32 bit single precision IEEE 754 binary floating point number 0 - 1101 1010 - 111 1110 0000 0000 0000 0000 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 1101 1010 - 111 1110 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:
1101 1010


The last 23 bits contain the mantissa:
111 1110 0000 0000 0000 0000

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

The exponent is allways a positive integer.

1101 1010(2) =


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


128 + 64 + 0 + 16 + 8 + 0 + 2 + 0 =


128 + 64 + 16 + 8 + 2 =


218(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 = 218 - 127 = 91


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 1110 0000 0000 0000 0000(2) =

1 × 2-1 + 1 × 2-2 + 1 × 2-3 + 1 × 2-4 + 1 × 2-5 + 1 × 2-6 + 0 × 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 =


0.5 + 0.25 + 0.125 + 0.062 5 + 0.031 25 + 0.015 625 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.5 + 0.25 + 0.125 + 0.062 5 + 0.031 25 + 0.015 625 =


0.984 375(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 + 0.984 375) × 291 =


1.984 375 × 291 =


4 913 074 530 913 852 966 005 899 264

Conclusion:

0 - 1101 1010 - 111 1110 0000 0000 0000 0000
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

4 913 074 530 913 852 966 005 899 264(10)

More operations of this kind:

0 - 1101 1010 - 111 1101 1111 1111 1111 1111 = ?

0 - 1101 1010 - 111 1110 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 - 1101 1010 - 111 1110 0000 0000 0000 0000 = ? Jan 26 18:14 UTC (GMT)
0 - 0101 0100 - 010 0000 0000 0000 0000 0000 = ? Jan 26 18:13 UTC (GMT)
0 - 0111 1110 - 101 0000 0000 0000 0000 0000 = ? Jan 26 18:13 UTC (GMT)
1 - 1111 1111 - 000 1001 0001 0011 1000 0101 = ? Jan 26 18:13 UTC (GMT)
1 - 0110 1010 - 101 1100 0101 0001 0101 1010 = ? Jan 26 18:13 UTC (GMT)
1 - 1110 0111 - 010 1111 1111 1111 1111 1111 = ? Jan 26 18:12 UTC (GMT)
0 - 0000 0110 - 011 0010 1111 1111 1111 1110 = ? Jan 26 18:11 UTC (GMT)
0 - 0000 0000 - 101 0000 0000 0000 0000 0000 = ? Jan 26 18:11 UTC (GMT)
1 - 0000 1001 - 100 1001 1001 1101 0010 1000 = ? Jan 26 18:11 UTC (GMT)
1 - 1000 1010 - 001 0101 0101 0101 0101 0101 = ? Jan 26 18:11 UTC (GMT)
0 - 0111 1110 - 101 0000 0000 0000 0000 0000 = ? Jan 26 18:09 UTC (GMT)
0 - 0001 0100 - 100 0001 0000 0000 0000 0111 = ? Jan 26 18:09 UTC (GMT)
0 - 1100 0000 - 100 1001 0000 1111 1110 0001 = ? Jan 26 18:08 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)