Binary ↘ Float: The 32 Bit Single Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 0000 0000 - 010 0000 1011 0011 1010 1011 Converted and Written as a Base Ten Decimal System Number (as a Float)

0 - 0000 0000 - 010 0000 1011 0011 1010 1011: 32 bit single precision IEEE 754 binary floating point standard representation number converted to decimal system (base ten)

1. Identify the elements that make up the binary representation of the number:

The first bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.
0

The next 8 bits contain the exponent:
0000 0000

The last 23 bits contain the mantissa:
010 0000 1011 0011 1010 1011


2. Reserved bitpattern.

We notice that all the bits that make up the exponent are on 0 (clear) and at least one bit of the mantissa is set on 1 (set).

This is one of the reserved bitpatterns of the special values of: Denormalized.

Denormalized numbers are too small to be correctly represented so they approximate to zero.

Depending on the sign bit, -0 and +0 are two distinct values though they both compare as equal (0).


3. Convert the exponent from binary (from base 2) to decimal (in base 10).

The exponent is allways a positive integer.

0000 0000(2) =

0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 0 × 20 =

0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =

0(10)

4. Adjust the exponent.

Subtract the excess bits: 2(8 - 1) - 1 = 127,

that is due to the 8 bit excess/bias notation.

The exponent, adjusted = 0 - 127 = -127


5. Convert the mantissa from binary (from base 2) to decimal (in base 10).

The mantissa represents the fractional part of the number (what comes after the whole part of the number, separated from it by a comma).


010 0000 1011 0011 1010 1011(2) =

0 × 2-1 + 1 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 1 × 2-8 + 0 × 2-9 + 1 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 2-14 + 1 × 2-15 + 1 × 2-16 + 0 × 2-17 + 1 × 2-18 + 0 × 2-19 + 1 × 2-20 + 0 × 2-21 + 1 × 2-22 + 1 × 2-23 =

0 + 0.25 + 0 + 0 + 0 + 0 + 0 + 0.003 906 25 + 0 + 0.000 976 562 5 + 0.000 488 281 25 + 0 + 0 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0 + 0.000 000 953 674 316 406 25 + 0 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =

0.25 + 0.003 906 25 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 000 953 674 316 406 25 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =

0.255 483 031 272 888 183 593 75(10)

6. Put all the numbers into expression to calculate the single precision floating point decimal value:

(-1)Sign × (1 + Mantissa) × 2(Adjusted exponent) =

(-1)0 × (1 + 0.255 483 031 272 888 183 593 75) × 2-127 =

1.255 483 031 272 888 183 593 75 × 2-127 =

0

0 - 0000 0000 - 010 0000 1011 0011 1010 1011 converted from a 32 bit single precision IEEE 754 binary floating point standard representation number to a decimal system number, written in base ten (float) = 0(10)

Spaces were used to group digits: for binary, by 4, for decimal, by 3.

More operations with 32 bit single precision IEEE 754 binary floating point standard representation numbers:

» Number 0 - 0000 0000 - 010 0000 1011 0011 1010 1010 converted from 32 bit single precision IEEE 754 binary floating point standard representation to decimal system written in base ten (float) = ?

» Number 0 - 0000 0000 - 010 0000 1011 0011 1010 1100 converted from 32 bit single precision IEEE 754 binary floating point standard representation to decimal system written in base ten (float) = ?

The latest 32 bit single precision IEEE 754 floating point binary standard numbers converted and written as decimal system numbers (in base ten, float)

The number 1 - 0100 0000 - 100 0011 1111 1111 1111 1011 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 1 - 1000 0101 - 000 0110 1100 0001 1100 1000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 1 - 0100 0101 - 110 1110 1000 0000 0011 1010 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 0 - 1000 0001 - 010 0110 0110 0110 0111 1110 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 0 - 1000 0011 - 110 1001 1111 1111 1111 1011 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 0 - 0100 1100 - 110 0110 0110 0110 0110 0011 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 1 - 1000 0101 - 000 0110 1100 0001 1100 1000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 09:00 UTC (GMT)
The number 1 - 0100 0101 - 110 1110 1000 0000 0011 1010 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 08:59 UTC (GMT)
The number 0 - 1000 0001 - 010 0110 0110 0110 0111 1110 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 08:59 UTC (GMT)
The number 0 - 0000 0000 - 010 0011 1101 0111 0001 1010 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 12 08:57 UTC (GMT)
All 32 bit single precision IEEE 754 binary floating point representation numbers converted to base ten decimal numbers (float)

How to convert numbers from 32 bit single precision IEEE 754 binary floating point standard to decimal system in base 10

Follow the steps below to convert a number from 32 bit single precision IEEE 754 binary floating point representation to base 10 decimal system:

  • 1. Identify the three elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 8 bits contain the exponent.
    The last 23 bits contain the mantissa.
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
  • 3. Adjust the exponent, subtract the excess bits, 2(8 - 1) - 1 = 127, that is due to the 8 bit excess/bias notation.
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
  • 5. Put all the numbers into expression to calculate the single precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted)

Example: convert the number 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 from 32 bit single precision IEEE 754 binary floating point system to base 10 decimal system (float):

  • 1. Identify the elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 8 bits contain the exponent: 1000 0001
    The last 23 bits contain the mantissa: 100 0001 0000 0010 0000 0000
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):
    1000 0001(2) =
    1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 =
    128 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =
    128 + 1 =
    129(10)
  • 3. Adjust the exponent, subtract the excess bits, 2(8 - 1) - 1 = 127, that is due to the 8 bit excess/bias notation:
    Exponent adjusted = 129 - 127 = 2
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10):
    100 0001 0000 0010 0000 0000(2) =
    1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 1 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 =
    0.5 + 0 + 0 + 0 + 0 + 0 + 0.007 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =
    0.5 + 0.007 812 5 + 0.000 061 035 156 25 =
    0.507 873 535 156 25(10)
  • 5. Put all the numbers into expression to calculate the single precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =
    (-1)1 × (1 + 0.507 873 535 156 25) × 22 =
    -1.507 873 535 156 25 × 22 =
    -6.031 494 140 625

  • 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point representation to decimal number (float) in decimal system (in base 10) = -6.031 494 140 625(10)