Binary ↘ Float: The 32 Bit Single Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 0000 0000 - 000 0110 0111 0010 0000 0101 Converted and Written as a Base Ten Decimal System Number (as a Float)

0 - 0000 0000 - 000 0110 0111 0010 0000 0101: 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:
000 0110 0111 0010 0000 0101


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).


000 0110 0111 0010 0000 0101(2) =

0 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 1 × 2-5 + 1 × 2-6 + 0 × 2-7 + 0 × 2-8 + 1 × 2-9 + 1 × 2-10 + 1 × 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 + 1 × 2-21 + 0 × 2-22 + 1 × 2-23 =


0 + 0 + 0 + 0 + 0.031 25 + 0.015 625 + 0 + 0 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 25 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0.000 000 119 209 289 550 781 25 =


0.031 25 + 0.015 625 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 061 035 156 25 + 0.000 000 476 837 158 203 125 + 0.000 000 119 209 289 550 781 25 =


0.050 354 599 952 697 753 906 25(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 + 0.050 354 599 952 697 753 906 25) × 2-127 =


1.050 354 599 952 697 753 906 25 × 2-127 =


0

0 - 0000 0000 - 000 0110 0111 0010 0000 0101 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.

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 0 - 0000 0000 - 000 0110 0111 0010 0000 0101 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:43 UTC (GMT)
The number 0 - 0110 1000 - 110 0100 0100 0000 0000 0001 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:43 UTC (GMT)
The number 0 - 1000 0110 - 101 0110 1101 1000 1111 1001 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:43 UTC (GMT)
The number 0 - 1001 1010 - 001 1010 1000 1001 1111 1111 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:42 UTC (GMT)
The number 0 - 1001 1000 - 000 0000 0011 0000 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:41 UTC (GMT)
The number 1 - 0100 0111 - 100 1011 1100 0000 0100 1100 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:41 UTC (GMT)
The number 0 - 1111 1010 - 010 1101 1111 1111 1110 1011 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Mar 29 04:41 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)