32 bit single precision IEEE 754 binary floating point number 0 - 1110 1011 - 111 1011 1000 1111 1100 0000 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 1110 1011 - 111 1011 1000 1111 1100 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:
1110 1011


The last 23 bits contain the mantissa:
111 1011 1000 1111 1100 0000

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

The exponent is allways a positive integer.

1110 1011(2) =


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


128 + 64 + 32 + 0 + 8 + 0 + 2 + 1 =


128 + 64 + 32 + 8 + 2 + 1 =


235(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 = 235 - 127 = 108


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)

111 1011 1000 1111 1100 0000(2) =

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


0.5 + 0.25 + 0.125 + 0.062 5 + 0 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0 + 0 + 0 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0 + 0 + 0 =


0.5 + 0.25 + 0.125 + 0.062 5 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 =


0.965 324 401 855 468 75(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 + 0.965 324 401 855 468 75) × 2108 =


1.965 324 401 855 468 75 × 2108 =


637 784 232 359 749 346 867 479 001 956 352

Conclusion:

0 - 1110 1011 - 111 1011 1000 1111 1100 0000
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

637 784 232 359 749 346 867 479 001 956 352(10)

More operations of this kind:

0 - 1110 1011 - 111 1011 1000 1111 1011 1111 = ?

0 - 1110 1011 - 111 1011 1000 1111 1100 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 - 1110 1011 - 111 1011 1000 1111 1100 0000 = ? Jan 16 05:29 UTC (GMT)
0 - 0001 1000 - 001 1000 1001 0000 0000 0001 = ? Jan 16 05:29 UTC (GMT)
1 - 0010 1100 - 010 0000 0000 0000 0000 0001 = ? Jan 16 05:29 UTC (GMT)
1 - 1100 0000 - 001 0101 1101 0011 1110 1001 = ? Jan 16 05:29 UTC (GMT)
0 - 0001 0111 - 000 0000 0010 1111 1100 1011 = ? Jan 16 05:28 UTC (GMT)
0 - 0001 0101 - 001 0011 0011 0011 0011 0111 = ? Jan 16 05:28 UTC (GMT)
0 - 0001 0100 - 100 0000 0000 0000 0000 0000 = ? Jan 16 05:28 UTC (GMT)
0 - 1000 0010 - 110 1000 0000 0000 0000 0100 = ? Jan 16 05:28 UTC (GMT)
0 - 0000 1101 - 110 0101 0111 1111 1111 1111 = ? Jan 16 05:27 UTC (GMT)
0 - 0000 1010 - 000 0000 0000 1111 1111 1110 = ? Jan 16 05:27 UTC (GMT)
0 - 0000 1001 - 101 0010 0100 0010 1010 0111 = ? Jan 16 05:27 UTC (GMT)
0 - 0111 1110 - 010 0111 1010 1110 0001 0100 = ? Jan 16 05:27 UTC (GMT)
0 - 0000 1000 - 101 1100 0000 0000 0000 0000 = ? Jan 16 05:27 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)