Binary ↘ Float: The 32 Bit Single Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 0001 1101 - 101 0000 0011 0100 1101 0001 Converted and Written as a Base Ten Decimal System Number (as a Float)

0 - 0001 1101 - 101 0000 0011 0100 1101 0001: 32 bit single precision IEEE 754 binary floating point standard representation number converted to decimal system (base ten)

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:
0001 1101

The last 23 bits contain the mantissa:
101 0000 0011 0100 1101 0001


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

The exponent is allways a positive integer.

0001 1101(2) =

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

0 + 0 + 0 + 16 + 8 + 4 + 0 + 1 =

16 + 8 + 4 + 1 =

29(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 = 29 - 127 = -98


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).


101 0000 0011 0100 1101 0001(2) =

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

0.5 + 0 + 0.125 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 976 562 5 + 0.000 488 281 25 + 0 + 0.000 122 070 312 5 + 0 + 0 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0 + 0.000 001 907 348 632 812 5 + 0 + 0 + 0 + 0.000 000 119 209 289 550 781 25 =

0.5 + 0.125 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 122 070 312 5 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0.000 001 907 348 632 812 5 + 0.000 000 119 209 289 550 781 25 =

0.626 611 828 804 016 113 281 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 + 0.626 611 828 804 016 113 281 25) × 2-98 =

1.626 611 828 804 016 113 281 25 × 2-98 =

0.000 000 000 000 000 000 000 000 000 005 132 681 91

0 - 0001 1101 - 101 0000 0011 0100 1101 0001 converted from a 32 bit single precision IEEE 754 binary floating point standard representation number to a decimal system number, written in base ten (float) = 0.000 000 000 000 000 000 000 000 000 005 132 681 91(10)

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

More operations with 32 bit single precision IEEE 754 binary floating point standard representation numbers:

» Number 0 - 0001 1101 - 101 0000 0011 0100 1101 0000 converted from 32 bit single precision IEEE 754 binary floating point standard representation to decimal system written in base ten (float) = ?

» Number 0 - 0001 1101 - 101 0000 0011 0100 1101 0010 converted from 32 bit single precision IEEE 754 binary floating point standard representation to decimal system written in base ten (float) = ?

The latest 32 bit single precision IEEE 754 floating point binary standard numbers converted and written as decimal system numbers (in base ten, float)

The number 0 - 0111 1110 - 100 1011 0000 1111 0010 0100 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:42 UTC (GMT)
The number 0 - 0101 0011 - 010 1011 1010 1010 0100 0111 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:42 UTC (GMT)
The number 1 - 1000 0000 - 110 0001 0010 0100 0110 1011 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:42 UTC (GMT)
The number 0 - 0100 1111 - 100 1000 0000 0000 0001 0000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:41 UTC (GMT)
The number 0 - 1000 0100 - 011 0111 0010 0000 0000 0111 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:41 UTC (GMT)
The number 0 - 0011 0001 - 111 1000 0010 1010 0001 0010 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:40 UTC (GMT)
The number 1 - 1110 0110 - 101 0000 0000 0000 0001 0001 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:40 UTC (GMT)
The number 0 - 1000 0111 - 100 1000 0100 0000 0000 1001 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:39 UTC (GMT)
The number 0 - 1000 0110 - 010 1111 0111 1111 1111 1110 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:38 UTC (GMT)
The number 0 - 0110 1101 - 010 0110 0110 1101 1001 1011 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? May 02 08:36 UTC (GMT)
All 32 bit single precision IEEE 754 binary floating point representation numbers converted to base ten decimal numbers (float)

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)