2.87 Floating-Point II
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Problem:
Consider a 16-bit floating-point representation based on the IEEE floating-point format, with one sign bit, seven exponenet bits (k=7), and eight fraction bits (n=8). The exponent bias is 2^6-1=63.
Fill in the table that follows for each of the numbers given, with the following instructions for each column:
Hex: The four hexadecimal digits describing the encoded form.
M: The value of the significand. This should be a number of the form x or x/y, where x is an integer, and y is an integral power of 2. Examples include: 0, 67/64, and 1/256.
E: The integer value of the exponent.
V: The numeric value represented. Use the notation x or x*2^z, where x and z are integers.
As an example, to represent the number 7/8, we would have s=0, M=7/4, and E=-1. Our number would therefore have an exponent field of 0x3E (decimal value 63-1=62) and a signifcand field 0xC0 (binary 11000000), giving a hex representation 3EC0.
You need not fill in entries marked "--".
Desc | Hex | M | E | V | D |
-0 | 0x8000 | 0 | -14 | -0 | -0.0 |
Smallest value >2 | 0x4001 | 1025/1024 | 1 | 1025/512 | 2.001953125 |
512 | 0x6000 | 1 | 9 | 512 | 512.0 |
Largest denormalized | 0x03FF | 1023/1024 | -14 | 1023/(1^24) | 0.000060975551605 |
−∞ | 0xFC00 | -- | -- | -- | -- |
Number with hex representation | -- | 123/64 | -1 | 123/128 | 0.9609375 |
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