Python doesn't have builtin unsigned types. You can use mathematical operations to compute a new int representing the value you would get in C, but there is no "unsigned value" of a Python int. The Python int is an abstraction of an integer value, not a direct access to a fixed-byte-size integer.
You could use the struct
Python built-in library:
Encode:
import struct
i = -6884376
print('{0:b}'.format(i))
packed = struct.pack('>l', i) # Packing a long number.
unpacked = struct.unpack('>L', packed)[0] # Unpacking a packed long number to unsigned long
print(unpacked)
print('{0:b}'.format(unpacked))
Out:
-11010010000110000011000
4288082920
11111111100101101111001111101000
Decode:
dec_pack = struct.pack('>L', unpacked) # Packing an unsigned long number.
dec_unpack = struct.unpack('>l', dec_pack)[0] # Unpacking a packed unsigned long number to long (revert action).
print(dec_unpack)
Out:
-6884376
[NOTE]:
>
is BigEndian operation.l
is long.L
is unsigned long.amd64
architecture int
and long
are 32bit, So you could use i
and I
instead of l
and L
respectively. To get the value equivalent to your C cast, just bitwise and with the appropriate mask. e.g. if unsigned long
is 32 bit:
>>> i = -6884376
>>> i & 0xffffffff
4288082920
or if it is 64 bit:
>>> i & 0xffffffffffffffff
18446744073702667240
Do be aware though that although that gives you the value you would have in C, it is still a signed value, so any subsequent calculations may give a negative result and you'll have to continue to apply the mask to simulate a 32 or 64 bit calculation.
This works because although Python looks like it stores all numbers as sign and magnitude, the bitwise operations are defined as working on two's complement values. C stores integers in twos complement but with a fixed number of bits. Python bitwise operators act on twos complement values but as though they had an infinite number of bits: for positive numbers they extend leftwards to infinity with zeros, but negative numbers extend left with ones. The &
operator will change that leftward string of ones into zeros and leave you with just the bits that would have fit into the C value.
Displaying the values in hex may make this clearer (and I rewrote to string of f's as an expression to show we are interested in either 32 or 64 bits):
>>> hex(i)
'-0x690c18'
>>> hex (i & ((1 << 32) - 1))
'0xff96f3e8'
>>> hex (i & ((1 << 64) - 1)
'0xffffffffff96f3e8L'
For a 32 bit value in C, positive numbers go up to 2147483647 (0x7fffffff), and negative numbers have the top bit set going from -1 (0xffffffff) down to -2147483648 (0x80000000). For values that fit entirely in the mask, we can reverse the process in Python by using a smaller mask to remove the sign bit and then subtracting the sign bit:
>>> u = i & ((1 << 32) - 1)
>>> (u & ((1 << 31) - 1)) - (u & (1 << 31))
-6884376
Or for the 64 bit version:
>>> u = 18446744073702667240
>>> (u & ((1 << 63) - 1)) - (u & (1 << 63))
-6884376
This inverse process will leave the value unchanged if the sign bit is 0, but obviously it isn't a true inverse because if you started with a value that wouldn't fit within the mask size then those bits are gone.
just use abs for converting unsigned to signed in python
a=-12
b=abs(a)
print(b)
Output: 12
Since version 3.2 :
def toSigned(n, byte_count):
return int.from_bytes(n.to_bytes(byte_count, 'little'), 'little', signed=True)
output :
In [8]: toSigned(5, 1)
Out[8]: 5
In [9]: toSigned(0xff, 1)
Out[9]: -1
Source: Stackoverflow.com