Let\'s say I have this number i = -6884376
.
How do I refer to it as to an unsigned variable?
Something like (unsigned long)i
in C.
just use abs for converting unsigned to signed in python
a=-12
b=abs(a)
print(b)
Output: 12
Assuming:
(unsigned long)
you mean unsigned 32-bit integer,then you just need to add 2**32 (or 1 << 32)
to the negative value.
For example, apply this to -1:
>>> -1
-1
>>> _ + 2**32
4294967295L
>>> bin(_)
'0b11111111111111111111111111111111'
Assumption #1 means you want -1 to be viewed as a solid string of 1 bits, and assumption #2 means you want 32 of them.
Nobody but you can say what your hidden assumptions are, though. If, for example, you have 1's-complement representations in mind, then you need to apply the ~
prefix operator instead. Python integers work hard to give the illusion of using an infinitely wide 2's complement representation (like regular 2's complement, but with an infinite number of "sign bits").
And to duplicate what the platform C compiler does, you can use the ctypes
module:
>>> import ctypes
>>> ctypes.c_ulong(-1) # stuff Python's -1 into a C unsigned long
c_ulong(4294967295L)
>>> _.value
4294967295L
C's unsigned long
happens to be 4 bytes on the box that ran this sample.
python does not have any builtin to convert int to unsigned int.Instead it is having long for longer range.
>>> val = 9223372036854775807 (maximum value of int 64)
>>> type(val)
<type 'int'>
>>> val += 1
>>> type(val)
<type 'long'>
By increasing the value of val by 1, I exceed the limit of a signed 64 bit integer and the value is converted to a long. If Python had used or converted to an unsigned integer, val would still have been an int. Or, not long.
Signed integers are represented by a bit, usually the most significant bit, being set to 0 for positive numbers or 1 for negative numbers. What val & 0xff does is actually val & 0x000000ff (on a 32 bit machine). In other words, the signed bit is set to 0 and an unsigned value is emulated.
An example:
>>> val = -1
>>> val & 0xff
255
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.
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
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.