32 bit single precision IEEE 754 binary floating point number 0 - 0000 0000 - 000 0000 1000 0000 1011 1010 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 0 - 0000 0000 - 000 0000 1000 0000 1011 1010 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:
0000 0000


The last 23 bits contain the mantissa:
000 0000 1000 0000 1011 1010

2. Reserved bitpattern.

Notice that all the exponent bits are on 0 (clear) and at least one of the mantissa bits is on 1 (set).

This is one of the reserved bitpatterns of the special values of: Denormalized.

Denormalized numbers are too small to be correctly represented so they approximate to zero. Depending on the sign bit, -0 and +0 are two distinct values though they both compare as equal (0).

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

The exponent is allways a positive integer.

0000 0000(2) =


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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0(10)

4. Adjust the exponent.

Subtract the excess bits: 2(8 - 1) - 1 = 127, that is due to the 8 bit excess/bias notation:

Exponent adjusted = 0 - 127 = -127


5. 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)

000 0000 1000 0000 1011 1010(2) =

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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.003 906 25 + 0 + 0 + 0 + 0 + 0 + 0 + 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 + 0.000 000 238 418 579 101 562 5 + 0 =


0.003 906 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 + 0.000 000 238 418 579 101 562 5 =


0.003 928 422 927 856 445 312 5(10)

6. 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 + 0.003 928 422 927 856 445 312 5) × 2-127 =


1.003 928 422 927 856 445 312 5 × 2-127 =


0

0 - 0000 0000 - 000 0000 1000 0000 1011 1010 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
0(10)

More operations of this kind:

0 - 0000 0000 - 000 0000 1000 0000 1011 1001 = ?

0 - 0000 0000 - 000 0000 1000 0000 1011 1011 = ?


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 - 0000 0000 - 000 0000 1000 0000 1011 1010 = ? Apr 18 09:48 UTC (GMT)
0 - 0111 1100 - 010 0110 0110 0110 0110 0001 = ? Apr 18 09:47 UTC (GMT)
0 - 1001 1001 - 100 1101 0011 1011 0010 0011 = ? Apr 18 09:47 UTC (GMT)
0 - 1111 1111 - 111 1001 1110 1110 0001 1110 = ? Apr 18 09:46 UTC (GMT)
1 - 0010 1101 - 011 0000 0000 0000 0000 0010 = ? Apr 18 09:45 UTC (GMT)
1 - 1000 0101 - 010 0000 0000 0000 0000 0000 = ? Apr 18 09:44 UTC (GMT)
1 - 1000 0111 - 000 1110 1101 1111 1111 1000 = ? Apr 18 09:43 UTC (GMT)
0 - 0000 0100 - 011 1101 1000 0000 0000 0000 = ? Apr 18 09:43 UTC (GMT)
0 - 0110 1011 - 101 1101 1101 1100 1000 1001 = ? Apr 18 09:43 UTC (GMT)
1 - 0100 0101 - 001 0101 0101 0111 1110 1011 = ? Apr 18 09:42 UTC (GMT)
1 - 1100 1001 - 010 0001 1000 0000 0000 0000 = ? Apr 18 09:42 UTC (GMT)
0 - 1001 1001 - 110 1101 0011 0011 0010 0011 = ? Apr 18 09:42 UTC (GMT)
0 - 0101 0011 - 001 0110 0010 1011 0101 0101 = ? Apr 18 09:42 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)