32bit IEEE 754: Single Precision Floating Point Binary -> Float: 1 - 1000 0100 - 111 0100 1101 1011 0001 1111 The 32 Bit Single Precision IEEE 754 Binary Floating Point Standard Representation Number Converted and Written as a Base Ten Decimal System Number (Float)

1 - 1000 0100 - 111 0100 1101 1011 0001 1111: 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.
1


The next 8 bits contain the exponent:
1000 0100


The last 23 bits contain the mantissa:
111 0100 1101 1011 0001 1111


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

The exponent is allways a positive integer.

1000 0100(2) =


1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 0 × 20 =


128 + 0 + 0 + 0 + 0 + 4 + 0 + 0 =


128 + 4 =


132(10)

3. 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 = 132 - 127 = 5


4. 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).


111 0100 1101 1011 0001 1111(2) =

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


0.5 + 0.25 + 0.125 + 0 + 0.031 25 + 0 + 0 + 0.003 906 25 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0 + 0 + 0 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =


0.5 + 0.25 + 0.125 + 0.031 25 + 0.003 906 25 + 0.001 953 125 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =


0.912 937 045 097 351 074 218 75(10)

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

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


(-1)1 × (1 + 0.912 937 045 097 351 074 218 75) × 25 =


-1.912 937 045 097 351 074 218 75 × 25 =


-61.213 985 443 115 234 375

1 - 1000 0100 - 111 0100 1101 1011 0001 1111 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) = -61.213 985 443 115 234 375(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 1 - 1000 0100 - 111 0100 1101 1011 0001 1111 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:05 UTC (GMT)
The number 0 - 1000 1111 - 001 0111 1000 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 = ? Feb 27 03:05 UTC (GMT)
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The number 1 - 1000 1010 - 000 1010 0100 0000 0101 0100 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:05 UTC (GMT)
The number 0 - 1000 1110 - 111 1111 1110 0000 0000 0010 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:04 UTC (GMT)
The number 0 - 1000 1000 - 111 0110 1110 0111 1010 1100 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:04 UTC (GMT)
The number 0 - 1000 0100 - 110 0110 1000 0000 0100 0111 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:04 UTC (GMT)
The number 0 - 1000 0011 - 100 0001 1111 1111 1011 0110 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:04 UTC (GMT)
The number 0 - 0000 0011 - 100 0000 0000 0000 0101 0000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Feb 27 03:04 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)