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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 References
Step1:  Binary representation of an integer
With formatting
Step2:  with bin()
Step3:  Binary representation of a float
radix point
Step4:  Representation errors
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Python Code:
import sys
import math
import ctypes
import struct
sys.float_info
sys.int_info
format(0.125, '.12g')
format(0.101, '.12g')
0.3
0.1
0.1 + 0.1 + 0.1 == 0.3
f"{0x1234:b}"
Explanation: References:
- https://en.wikipedia.org/wiki/IEEE_754
- https://en.wikipedia.org/wiki/Double-precision_floating-point_format
- https://docs.python.org/3/library/functions.html#round
- https://docs.python.org/3/tutorial/floatingpoint.html
- https://docs.python.org/3/library/functions.html#bin
End of explanation
a = 1
"{0:b}".format(a)
Explanation: Binary representation of an integer
With formatting
End of explanation
bin(a)
Explanation: with bin()
End of explanation
sys.float_info
a = 1.0
a
bin(ctypes.c_uint.from_buffer(ctypes.c_float(a)).value)
bin(struct.unpack('!i',struct.pack('!f',a))[0])
a = 2.0
a
bin(ctypes.c_uint.from_buffer(ctypes.c_float(a)).value)
bin(struct.unpack('!i',struct.pack('!f',a))[0])
a = 4.0
a
bin(ctypes.c_uint.from_buffer(ctypes.c_float(a)).value)
bin(struct.unpack('!i',struct.pack('!f',a))[0])
a = 8.0
a
bin(ctypes.c_uint.from_buffer(ctypes.c_float(a)).value)
bin(struct.unpack('!i',struct.pack('!f',a))[0])
a = 2.0**500
a
bin(ctypes.c_uint.from_buffer(ctypes.c_float(a)).value)
bin(struct.unpack('!i',struct.pack('!f',a))[0])
a = -2.0
a
bin(ctypes.c_uint.from_buffer(ctypes.c_float(a)).value)
bin(struct.unpack('!i',struct.pack('!f',a))[0])
Explanation: Binary representation of a float
radix point: the symbol used in numerical representations to separate the integer part of a number (to the left of the radix point) from its fractional part (to the right of the radix point)
Double precision:
- sign (+/-): 1 bit
- exponent: 11 bits
- significand precision: 53 bits (52 explicitly stored)
$e$ biased exponent
$(-1)^{sign} \left( 1.b_{p-1} b_{p-2} ... b_{0} \right)_{2} \cdot 2^{e-1023}$
$(-1)^{sign} \left( 1 + \sum_{i=1}^{p} b_{52-i} 2^{-i} \right) \cdot 2^{e-1023}$
End of explanation
sum([0.1] * 10) == 1.0
math.fsum([0.1] * 10) == 1.0
Explanation: Representation errors
End of explanation
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