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'''Single-precision floating-point format ''' is a [[computer number format]] that occupies 4 bytes (32 bits) in computer memory and represents a wide dynamic range of values by using a [[floating point]].


In [[IEEE 754-2008]] the 32-bit base 2 format is officially referred to as '''binary32'''. It was called '''single''' in [[IEEE 754-1985]].
In older computers, other floating-point formats of 4 bytes were used.


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One of the first [[programming language]]s to provide single- and double-precision floating-point data types was [[Fortran]].
Before the widespread adoption of [[IEEE 754-1985]], the representation and properties of  the double float data type depended on the computer manufacturer and computer model.
 
Single-precision binary floating-point is used due to its wider range over [[Fixed-point arithmetic|fixed point]] (of the same bit-width), even if at the cost of precision.
 
Single precision is known as '''REAL''' in [[Fortran]],<ref>http://scc.ustc.edu.cn/zlsc/sugon/intel/compiler_f/main_for/lref_for/source_files/rfreals.htm</ref> as '''float''' in [[C (programming language)|C]], [[C++ (programming language)|C++]], [[C Sharp (programming language)|C#]], [[Java (programming language)|Java]]<ref>http://java.sun.com/docs/books/tutorial/java/nutsandbolts/datatypes.html</ref> and [[Haskell (programming language)|Haskell]], and as '''single''' in [[Object Pascal|Delphi (Pascal)]], Visual Basic, and [[MATLAB]]. However, '''float''' in [[Python (programming language)|Python]], [[Ruby]], [[PHP]], and [[OCaml]] and '''single''' in versions of [[GNU Octave|Octave]] prior to 3.2 refer to [[Double-precision floating-point format|double-precision]] numbers. In [[PostScript]] the only floating-point precision is single.
 
{{Floating-point}}
 
== IEEE 754 single-precision binary floating-point format: binary32 ==
 
The IEEE 754 standard specifies a '''binary32''' as having:
* [[Sign bit]]: 1 bit
* [[Exponent]] width: 8 bits
* [[Significand]] [[precision (arithmetic)|precision]]: 24 bits (23 explicitly stored)
<!-- "significand", with a d at the end, is a technical term, please do not confuse with "significant" -->
 
This gives from 6 to 9 significant decimal digits precision (if a decimal string with at most 6 significant decimal is converted to IEEE 754 single precision and then converted back to the same number of significant decimal, then the final string should match the original; and if an IEEE 754 single precision is converted to a decimal string with at least 9 significant decimal and then converted back to single, then the final number must match the original <ref name=whyieee>{{cite web|url=http://www.cs.berkeley.edu/~wkahan/ieee754status/IEEE754.PDF|title=Lecture Notes on the Status of IEEE Standard 754 for Binary Floating-Point Arithmetic| author=William Kahan |date=1 October 1987}}</ref>).
 
Sign bit determines the sign of the number, which is the sign of the significand as well. Exponent is either an 8 bit signed integer from −128 to 127 (2's Complement) or an 8 bit unsigned integer from 0 to 255 which is the accepted biased form in IEEE 754 binary32 definition. For this case an exponent value of 127 represents the actual zero.
The true significand includes 23 fraction bits to the right of the binary point and an implicit leading bit (to the left of the binary point) with value 1 unless the exponent is stored with all zeros. Thus only 23 fraction bits of the [[significand]] appear in the memory format but the total precision is 24 bits (equivalent to log<sub>10</sub>(2<sup>24</sup>) ≈  7.225 decimal digits). The bits are laid out as follows:
 
[[Image:Float example.svg]]
 
The real value assumed by a given 32 bit '''binary32''' data with a given biased exponent '''e'''
and a '''23 bit fraction''' is <math> = (-1)^\text{sign}(1.b_{22}b_{21}...b_{0})_2 \times 2^{e-127} </math>
where more precisely we have <math> \text{value} = (-1)^\text{sign}\left(1 + \sum_{i=1}^{23} b_{23-i} 2^{-i} \right)\times 2^{(e-127)}</math>.
 
In this example:
* <math>\text{sign} = 0 </math>
* <math>1 + \sum_{i=1}^{23} b_{23-i} 2^{-i} = 1 + 2^{-2} = 1.25 </math>
* <math>2^{(e-127)} = 2^{124-127} = 2^{-3} </math>
 
thus:
* <math> \text{value} = 1.25 \times 2^{-3} = 0.15625 </math>
 
=== Exponent encoding ===
 
The single-precision binary floating-point exponent is encoded using an [[offset-binary]] representation, with the zero offset being 127; also known as exponent bias in the IEEE 754 standard.
 
* E<sub>min</sub> = 01<sub>H</sub>−7F<sub>H</sub> = −126
* E<sub>max</sub> = FE<sub>H</sub>−7F<sub>H</sub> = 127
* [[Exponent bias]] = 7F<sub>H</sub> = 127
 
Thus, in order to get the true exponent as defined by the offset binary representation, the offset of 127 has to be subtracted from the stored exponent.
 
The stored exponents 00<sub>H</sub> and FF<sub>H</sub> are interpreted specially.
 
{|class="wikitable" style="text-align:center"
! Exponent !! Significand zero !! Significand non-zero !! Equation
|-
| 00<sub>H</sub> || [[0 (number)|zero]], [[−0]] || [[denormal numbers]] || (−1)<sup>signbits</sup>×2<sup>−126</sup>× 0.significandbits
|-
| 01<sub>H</sub>, ..., FE<sub>H</sub> ||colspan=2| normalized value || (−1)<sup>signbits</sup>×2<sup>exponentbits−127</sup>× 1.significandbits
|-
| FF<sub>H</sub> || ±[[infinity]] || [[NaN]] (quiet, signalling)
|}
 
The minimum positive (denormal) value is
2<sup>−149</sup> ≈ 1.4 × 10<sup>−45</sup>.
The minimum positive normal value is 2<sup>−126</sup> ≈ 1.18 × 10<sup>−38</sup>.
The maximum representable value is (2−2<sup>−23</sup>) × 2<sup>127</sup> ≈ 3.4 × 10<sup>38</sup>.
 
=== Converting from decimal representation to binary32 format===
In general refer to the IEEE 754 standard itself for the strict conversion (including the rounding behaviour)
of a real number into its equivalent binary32 format.
Here we can show how to convert a base 10 real number into an IEEE 754 binary32 format
using the following outline:
 
* consider a real number with an integer and a fraction part such as 12.375
* convert and [[normalized number|normalize]] the integer part into [[binary numeral system|binary]]
* convert the fraction part using the following technique as shown here
* add the two results and adjust them to produce a proper final conversion
'''Conversion of the fractional part:'''
 
consider 0.375, the fractional part of 12.375. To convert it into a binary fraction, multiply the fraction by 2, take the integer part and re-multiply new fraction by 2 until a fraction of zero is found or until the precision limit is reached which is 23 fraction digits for IEEE 754 binary32 format.
 
0.375 x 2 = 0.750 = 0 + 0.750 => b<sub><small>−1</sub></small> = 0, the integer part represents the binary fraction digit. Re-multiply 0.750 by 2 to proceed
 
0.750 x 2 = 1.500 = 1 + 0.500 => b<sub><small>−2</sub></small> = 1
 
0.500 x 2 = 1.000 = 1 + 0.000 => b<sub><small>−3</sub></small> = 1, fraction = 0.000, terminate
 
We see that (0.375)<sub><small>10</sub></small> can be exactly represented in binary as (0.011)<sub><small>2</sub></small>. Not all decimal fractions can be represented in a finite digit binary fraction. For example decimal 0.1 cannot be represented in binary exactly. So it is only approximated.
 
Therefore (12.375)<sub>10</sub> = (12)<sub>10</sub> + (0.375)<sub>10</sub> = (1100)<sub>2</sub> + (0.011)<sub>2</sub> = (1100.011)<sub>2</sub>
 
Also IEEE 754 binary32 format requires that you represent real values in <math> (1.x_1x_2...x_{23})_2 \times 2^{e}</math> format, (see [[Normalized number]], [[Denormalized number]]) so that 1100.011 is shifted to the right by 3 digits to become <math> (1.100011)_2 \times 2^{3} </math>
 
Finally we can see that:  <math> (12.375)_{10} =(1.100011)_2 \times 2^{3} </math>
 
From which we deduce:
*  The exponent is 3 (and in the biased form it is therefore 130 = 1000 0010)
*  The fraction is 100011 (looking to the right of the binary point)
 
From these we can form the resulting 32 bit IEEE 754 binary32 format representation of
12.375 as: 0-10000010-10001100000000000000000 = 41460000<sub>H</sub>
 
'''Note:''' consider converting 68.123 into IEEE 754 binary32 format:
Using the above procedure you expect to get 42883EF9<sub>H</sub> with the last 4 bits being 1001
However due to the default rounding behaviour of IEEE 754 format what you get is 42883EFA<sub>H</sub> whose last 4 bits are 1010 .
 
'''Ex 1:'''
Consider decimal 1
We can see that:  <math> (1)_{10} =(1.0)_2 \times 2^{0} </math>
 
From which we deduce:
*  The exponent is 0 (and in the biased form it is therefore 127 = 0111 1111 )
*  The fraction is 0 (looking to the right of the binary point in 1.0 is all 0 = 000...0)
 
From these we can form the resulting 32 bit IEEE 754 binary32 format representation of
real number 1 as: 0-01111111-00000000000000000000000 = 3f800000<sub>H</sub>
 
'''Ex 2:'''
Consider a value 0.25 .
We can see that: <math> (0.25)_{10} =(1.0)_2 \times 2^{-2} </math>
 
From which we deduce:
* The exponent is −2 (and in the biased form it is 127+(−2)= 125 = 0111 1101 )
* The fraction is  0 (looking to the right of binary point in 1.0 is all zeros)
 
From these we can form the resulting 32 bit IEEE 754 binary32 format representation of  
real number 0.25 as: 0-01111101-00000000000000000000000 = 3e800000<sub>H</sub>
 
'''Ex 3:'''
Consider a value of 0.375 We saw that <math> 0.375 = {(1.1)_2}\times 2^{-2} </math>
 
Hence after determining a representation of 0.375 as <math>{(1.1)_2}\times 2^{-2} </math>
we can proceed as above:
 
* The exponent is −2 (and in the biased form it is 127+(−2)= 125 = 0111 1101 )
* The fraction is  1 (looking to the right of binary point in 1.1 is a single 1 = x<sub>1</sub>)
 
From these we can form the resulting 32 bit IEEE 754 binary32 format representation of
real number 0.375 as: 0-01111101-10000000000000000000000 = 3ec00000<sub>H</sub>
 
=== Single-precision examples ===
 
These examples are given in bit ''representation'', in [[hexadecimal]],
of the floating-point value. This includes the sign, (biased) exponent, and significand.
 
3f80 0000  = 1
c000 0000  = −2
7f7f ffff  ≈ 3.4028234 × 10<sup>38</sup> (max single precision)
0000 0000  = 0
8000 0000  = −0
7f80 0000  = infinity
ff80 0000  = −infinity
3eaa aaab  ≈ 1/3
 
By default, 1/3 rounds up, instead of down like [[double precision]], because of the even number of bits in the significand.
So the bits beyond the rounding point are <code>1010...</code> which is more than 1/2 of a [[unit in the last place]].
 
=== Converting from single-precision binary to decimal ===
 
We start with the hexadecimal representation of the value, 41c80000, in this example, and convert it to binary
 
41c8 0000<sub>16</sub> = 0100 0001 1100 1000 0000 0000 0000 0000<sub>2</sub>
 
then we break it down into three parts; sign bit, exponent and significand.
 
Sign bit: 0
Exponent: 1000 0011<sub>2</sub> = 83<sub>16</sub> = 131
Significand: 100 1000 0000 0000 0000 0000<sub>2</sub> = 480000<sub>16</sub>
 
We then add the implicit 24th bit to the significand
 
Significand: '''1'''100 1000 0000 0000 0000 0000<sub>2</sub> = C80000<sub>16</sub>
 
and decode the exponent value by subtracting 127
 
Raw exponent: 83<sub>16</sub> = 131
Decoded exponent: 131 − 127 = '''4'''
 
Each of the 24 bits of the significand (including the implicit 24th bit), bit 23 to bit 0, represents a value, starting at 1 and halves for each bit, as follows
 
bit 23 = 1
bit 22 = 0.5
bit 21 = 0.25
bit 20 = 0.125
bit 19 = 0.0625
.
.
bit  0 = 0.00000011920928955078125
 
The significand in this example has three bits set, bit 23, bit 22 and bit 19. We can now decode the significand
by adding the values represented by these bits.
 
Decoded significand: 1 + 0.5 + 0.0625 = 1.5625 = C80000/2<sup>23</sup>
 
Then we need to multiply with the base, 2, to the power of the exponent to get the final result
 
1.5625 × 2<sup>4</sup> = '''25'''
 
Thus
 
41c8 0000  = 25
 
This is equivalent to:
<math>n = (-1)^s \times
          (1+m*2^{-23})\times
          2^{x - 127}</math>
where <math>s</math> is the sign bit, <math>x</math> is the exponent, and <math>m</math> is the significand.
 
=== Trivia ===
 
A fascinating example of how the floating-point format can be misused in a good way is shown in the [[Fast inverse square root]] implementation, where the complex calculation of square root and inversion are replaced (approximately) by a bit-shift and subtraction operated on the 32-bits of the floating point encoding as if it were an integer.
 
== See also ==
* [[IEEE 754-2008|IEEE Standard for Floating-Point Arithmetic (IEEE 754)]]
* [[ISO/IEC 10967]], language independent arithmetic
* [[Primitive data type]]
* [[Numerical stability]]
* [[Double-precision floating-point format]]
 
==External links==
* [http://www.h-schmidt.net/FloatApplet/IEEE754.html Online calculator]
* [http://www.binaryconvert.com/convert_float.html Online converter for IEEE 754 numbers with single precision]
* [http://www.mathworks.com/matlabcentral/fileexchange/23173 C source code to convert between IEEE double, single, and half precision]
 
==References==
<references/>
 
[[Category:Binary arithmetic]]
[[Category:Computer arithmetic]]
[[Category:Data types]]

Revision as of 22:01, 23 May 2013

Single-precision floating-point format is a computer number format that occupies 4 bytes (32 bits) in computer memory and represents a wide dynamic range of values by using a floating point.

In IEEE 754-2008 the 32-bit base 2 format is officially referred to as binary32. It was called single in IEEE 754-1985. In older computers, other floating-point formats of 4 bytes were used.

One of the first programming languages to provide single- and double-precision floating-point data types was Fortran. Before the widespread adoption of IEEE 754-1985, the representation and properties of the double float data type depended on the computer manufacturer and computer model.

Single-precision binary floating-point is used due to its wider range over fixed point (of the same bit-width), even if at the cost of precision.

Single precision is known as REAL in Fortran,[1] as float in C, C++, C#, Java[2] and Haskell, and as single in Delphi (Pascal), Visual Basic, and MATLAB. However, float in Python, Ruby, PHP, and OCaml and single in versions of Octave prior to 3.2 refer to double-precision numbers. In PostScript the only floating-point precision is single.

Template:Floating-point

IEEE 754 single-precision binary floating-point format: binary32

The IEEE 754 standard specifies a binary32 as having:

This gives from 6 to 9 significant decimal digits precision (if a decimal string with at most 6 significant decimal is converted to IEEE 754 single precision and then converted back to the same number of significant decimal, then the final string should match the original; and if an IEEE 754 single precision is converted to a decimal string with at least 9 significant decimal and then converted back to single, then the final number must match the original [3]).

Sign bit determines the sign of the number, which is the sign of the significand as well. Exponent is either an 8 bit signed integer from −128 to 127 (2's Complement) or an 8 bit unsigned integer from 0 to 255 which is the accepted biased form in IEEE 754 binary32 definition. For this case an exponent value of 127 represents the actual zero.

The true significand includes 23 fraction bits to the right of the binary point and an implicit leading bit (to the left of the binary point) with value 1 unless the exponent is stored with all zeros. Thus only 23 fraction bits of the significand appear in the memory format but the total precision is 24 bits (equivalent to log10(224) ≈ 7.225 decimal digits). The bits are laid out as follows:

File:Float example.svg

The real value assumed by a given 32 bit binary32 data with a given biased exponent e and a 23 bit fraction is =(1)sign(1.b22b21...b0)2×2e127 where more precisely we have value=(1)sign(1+i=123b23i2i)×2(e127).

In this example:

thus:

Exponent encoding

The single-precision binary floating-point exponent is encoded using an offset-binary representation, with the zero offset being 127; also known as exponent bias in the IEEE 754 standard.

  • Emin = 01H−7FH = −126
  • Emax = FEH−7FH = 127
  • Exponent bias = 7FH = 127

Thus, in order to get the true exponent as defined by the offset binary representation, the offset of 127 has to be subtracted from the stored exponent.

The stored exponents 00H and FFH are interpreted specially.

Exponent Significand zero Significand non-zero Equation
00H zero, −0 denormal numbers (−1)signbits×2−126× 0.significandbits
01H, ..., FEH normalized value (−1)signbits×2exponentbits−127× 1.significandbits
FFH ±infinity NaN (quiet, signalling)

The minimum positive (denormal) value is 2−149 ≈ 1.4 × 10−45. The minimum positive normal value is 2−126 ≈ 1.18 × 10−38. The maximum representable value is (2−2−23) × 2127 ≈ 3.4 × 1038.

Converting from decimal representation to binary32 format

In general refer to the IEEE 754 standard itself for the strict conversion (including the rounding behaviour) of a real number into its equivalent binary32 format.

Here we can show how to convert a base 10 real number into an IEEE 754 binary32 format using the following outline:

  • consider a real number with an integer and a fraction part such as 12.375
  • convert and normalize the integer part into binary
  • convert the fraction part using the following technique as shown here
  • add the two results and adjust them to produce a proper final conversion

Conversion of the fractional part:

consider 0.375, the fractional part of 12.375. To convert it into a binary fraction, multiply the fraction by 2, take the integer part and re-multiply new fraction by 2 until a fraction of zero is found or until the precision limit is reached which is 23 fraction digits for IEEE 754 binary32 format.

0.375 x 2 = 0.750 = 0 + 0.750 => b−1 = 0, the integer part represents the binary fraction digit. Re-multiply 0.750 by 2 to proceed

0.750 x 2 = 1.500 = 1 + 0.500 => b−2 = 1

0.500 x 2 = 1.000 = 1 + 0.000 => b−3 = 1, fraction = 0.000, terminate

We see that (0.375)10 can be exactly represented in binary as (0.011)2. Not all decimal fractions can be represented in a finite digit binary fraction. For example decimal 0.1 cannot be represented in binary exactly. So it is only approximated.

Therefore (12.375)10 = (12)10 + (0.375)10 = (1100)2 + (0.011)2 = (1100.011)2

Also IEEE 754 binary32 format requires that you represent real values in (1.x1x2...x23)2×2e format, (see Normalized number, Denormalized number) so that 1100.011 is shifted to the right by 3 digits to become (1.100011)2×23

Finally we can see that: (12.375)10=(1.100011)2×23

From which we deduce:

  • The exponent is 3 (and in the biased form it is therefore 130 = 1000 0010)
  • The fraction is 100011 (looking to the right of the binary point)

From these we can form the resulting 32 bit IEEE 754 binary32 format representation of 12.375 as: 0-10000010-10001100000000000000000 = 41460000H

Note: consider converting 68.123 into IEEE 754 binary32 format: Using the above procedure you expect to get 42883EF9H with the last 4 bits being 1001 However due to the default rounding behaviour of IEEE 754 format what you get is 42883EFAH whose last 4 bits are 1010 .

Ex 1: Consider decimal 1 We can see that: (1)10=(1.0)2×20

From which we deduce:

  • The exponent is 0 (and in the biased form it is therefore 127 = 0111 1111 )
  • The fraction is 0 (looking to the right of the binary point in 1.0 is all 0 = 000...0)

From these we can form the resulting 32 bit IEEE 754 binary32 format representation of real number 1 as: 0-01111111-00000000000000000000000 = 3f800000H

Ex 2: Consider a value 0.25 . We can see that: (0.25)10=(1.0)2×22

From which we deduce:

  • The exponent is −2 (and in the biased form it is 127+(−2)= 125 = 0111 1101 )
  • The fraction is 0 (looking to the right of binary point in 1.0 is all zeros)

From these we can form the resulting 32 bit IEEE 754 binary32 format representation of real number 0.25 as: 0-01111101-00000000000000000000000 = 3e800000H

Ex 3: Consider a value of 0.375 . We saw that 0.375=(1.1)2×22

Hence after determining a representation of 0.375 as (1.1)2×22 we can proceed as above:

  • The exponent is −2 (and in the biased form it is 127+(−2)= 125 = 0111 1101 )
  • The fraction is 1 (looking to the right of binary point in 1.1 is a single 1 = x1)

From these we can form the resulting 32 bit IEEE 754 binary32 format representation of real number 0.375 as: 0-01111101-10000000000000000000000 = 3ec00000H

Single-precision examples

These examples are given in bit representation, in hexadecimal, of the floating-point value. This includes the sign, (biased) exponent, and significand.

3f80 0000   = 1
c000 0000   = −2

7f7f ffff   ≈ 3.4028234 × 1038  (max single precision)

0000 0000   = 0
8000 0000   = −0

7f80 0000   = infinity
ff80 0000   = −infinity
				
3eaa aaab   ≈ 1/3

By default, 1/3 rounds up, instead of down like double precision, because of the even number of bits in the significand. So the bits beyond the rounding point are 1010... which is more than 1/2 of a unit in the last place.

Converting from single-precision binary to decimal

We start with the hexadecimal representation of the value, 41c80000, in this example, and convert it to binary

41c8 000016 = 0100 0001 1100 1000 0000 0000 0000 00002

then we break it down into three parts; sign bit, exponent and significand.

Sign bit: 0
Exponent: 1000 00112 = 8316 = 131
Significand: 100 1000 0000 0000 0000 00002 = 48000016

We then add the implicit 24th bit to the significand

Significand: 1100 1000 0000 0000 0000 00002 = C8000016

and decode the exponent value by subtracting 127

Raw exponent: 8316 = 131
Decoded exponent: 131 − 127 = 4

Each of the 24 bits of the significand (including the implicit 24th bit), bit 23 to bit 0, represents a value, starting at 1 and halves for each bit, as follows

bit 23 = 1
bit 22 = 0.5
bit 21 = 0.25
bit 20 = 0.125
bit 19 = 0.0625
.
.
bit  0 = 0.00000011920928955078125

The significand in this example has three bits set, bit 23, bit 22 and bit 19. We can now decode the significand by adding the values represented by these bits.

Decoded significand: 1 + 0.5 + 0.0625 = 1.5625 = C80000/223

Then we need to multiply with the base, 2, to the power of the exponent to get the final result

1.5625 × 24 = 25

Thus

41c8 0000   = 25

This is equivalent to:

n=(1)s×(1+m*223)×2x127

where s is the sign bit, x is the exponent, and m is the significand.

Trivia

A fascinating example of how the floating-point format can be misused in a good way is shown in the Fast inverse square root implementation, where the complex calculation of square root and inversion are replaced (approximately) by a bit-shift and subtraction operated on the 32-bits of the floating point encoding as if it were an integer.

See also

External links

References