32 bit single precision IEEE 754 binary floating point number 0 - 1111 1111 - 101 0101 0101 0101 0100 1111 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 1111 1111 - 101 0101 0101 0101 0100 1111
to decimal system (base ten)

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

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


The next 8 bits contain the exponent:
1111 1111


The last 23 bits contain the mantissa:
101 0101 0101 0101 0100 1111

2. Reserved bitpattern.

Notice that all the exponent bits are on 1 (set) and the first mantissa bit (the most significant) is on 1 (set).

This is one of the reserved bitpatterns of the special values of: QNaN (Quiet Not a Number).

A QNaN is generated by an operation when the result is not mathematically defined. A QNaN is a category of NaN (Not A Number). Generally, a NaN is used to represent a value that is not a number.

Conclusion:

0 - 1111 1111 - 101 0101 0101 0101 0100 1111
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

QNaN, Quiet Not a Number

More operations of this kind:

0 - 1111 1111 - 101 0101 0101 0101 0100 1110 = ?

0 - 1111 1111 - 101 0101 0101 0101 0101 0000 = ?


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

32 bit single precision IEEE 754 binary floating point standard representation of numbers requires three building blocks: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits), mantissa (23 bits)

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

0 - 1111 1111 - 101 0101 0101 0101 0100 1111 = ? Jan 16 05:32 UTC (GMT)
1 - 1001 1011 - 011 0001 1001 1011 0111 1111 = ? Jan 16 05:32 UTC (GMT)
0 - 0101 1000 - 100 0110 0000 0000 0000 0000 = ? Jan 16 05:32 UTC (GMT)
0 - 0101 0101 - 010 1010 1000 0000 0000 0000 = ? Jan 16 05:32 UTC (GMT)
0 - 0101 0000 - 010 1111 1111 1111 1111 1110 = ? Jan 16 05:31 UTC (GMT)
1 - 1011 0000 - 101 1011 1000 0111 1111 1111 = ? Jan 16 05:31 UTC (GMT)
0 - 0101 0000 - 000 0000 0000 0000 0000 0010 = ? Jan 16 05:31 UTC (GMT)
0 - 0100 0111 - 010 0100 1110 0001 1000 1011 = ? Jan 16 05:31 UTC (GMT)
0 - 1000 1000 - 111 1010 0000 0001 0111 1000 = ? Jan 16 05:31 UTC (GMT)
0 - 0100 0010 - 011 0010 1100 1100 1100 1100 = ? Jan 16 05:31 UTC (GMT)
1 - 1000 0000 - 110 0000 0000 0000 0000 0000 = ? Jan 16 05:31 UTC (GMT)
0 - 0100 0001 - 011 1110 0010 0000 0000 0001 = ? Jan 16 05:31 UTC (GMT)
0 - 0100 0000 - 101 1000 0111 1111 1111 1101 = ? Jan 16 05:31 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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)