32 bit single precision IEEE 754 binary floating point number 0 - 0000 0000 - 000 0000 0000 0000 0011 0100 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 0000 0000 - 000 0000 0000 0000 0011 0100.

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 0000 0000 0011 0100

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).

Proof:

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

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)

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 = 0 - 127 = -127

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):

000 0000 0000 0000 0011 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 + 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 + 1 × 2-18 + 1 × 2-19 + 0 × 2-20 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 =

0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 =

0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 476 837 158 203 125 =

0.000 006 198 883 056 640 625(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 + 0.000 006 198 883 056 640 625) × 2-127 =

1.000 006 198 883 056 640 625 × 2-127 =

0

0 - 0000 0000 - 000 0000 0000 0000 0011 0100
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

0(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)

0 - 0000 0000 - 000 0000 0000 0000 0011 0100 = 0 Dec 06 18:59 UTC (GMT)
1 - 0111 1110 - 100 0100 0000 0000 0000 0000 = -0.765 625 Dec 06 18:58 UTC (GMT)
1 - 1000 0001 - 001 1100 0000 0000 0000 0000 = -4.875 Dec 06 18:56 UTC (GMT)
1 - 0000 0010 - 011 1100 1011 0100 0100 0001 = -0.000 000 000 000 000 000 000 000 000 000 000 000 02 Dec 06 18:56 UTC (GMT)
0 - 1000 1001 - 010 0100 0010 1010 0000 0000 = 1 313.312 5 Dec 06 18:52 UTC (GMT)
0 - 1000 0100 - 101 1111 0000 0000 0000 0000 = 55.75 Dec 06 18:51 UTC (GMT)
1 - 1010 1010 - 000 1110 1010 1001 0000 0000 = -9 803 531 288 576 Dec 06 18:50 UTC (GMT)
0 - 1000 0111 - 110 0011 0000 0000 0000 0000 = 454 Dec 06 18:50 UTC (GMT)
1 - 1000 0110 - 001 1111 0001 0100 1101 0000 = -159.081 298 828 125 Dec 06 18:49 UTC (GMT)
0 - 1000 0011 - 000 1000 0000 0000 0000 0000 = 17 Dec 06 18:49 UTC (GMT)
0 - 1111 1111 - 011 1100 0011 1011 1111 1000 = SNaN, Signalling Not a Number Dec 06 18:49 UTC (GMT)
0 - 0000 1111 - 000 1111 0101 0000 0010 1010 = 0.000 000 000 000 000 000 000 000 000 000 000 215 63 Dec 06 18:49 UTC (GMT)
0 - 1000 0110 - 001 0110 0010 0101 1000 1001 = 150.146 621 704 101 562 5 Dec 06 18:49 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)