32 bit single precision IEEE 754 binary floating point number 0 - 1010 1010 - 111 1001 0101 0101 0101 0101 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 1010 1010 - 111 1001 0101 0101 0101 0101.

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:
1010 1010


The last 23 bits contain the mantissa:
111 1001 0101 0101 0101 0101

2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):

1010 1010(2) =


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


128 + 0 + 32 + 0 + 8 + 0 + 2 + 0 =


128 + 32 + 8 + 2 =


170(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 = 170 - 127 = 43

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):

111 1001 0101 0101 0101 0101(2) =

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


0.5 + 0.25 + 0.125 + 0.062 5 + 0 + 0 + 0.007 812 5 + 0 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0 + 0.000 122 070 312 5 + 0 + 0.000 030 517 578 125 + 0 + 0.000 007 629 394 531 25 + 0 + 0.000 001 907 348 632 812 5 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0.000 000 119 209 289 550 781 25 =


0.5 + 0.25 + 0.125 + 0.062 5 + 0.007 812 5 + 0.001 953 125 + 0.000 488 281 25 + 0.000 122 070 312 5 + 0.000 030 517 578 125 + 0.000 007 629 394 531 25 + 0.000 001 907 348 632 812 5 + 0.000 000 476 837 158 203 125 + 0.000 000 119 209 289 550 781 25 =


0.947 916 626 930 236 816 406 25(10)

Conclusion:

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 + 0.947 916 626 930 236 816 406 25) × 243 =


1.947 916 626 930 236 816 406 25 × 243 =


17 134 055 849 984

0 - 1010 1010 - 111 1001 0101 0101 0101 0101
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =


17 134 055 849 984(10)

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)

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)