld would need 30 bits to represent this number exactly.
In this example, usin...
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This note was uploaded on 02/12/2014 for the course MATH 4800 taught by Professor Lie during the Spring '09 term at Rensselaer Polytechnic Institute.
- Spring '09