32 bit single precision IEEE 754 binary floating point number 1 - 1001 1100 - 101 1110 1101 0000 0011 0010 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
1 - 1001 1100 - 101 1110 1101 0000 0011 0010
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 1100


The last 23 bits contain the mantissa:
101 1110 1101 0000 0011 0010

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

The exponent is allways a positive integer.

1001 1100(2) =


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


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


128 + 16 + 8 + 4 =


156(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 = 156 - 127 = 29


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)

101 1110 1101 0000 0011 0010(2) =

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


0.5 + 0 + 0.125 + 0.062 5 + 0.031 25 + 0.015 625 + 0 + 0.003 906 25 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 0 =


0.5 + 0.125 + 0.062 5 + 0.031 25 + 0.015 625 + 0.003 906 25 + 0.001 953 125 + 0.000 488 281 25 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 238 418 579 101 562 5 =


0.740 728 616 714 477 539 062 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 + 0.740 728 616 714 477 539 062 5) × 229 =


-1.740 728 616 714 477 539 062 5 × 229 =


-934 546 560

Conclusion:

1 - 1001 1100 - 101 1110 1101 0000 0011 0010
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

-934 546 560(10)

More operations of this kind:

1 - 1001 1100 - 101 1110 1101 0000 0011 0001 = ?

1 - 1001 1100 - 101 1110 1101 0000 0011 0011 = ?


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 1100 - 101 1110 1101 0000 0011 0010 = ? Jan 23 03:55 UTC (GMT)
0 - 1000 1100 - 001 1100 0011 1111 1111 1111 = ? Jan 23 03:55 UTC (GMT)
0 - 1110 0100 - 010 1110 1110 0000 0111 0010 = ? Jan 23 03:54 UTC (GMT)
0 - 1000 1101 - 110 1110 0100 1111 1111 1101 = ? Jan 23 03:54 UTC (GMT)
0 - 1011 1110 - 010 0011 1101 1000 1010 0000 = ? Jan 23 03:53 UTC (GMT)
1 - 1001 0011 - 011 0000 0000 0000 0000 0000 = ? Jan 23 03:52 UTC (GMT)
0 - 0111 1010 - 001 1100 0101 1111 0000 1001 = ? Jan 23 03:52 UTC (GMT)
1 - 1000 0000 - 010 0010 0010 0010 0010 0001 = ? Jan 23 03:50 UTC (GMT)
1 - 1101 0111 - 110 1010 1011 1011 0101 0110 = ? Jan 23 03:50 UTC (GMT)
0 - 1000 0011 - 010 0100 0000 0000 0000 0000 = ? Jan 23 03:49 UTC (GMT)
0 - 1000 0010 - 101 0000 1100 0010 1000 1111 = ? Jan 23 03:49 UTC (GMT)
0 - 0000 0000 - 011 1110 1000 0000 0000 0001 = ? Jan 23 03:48 UTC (GMT)
0 - 1000 0001 - 101 0010 0000 0000 0000 1010 = ? Jan 23 03:48 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)