32 bit single precision IEEE 754 binary floating point number 1 - 0000 0001 - 010 1011 1100 0011 0100 0101 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 1 - 0000 0001 - 010 1011 1100 0011 0100 0101 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 0001


The last 23 bits contain the mantissa:
010 1011 1100 0011 0100 0101

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

The exponent is allways a positive integer.

0000 0001(2) =


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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =


1 =


1(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 = 1 - 127 = -126


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)

010 1011 1100 0011 0100 0101(2) =

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


0 + 0.25 + 0 + 0.062 5 + 0 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0.000 000 119 209 289 550 781 25 =


0.25 + 0.062 5 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 007 629 394 531 25 + 0.000 000 476 837 158 203 125 + 0.000 000 119 209 289 550 781 25 =


0.341 896 653 175 354 003 906 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 + 0.341 896 653 175 354 003 906 25) × 2-126 =


-1.341 896 653 175 354 003 906 25 × 2-126 =


-0

1 - 0000 0001 - 010 1011 1100 0011 0100 0101 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
-0(10)

More operations of this kind:

1 - 0000 0001 - 010 1011 1100 0011 0100 0100 = ?

1 - 0000 0001 - 010 1011 1100 0011 0100 0110 = ?


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 - 0000 0001 - 010 1011 1100 0011 0100 0101 = ? May 06 19:36 UTC (GMT)
0 - 0111 1100 - 010 0011 0010 0010 0010 1011 = ? May 06 19:35 UTC (GMT)
1 - 1000 0101 - 110 1111 1000 0000 0000 0000 = ? May 06 19:35 UTC (GMT)
0 - 1010 1101 - 101 0100 0000 0000 0000 0000 = ? May 06 19:35 UTC (GMT)
1 - 0111 0111 - 101 1001 1111 1111 1111 1011 = ? May 06 19:35 UTC (GMT)
1 - 1000 1111 - 010 0100 1110 0010 0000 0001 = ? May 06 19:35 UTC (GMT)
0 - 0000 0100 - 111 0000 0000 0000 0100 0101 = ? May 06 19:34 UTC (GMT)
0 - 0111 1110 - 101 0000 0000 0000 0000 0000 = ? May 06 19:34 UTC (GMT)
1 - 0101 0001 - 101 0010 0000 1111 0001 0100 = ? May 06 19:33 UTC (GMT)
0 - 0000 0010 - 010 0000 0000 0000 0000 0000 = ? May 06 19:32 UTC (GMT)
1 - 1000 0110 - 000 0000 0011 1111 1111 1111 = ? May 06 19:32 UTC (GMT)
0 - 0110 0101 - 011 1111 1110 0000 0000 0000 = ? May 06 19:30 UTC (GMT)
0 - 0101 0101 - 000 0000 0000 0000 0000 0001 = ? May 06 19:30 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)