Question

In: Computer Science

The biggest mysteries of the IEEE 754 Floating-Point Representation are “hidden bit” and “Bias. Can someone...

The biggest mysteries of the IEEE 754 Floating-Point Representation are “hidden bit” and “Bias.

Can someone explain to me why the "hidden bits" and "bias" are considered to be mysteries for the IEEE 754 floating point representation

Solutions

Expert Solution

FLOATING POINT REPRESENTATIONS:
It is assumed that the student is familiar with the discussion in Appendix B of the text by A. Tanenbaum,
Structured Computer Organization, 4th Ed., Prentice-Hall, 1999. Here we are concerned in particular with
the discussion of floating-point numbers and normalization in pp. 643-651.In the notation used here
floating point representations for real numbers have the form:
+-M x R^+-E

Here M is a signed mantissa, R is the radix of normalization, and E is a signed exponent. It is the signed
values M and E that are typically packed into computer \words" of varying length to encode single, double,
and extended precision (short, long, and extended) floating point numbers. The number of bits used in the
representation of M determines the precision of the representation. The number of bits used to represent
the signed value E determines the range of the representation. Only a finite number of values among those
in the representable range can be encoded in the n bits available, and this number is determined by the
precision.


Positive and negative floating point numbers are usually represented in one of two ways depending on the
computer's choice of arithmetic circuitry and representation for fixed point numbers. Machines that use one's
complement representations for fixed point numbers and have one's complement adders in their arithmetic units typically pack the positive floating point number into one (or more) word(s) and then complement all bits in this (these) word(s) to represent the negative floating point number.
Examples of these one's complement representations include the UNIVAC 1100 series machines and the CDC 6600, CYBER 70 series
and successor machines.


IEEE Floating Point Standard 754:

The IEEE Floating Point Standard uses four (or eight) 8-bit bytes to represent floating point numbers in machines whose fixed-point data use a two's complement representation for negative integers and that have
two's complement arithmetic circuitry.
A single precision floating point datum is packed into four consecutive 8-bit bytes viewed as a single 32-bit representation.
A double precision floating point datum is packed into eight consecutive 8-bit bytes viewed as a single 64-bit representation, where the bits in the second 32-bits (or second set of four bytes) is continuation of the low order bits in the mantissa (called the significant).
The mantissa (or significand) is a binary normalized mixed number (with integer and fraction parts) of the form 1:xxxx2 .
A single precision datum has an 8-bit characteristic that represents a signed exponent with bias (excess) (128 1)10 = 12710, and a double precision datum has an 11-bit characteristic that represents the signed exponent excess (1024 1)10 = 102310. Negative floating-point numbers are represented in a sign-magnitude format by packing the positive representation into the single (or double) precision format and then setting the SM bit only to one. Of the 32-bits used to represent single precision numbers only 23 remain in which to represent the binary normalized mixed number. Because every number is assumed normalized in the form 1:xxxx2 2^+-E, there is no need to represent the leading one and so only the x's in the fraction are put into the 23 remaining bits in the word, resulting in 24 bits of precision.
We imagine that the leading one lies under and is hidden by an opaque characteristic field; it is therefore, called a "hiddin bit"
When unpacking a floating-point number for output conversion or for arithmetic manipulation, the hidden bit must be inserted at its rightful place in order to become visible again.
A 32-bit (or 64-bit) floating-point word comprising four (or eight) bytes with all zeros is the representation for floating-point +0 by definition.
In addition to defining formats for normalized floating point representations and zero, the standard also specifies valid formats for denormalized numbers, infinity, and quantities that are not a number
(e.g., see A.Tanenbaum, Structured Computer Organization, 3rd Ed., Prentice-Hall, 1990, pp.565{572).

Precision of A floating-point representation:
In the IEEE-754 single precision floating-point representation, the mantissa is 23 bits long. This means that any two numbers in this representation cannot be closer than

1/2^23 = 1.1920928955078125*10^-7.

In double precision, this difference reduces to

1/2^52 = 2.220446049250313080847263336181640625*10^-16.


Related Solutions

Convert the following number into 32bit IEEE 754 floating point representation. 0.000101
Convert the following number into 32bit IEEE 754 floating point representation. 0.000101
Consider the following 32-bit floating point representation based on the IEEE floating point standard: There is...
Consider the following 32-bit floating point representation based on the IEEE floating point standard: There is a sign bit in the most significant bit. The next eight bits are the exponent, and the exponent bias is 28-1-1 = 127. The last 23 bits are the fraction bits. The representation encodes number of the form V = (-1)S x M x 2E, where S is the sign, M is the significand, and E is the biased exponent. The rules for the...
verilog code to implement 32 bit Floating Point Adder in Verilog using IEEE 754 floating point...
verilog code to implement 32 bit Floating Point Adder in Verilog using IEEE 754 floating point representation.
Convert 1.67e14 to the 32-bit IEEE 754 Floating Point Standard, with the following layout: first bit...
Convert 1.67e14 to the 32-bit IEEE 754 Floating Point Standard, with the following layout: first bit is sign bit, next 8 bits is exponent field, and remaining 23 bits is mantissa field; result is to be in hexadecimal and not to be rounded up. answer choices 5717E27B 57172EB7 5717E2B7 C717E2B7 5771E2B7
For IEEE 754 single-precision floating point, what is the hexadecimal representation of 27.101562? A. 35CCD001 B....
For IEEE 754 single-precision floating point, what is the hexadecimal representation of 27.101562? A. 35CCD001 B. 2F5C10D0 C. 41D8D000 D. 7DCA1111 E. None of the above
Using IEEE 754 single precision floating point, write the hexadecimal representation for each of the following:...
Using IEEE 754 single precision floating point, write the hexadecimal representation for each of the following: a. Zero b. -2.0 (base 10) c. 256. 0078125 (base 10) d. Negative infinity
The number –11.375 (decimal) represented as a 32-bit floating-point binary number according to the IEEE 754...
The number –11.375 (decimal) represented as a 32-bit floating-point binary number according to the IEEE 754 standard is
Convert 0.875 to an IEEE 754 single-precision floating-point number. Show the sign bit, the exponent, and...
Convert 0.875 to an IEEE 754 single-precision floating-point number. Show the sign bit, the exponent, and the fraction. Convert -3.875 to an IEEE 754 double-precision floating-point number. Show the sign bit, the exponent, and the fraction Convert the IEEE 754 single-precision floating-point numbers 42E4800016 and 0080000016 to their corresponding decimal numbers.
Given the following 32-bit binary sequences representing single precision IEEE 754 floating point numbers: a =...
Given the following 32-bit binary sequences representing single precision IEEE 754 floating point numbers: a = 0100 0000 1101 1000 0000 0000 0000 0000 b = 1011 1110 1110 0000 0000 0000 0000 0000 Perform the following arithmetic and show the results in both normalized binary format and IEEE 754 single-precision format. Show your steps. a)     a + b b)     a × b
Represent the following decimal numbers using IEEE-754 floating point representation. A. -0.375 B. -Infinity C. 17...
Represent the following decimal numbers using IEEE-754 floating point representation. A. -0.375 B. -Infinity C. 17 D. 5.25
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT