Binary ↘ Float: The 32 Bit Single Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 0000 0000 - 000 0000 0100 0010 0100 0011 Converted and Written as a Base Ten Decimal System Number (as a Float)
0 - 0000 0000 - 000 0000 0100 0010 0100 0011: 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:
0000 0000
The last 23 bits contain the mantissa:
000 0000 0100 0010 0100 0011
2. Reserved bitpattern.
We notice that all the bits that make up the exponent are on 0 (clear) and at least one bit of the mantissa is set 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 (from base 2) to decimal (in 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.
The exponent, adjusted = 0 - 127 = -127
5. 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).
000 0000 0100 0010 0100 0011(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 + 0 × 2-21 + 1 × 2-22 + 1 × 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 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =
0.001 953 125 + 0.000 061 035 156 25 + 0.000 007 629 394 531 25 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =
0.002 022 147 178 649 902 343 75(10)
6. 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 + 0.002 022 147 178 649 902 343 75) × 2-127 =
1.002 022 147 178 649 902 343 75 × 2-127 =
0
0 - 0000 0000 - 000 0000 0100 0010 0100 0011 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(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: