Why must I use the ~ operator when clearing a bit?
This question already has an answer here:
If you want to set a bit at third place from the right :
Y : 01001000
1 << 2 : 00000100
Y | (1 << 2) : 01001100 The | is OR, bits are set to 1 if any is 1.
If you want to remove the bit :
1 << 2 : 00000100
~(1 << 2) : 11111011 The ~ is NOT, bits are inversed
Y : 01001100
Y & ~(1 << 2) : 01001000 The & is AND, bits are set to 1 if both are 1.
I suggest you read more about Bitwise operators
No, ~ has nothing to do with interpreting the number as negative: tilde ~ operator interprets the number as a pattern of bits, which it then inverts (ie replaces zeros with ones and ones with zeros). In fact, if you apply ~ to an unsigned value, the result would remain positive.
Recall that 1 << k expression produces a pattern of all zeros and a single 1 at the position designated by k . This is a bit mask that can be used to force bit at position k to 1 by applying OR operation.
Now consider what happens when you apply ~ to it: all 0 s would become 1 s, and the only 1 would become zero. Hence, the result is a bit mask suitable for forcing a single bit to zero by applying AND operation.
The ~ operator turns all of the 0's to 1's and all of the 1's to 0's. In order to clear the bint in position n you want to and it will all ones and a zero in the nth position so shift a one to the nth position and ~ invert all the bits.
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