0 - 0111 1110 - 110 0101 1101 0010 0001 0101 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

0 - 0111 1110 - 110 0101 1101 0010 0001 0101: 32 bit single precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
0 - 0111 1110 - 110 0101 1101 0010 0001 0101, a 32 bit single precision IEEE 754 binary floating point representation standard to decimal?

1. Identify the elements that make up the binary representation of the number:

The first bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.
0

The next 8 bits contain the exponent:
0111 1110

The last 23 bits contain the mantissa:
110 0101 1101 0010 0001 0101


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

The exponent is allways a positive integer.

0111 1110(2) =

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

0 + 64 + 32 + 16 + 8 + 4 + 2 + 0 =

64 + 32 + 16 + 8 + 4 + 2 =

126(10)

3. Adjust the exponent.

Subtract the excess bits: 2(8 - 1) - 1 = 127,

that is due to the 8 bit excess/bias notation.

The exponent, adjusted = 126 - 127 = -1


4. Convert the mantissa from binary (from base 2) to decimal (in base 10).

The mantissa represents the fractional part of the number (what comes after the whole part of the number, separated from it by a comma).


110 0101 1101 0010 0001 0101(2) =

1 × 2-1 + 1 × 2-2 + 0 × 2-3 + 0 × 2-4 + 1 × 2-5 + 0 × 2-6 + 1 × 2-7 + 1 × 2-8 + 1 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 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 + 0 + 0.031 25 + 0 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 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 + 0.031 25 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 488 281 25 + 0.000 061 035 156 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 =

0.795 473 694 801 330 566 406 25(10)

5. Put all the numbers into expression to calculate the single precision floating point decimal value:

(-1)Sign × (1 + Mantissa) × 2(Adjusted exponent) =

(-1)0 × (1 + 0.795 473 694 801 330 566 406 25) × 2-1 =

1.795 473 694 801 330 566 406 25 × 2-1 = ...

= 0.897 736 847 400 665 283 203 125

0 - 0111 1110 - 110 0101 1101 0010 0001 0101, a 32 bit single precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (float) = 0.897 736 847 400 665 283 203 125(10)

Spaces were used to group digits: for binary, by 4, for decimal, by 3.

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How to convert 32 bit single precision IEEE 754 binary floating point standard numbers 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)