32 bit single precision IEEE 754 binary floating point number 0 - 0111 1110 - 100 0000 0000 0000 0000 0000 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point: 0 - 0111 1110 - 100 0000 0000 0000 0000 0000.

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 0000 0000 0000 0000 0000(2) =

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

 0 - 0111 1110 - 100 0000 0000 0000 0000 0000 = 0.75 Jun 27 12:39 UTC (GMT) 1 - 1000 0000 - 001 1110 1110 0110 0110 0110 = -2.482 812 404 632 568 359 375 Jun 27 12:37 UTC (GMT) 0 - 1001 1100 - 000 0001 0000 0000 0000 0000 = 541 065 216 Jun 27 12:37 UTC (GMT) 0 - 1111 1111 - 110 0000 0000 0000 0000 0000 = QNaN, Quiet Not a Number Jun 27 12:36 UTC (GMT) 0 - 0001 1010 - 111 1111 1101 1100 1111 1010 = 0.000 000 000 000 000 000 000 000 000 000 788 439 32 Jun 27 12:34 UTC (GMT) 0 - 1000 0111 - 100 1001 0000 0000 0000 0000 = 402 Jun 27 12:32 UTC (GMT) 0 - 1000 0011 - 101 1100 0111 1010 1110 0001 = 27.559 999 465 942 382 812 5 Jun 27 12:31 UTC (GMT) 0 - 0001 0001 - 110 1111 0100 1010 1110 0000 = 0.000 000 000 000 000 000 000 000 000 000 001 440 18 Jun 27 12:31 UTC (GMT) 0 - 0000 0000 - 000 0000 0000 0000 0000 0000 = 0 Jun 27 12:31 UTC (GMT) 0 - 0000 0000 - 111 1000 0000 0000 0000 0000 = 0 Jun 27 12:30 UTC (GMT) 1 - 0110 1111 - 100 1000 1010 1000 1000 0000 = -0.000 023 920 321 837 067 604 064 941 406 25 Jun 27 12:29 UTC (GMT) 0 - 1000 0010 - 111 1111 1111 1111 1010 1111 = 15.999 922 752 380 371 093 75 Jun 27 12:29 UTC (GMT) 0 - 1000 0110 - 001 1001 1100 0000 0000 0000 = 153.75 Jun 27 12:27 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