32 bit single precision IEEE 754 binary floating point number 1 - 0111 1011 - 011 1000 1011 0000 0101 1100 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point: 1 - 0111 1011 - 011 1000 1011 0000 0101 1100to decimal system (base ten)

4. 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)

011 1000 1011 0000 0101 1100(2) =

Latest 32 bit single precision IEEE 754 floating point binary standard numbers converted to decimal base ten (float)

 1 - 0111 1011 - 011 1000 1011 0000 0101 1100 = -0.090 180 128 812 789 916 992 187 5 Nov 25 20:18 UTC (GMT) 1 - 0000 0010 - 000 0000 0000 0000 0000 0001 = -0 Nov 25 20:17 UTC (GMT) 1 - 1000 0111 - 100 1100 1100 1100 1100 1101 = -409.600 006 103 515 625 Nov 25 20:16 UTC (GMT) 0 - 1000 1100 - 000 1000 1011 1100 1000 1010 = 8 751.134 765 625 Nov 25 20:16 UTC (GMT) 0 - 1000 1101 - 111 1101 1100 0000 0000 0000 = 32 480 Nov 25 20:16 UTC (GMT) 0 - 1000 0010 - 011 0001 1001 1001 1001 1001 = 11.099 999 427 795 410 156 25 Nov 25 20:15 UTC (GMT) 1 - 0001 1000 - 010 0111 1111 1111 1111 1110 = -0 Nov 25 20:13 UTC (GMT) 0 - 0111 1110 - 100 0001 0100 1010 0001 0110 = 0.755 036 711 692 810 058 593 75 Nov 25 20:13 UTC (GMT) 0 - 1111 1111 - 000 0000 0000 0000 0000 0000 = +∞ (Infinity, positive) Nov 25 20:13 UTC (GMT) 0 - 0111 1111 - 000 0000 0001 1111 1111 1110 = 1.000 976 324 081 420 898 437 5 Nov 25 20:13 UTC (GMT) 0 - 1000 0010 - 011 0110 0000 0000 0000 0000 = 11.375 Nov 25 20:13 UTC (GMT) 0 - 1000 0100 - 010 1000 1111 1111 1111 1111 = 42.249 996 185 302 734 375 Nov 25 20:11 UTC (GMT) 0 - 1010 1010 - 000 0000 0000 0000 0000 0000 = 8 796 093 022 208 Nov 25 20:11 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