32 bit single precision IEEE 754 binary floating point number 1 - 1001 0001 - 011 0001 1111 1111 1111 1110 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
1 - 1001 0001 - 011 0001 1111 1111 1111 1110
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:
1001 0001


The last 23 bits contain the mantissa:
011 0001 1111 1111 1111 1110

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

The exponent is allways a positive integer.

1001 0001(2) =


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


128 + 0 + 0 + 16 + 0 + 0 + 0 + 1 =


128 + 16 + 1 =


145(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 = 145 - 127 = 18


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 0001 1111 1111 1111 1110(2) =

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


0 + 0.25 + 0.125 + 0 + 0 + 0 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 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 + 0.000 003 814 697 265 625 + 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 =


0.25 + 0.125 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 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 + 0.000 003 814 697 265 625 + 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 =


0.390 624 761 581 420 898 437 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)1 × (1 + 0.390 624 761 581 420 898 437 5) × 218 =


-1.390 624 761 581 420 898 437 5 × 218 =


-364 543.937 5

Conclusion:

1 - 1001 0001 - 011 0001 1111 1111 1111 1110
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

-364 543.937 5(10)

More operations of this kind:

1 - 1001 0001 - 011 0001 1111 1111 1111 1101 = ?

1 - 1001 0001 - 011 0001 1111 1111 1111 1111 = ?


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)

1 - 1001 0001 - 011 0001 1111 1111 1111 1110 = ? Jan 24 11:36 UTC (GMT)
1 - 0101 0110 - 101 0000 0000 0000 0000 0001 = ? Jan 24 11:34 UTC (GMT)
0 - 0011 1110 - 001 1111 0001 1111 0001 1010 = ? Jan 24 11:33 UTC (GMT)
0 - 1000 1101 - 001 0001 0100 1010 0100 0010 = ? Jan 24 11:32 UTC (GMT)
1 - 0100 0101 - 110 1000 0000 0000 0000 0000 = ? Jan 24 11:32 UTC (GMT)
0 - 0111 0100 - 110 1000 0010 0101 1001 1101 = ? Jan 24 11:31 UTC (GMT)
0 - 1111 1100 - 011 0110 1101 1011 0111 0000 = ? Jan 24 11:30 UTC (GMT)
0 - 0111 1100 - 100 1100 1100 1100 1100 1110 = ? Jan 24 11:30 UTC (GMT)
0 - 0111 0101 - 100 0111 1100 0100 0100 1001 = ? Jan 24 11:30 UTC (GMT)
0 - 1000 0101 - 100 1101 1100 0110 1010 1001 = ? Jan 24 11:29 UTC (GMT)
0 - 1000 1010 - 101 0100 1101 0010 1101 1010 = ? Jan 24 11:28 UTC (GMT)
0 - 1000 1000 - 110 0110 0000 0000 0000 0001 = ? Jan 24 11:28 UTC (GMT)
1 - 1110 1100 - 000 0000 0000 0000 0100 0100 = ? Jan 24 11:28 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)