Converter from 32 bit single precision IEEE 754 binary floating point standard system: converting to base ten decimal numbers (float)

Convert 32 bit single precision IEEE 754 binary floating point standard numbers to base ten decimal float

Entered binary numbers length must be as indicated - or else extra bits of '0' value will be added to the end (to the right).

Latest 32 bit single precision IEEE 754 binary floating point standard converted to decimal numbers (float) in base ten


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 to base 10 decimal system:

  • 1. View 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.
    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 number 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 from 32 bit single precision IEEE 754 binary floating point to decimal number (float) in decimal system (in base 10):

  • 1. View 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 to decimal number (float) in decimal system (in base 10) = -6,031 494 140 625(10)