two's complment binary

[Skaperen]

d'rat ... the bin() function does not give the two's complement form of negative numbers. anyone know of something that does that?

[Gribouillis]

In python, integers conceptually have a semi-infinite number of bits, so for example if

>>> bin(37) '0b100101'

the integer 37 is represented by the infinite sequence ...000000100101. It means that the two's complement ~37 is a sequence starting with an infinite number of 1's, but of course it is not stored this way. Instead, it is stored as the negative number -38. For every integer, one has

assert ~x == -(x+1)

So you don't see the infinite sequence, but with regards to bitwise operations, you need to consider negative integers as a sequence of bits starting with an infinite number of 1s.

For example the sequence of bits of -23762 is the two's complement of the bits of 23761. You can confirm this by using a 32 bits mask for example

>>> mask = (1 << 32) - 1 >>> bin(mask) '0b11111111111111111111111111111111' >>> bin(23761) '0b101110011010001' >>> bin(-23762 & mask) '0b11111111111111111010001100101110'

[Skaperen]

now i see the complication. the leading sequence is infinite making my attempts to do things in just 8 bits not workable with 16 bit values. i was trying to get the low order 8 bits from the way a two's complement based machine would store a number that might need 16 bits and/or might be negative. once i realized i was not getting two's complement form from bin(), i tried to do it myself in just 8 bits. i was thinking that we should have a bin2() that would give a two's complement form for negative numbers. but how many 1's would they need in front? 32? 64? 65536?

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so now i made this:

bin2.py

def bin2(n,bits=64): return bin(n&((1<<bits)-1)) for i in range(-16,16): print(i,bin2(i,5))

i plan to keep the first 0b10 lines.

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now, how to convert in reverse? is '0b10000000' -128 or 128?