32 bit single precision IEEE 754 binary floating point number 0 - 0000 0000 - 000 0000 0100 0010 0100 0100 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 0000 0000 - 000 0000 0100 0010 0100 0100
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 0100 0010 0100 0100

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 0100 0010 0100 0100(2) =

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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.001 953 125 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 =


0.001 953 125 + 0.000 061 035 156 25 + 0.000 007 629 394 531 25 + 0.000 000 476 837 158 203 125 =


0.002 022 266 387 939 453 125(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 + 0.002 022 266 387 939 453 125) × 2-127 =


1.002 022 266 387 939 453 125 × 2-127 =


0

Conclusion:

0 - 0000 0000 - 000 0000 0100 0010 0100 0100
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 0100 0010 0100 0011 = ?

0 - 0000 0000 - 000 0000 0100 0010 0100 0101 = ?


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 0100 0010 0100 0100 = ? Jan 24 22:15 UTC (GMT)
0 - 1000 0101 - 001 1011 1111 1010 1101 1011 = ? Jan 24 22:15 UTC (GMT)
1 - 0001 1101 - 001 0100 0111 1010 1110 0010 = ? Jan 24 22:14 UTC (GMT)
0 - 0000 0000 - 100 1000 0000 0000 0000 0000 = ? Jan 24 22:14 UTC (GMT)
0 - 0000 0000 - 010 0001 0000 0000 0000 0000 = ? Jan 24 22:13 UTC (GMT)
1 - 0101 1010 - 010 1101 1010 1010 0000 0000 = ? Jan 24 22:13 UTC (GMT)
1 - 1000 0110 - 010 0000 0000 0000 0000 0011 = ? Jan 24 22:13 UTC (GMT)
0 - 0000 0000 - 011 0111 0000 0000 0000 0000 = ? Jan 24 22:12 UTC (GMT)
0 - 0000 0000 - 011 1101 1010 0110 0110 0111 = ? Jan 24 22:12 UTC (GMT)
0 - 0000 0000 - 000 0000 0011 1100 1011 1100 = ? Jan 24 22:12 UTC (GMT)
1 - 1100 0000 - 111 0111 1111 1111 1111 1111 = ? Jan 24 22:11 UTC (GMT)
1 - 1000 0000 - 100 0000 0001 1010 0001 1000 = ? Jan 24 22:11 UTC (GMT)
1 - 1111 0100 - 010 1101 0110 0000 0000 0000 = ? Jan 24 22:11 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)