32 bit single precision IEEE 754 binary floating point number 0 - 1000 0000 - 011 0011 0000 0000 0000 0000 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 0 - 1000 0000 - 011 0011 0000 0000 0000 0000 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:
1000 0000


The last 23 bits contain the mantissa:
011 0011 0000 0000 0000 0000

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

The exponent is allways a positive integer.

1000 0000(2) =


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


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


128 =


128(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 = 128 - 127 = 1


4. Convert the mantissa from binary (base 2) to decimal (base 10):

Mantissa represents the number's fractional part (the excess beyond the number's integer part, comma delimited)

011 0011 0000 0000 0000 0000(2) =

0 × 2-1 + 1 × 2-2 + 1 × 2-3 + 0 × 2-4 + 0 × 2-5 + 1 × 2-6 + 1 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 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 + 0.25 + 0.125 + 0 + 0 + 0.015 625 + 0.007 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.25 + 0.125 + 0.015 625 + 0.007 812 5 =


0.398 437 5(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)0 × (1 + 0.398 437 5) × 21 =


1.398 437 5 × 21 =


2.796 875

0 - 1000 0000 - 011 0011 0000 0000 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
2.796 875(10)

More operations of this kind:

0 - 1000 0000 - 011 0010 1111 1111 1111 1111 = ?

0 - 1000 0000 - 011 0011 0000 0000 0000 0001 = ?


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 - 1000 0000 - 011 0011 0000 0000 0000 0000 = ? Apr 18 08:21 UTC (GMT)
0 - 1000 1000 - 100 0000 0100 0000 0000 0000 = ? Apr 18 08:21 UTC (GMT)
1 - 1010 1000 - 010 0101 0000 1010 1010 1000 = ? Apr 18 08:21 UTC (GMT)
0 - 1000 0010 - 000 0100 0111 1010 1110 0010 = ? Apr 18 08:21 UTC (GMT)
1 - 1001 0100 - 100 0101 1001 0000 0001 1101 = ? Apr 18 08:20 UTC (GMT)
0 - 0110 1111 - 100 0000 1111 0000 0001 1110 = ? Apr 18 08:20 UTC (GMT)
1 - 1000 1000 - 101 1010 1101 1010 0000 1100 = ? Apr 18 08:20 UTC (GMT)
0 - 0101 0101 - 001 1111 1111 1111 1111 1110 = ? Apr 18 08:20 UTC (GMT)
0 - 0100 0011 - 011 1001 0011 0000 0000 0100 = ? Apr 18 08:20 UTC (GMT)
0 - 0000 1000 - 000 0000 0000 0011 0100 1001 = ? Apr 18 08:20 UTC (GMT)
1 - 0111 0000 - 001 0000 0000 0000 0000 0000 = ? Apr 18 08:20 UTC (GMT)
0 - 1000 0100 - 000 0100 0100 1011 0001 1110 = ? Apr 18 08:20 UTC (GMT)
1 - 1000 0111 - 000 1000 1000 0000 0000 0100 = ? Apr 18 08:19 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)