32 bit single precision IEEE 754 binary floating point number 0 - 1011 1100 - 100 0010 1011 0110 1011 1111 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 1011 1100 - 100 0010 1011 0110 1011 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:
1011 1100


The last 23 bits contain the mantissa:
100 0010 1011 0110 1011 1111

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

The exponent is allways a positive integer.

1011 1100(2) =


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


128 + 0 + 32 + 16 + 8 + 4 + 0 + 0 =


128 + 32 + 16 + 8 + 4 =


188(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 = 188 - 127 = 61


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)

100 0010 1011 0110 1011 1111(2) =

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


0.5 + 0 + 0 + 0 + 0 + 0.015 625 + 0 + 0.003 906 25 + 0 + 0.000 976 562 5 + 0.000 488 281 25 + 0 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 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 + 0.015 625 + 0.003 906 25 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 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 =


0.521 201 968 193 054 199 218 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 + 0.521 201 968 193 054 199 218 75) × 261 =


1.521 201 968 193 054 199 218 75 × 261 =


3 507 652 923 960 066 048

Conclusion:

0 - 1011 1100 - 100 0010 1011 0110 1011 1111
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

3 507 652 923 960 066 048(10)

More operations of this kind:

0 - 1011 1100 - 100 0010 1011 0110 1011 1110 = ?

0 - 1011 1100 - 100 0010 1011 0110 1100 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 - 1011 1100 - 100 0010 1011 0110 1011 1111 = ? Jan 24 20:45 UTC (GMT)
1 - 1100 0011 - 100 0100 0000 0000 0000 0010 = ? Jan 24 20:45 UTC (GMT)
0 - 0011 1011 - 101 0000 0000 0000 0000 0000 = ? Jan 24 20:44 UTC (GMT)
0 - 1000 0011 - 000 0110 0000 0000 0000 0110 = ? Jan 24 20:44 UTC (GMT)
0 - 0000 0000 - 000 0000 1000 0000 1011 1011 = ? Jan 24 20:44 UTC (GMT)
0 - 0000 0000 - 010 1000 0000 0000 0000 0000 = ? Jan 24 20:44 UTC (GMT)
0 - 0000 0000 - 100 1100 1100 1100 1100 0001 = ? Jan 24 20:44 UTC (GMT)
1 - 1001 1000 - 100 1100 0000 0000 0000 0101 = ? Jan 24 20:44 UTC (GMT)
0 - 0000 0000 - 101 1111 1100 0000 0000 0001 = ? Jan 24 20:43 UTC (GMT)
0 - 1110 0110 - 101 1111 1111 1111 1111 1111 = ? Jan 24 20:43 UTC (GMT)
1 - 1000 0110 - 111 1010 1111 1111 1111 1111 = ? Jan 24 20:43 UTC (GMT)
1 - 0101 0110 - 001 0010 1101 1100 1001 0010 = ? Jan 24 20:43 UTC (GMT)
0 - 0111 1101 - 010 1000 1111 0101 1100 0110 = ? Jan 24 20:43 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)