问题
Here is the python code snippet:
1 & -1 # 1
2 & -2 # 2
3 & -3 # 1
...
It seems any n & -n always return right most (last) bit, I don't really know why. Can someone help me to understand this?
回答1:
It's due to the way that negative numbers are represented in binary, which is called two's complement representation.
To create the two's complement of some number n (in other words, to create the representation of -n):
- Invert all the bits
- Add 1
So in other words, when you write 1 & -1 it really means 1 & ((~1)+1). The initial ~1 gives the value 1111110 and adding one gives 11111111. (Let's stick with 8 bits for these examples.) ANDing that values with 1 gives just 1.
In the next case, 2 & -2 means 2 & ((~2)+1). Inverting 2 gives 11111101 and adding one gives 11111110. Then AND with 2 (10 in binary) gives 2.
Finally 3 & -3 means 3 & ((~3)+1). Invert 3 gives 11111100, add 1 gives 11111101, and AND with 3 (11 binary) gives 1.
回答2:
~x = -1 -x
so
-x = ~x + 1
When you take the compliment of x (~x), all the 0-bits turn to 1 and all the 1 bits turn to zero. e.g. 101100 -> 010011.
When you add 1, the consecutive 1s on the right change to 0 and the first 0 bit gets set to 1: 010011 -> 010100
If you & that with the original, the 0 bits at the top that changed to 1 come out 0. The 1 bits at the bottom you flipped to 0 by adding come out 0. Only the rightmost 1 bit, which turned into the rightmost 0 bit in the complement and got reset to 1 by the addition, is 1 on both sides: 101100 & 010100 -> 000100
回答3:
The integers are stored in the memory in the binary form. The non-negative integers are stored as it is in as they are in their binary but the negative numbers are stored in the two's complement form. For example take any arbitrary number 158.
158 = 0000000010011110
while its negative ie.
-158 = 1111111101100010
Take for any number and bitwise AND with its negative then you will get rightmost set bit. This is because in the process of conversion of two's complement we start from right and put the bits as it is till we encounter our first set bit. That is the right most set bit is being written as it is. Then We flip the digits in the left starting from here. However Above procedure is just a shortcut method for calculating 2's complement. For actual process, you have to first take 1's complement of the number(flipping all bits set to unset and unset to set) and then add 1 to the whole result. Now here is insight of why does this shortcut works everytime Why does this two's complement shortcut work?
This will give you more insight https://www.geeksforgeeks.org/efficient-method-2s-complement-binary-string/
And take some numbers and work on examples and see yourself.
回答4:
The catch is in calculating the 2's complement not in performing bitwise AND operation, AND always needs same things to be same to give true.
What 2's complement is doing is, it is making the rightmost digit of both numbers taking part in bitwise AND operation, Look below how 2's complement operation is being calculated and you will find that we represent the number(decimal) in binary and then to calculate the 2's complement we start from right and copy all binary digits the same until we see the first 1, and when we see the first 1 then we make all left of it as opposite the the previous.
The catch is:
Say you want 2's complement of 4 i.e (-4) , so what it says is represent the decimal in binary and copy all bits(0) from right till you see the first 1 and after that reverse all 0s and 1s.
Example: we want 2's complement of 6 -> 0 1 1 0 = 0 0 1 0 , From right of 0110 we start and till we see the first 1 we copy exactly what is in first then we reverse all 0s and 1s.
Another operation with 2's complement
0100
1100 ( Bold are just same as above)
Now it is obvious that when you do bitwise AND then the right most digit will only make through the AND operation as you need both to be equal to get through AND.
来源：https://stackoverflow.com/questions/54375295/why-n-bitwise-and-n-always-return-the-right-most-bit-last-bit