How does the bitwise complement operator (~ tilde) work?

Question

Why is it that ~2 is equal to -3? How does ~ operator work?

Answer 1

int a=4; System.out.println(~a); Result would be :-5

'~' of any integer in java represents 1's complement of the no. for example i am taking ~4,which means in binary representation 0100. first , length of an integer is four bytes,i.e 4*8(8 bits for 1 byte)=32. So in system memory 4 is represented as 0000 0000 0000 0000 0000 0000 0000 0100 now ~ operator will perform 1's complement on the above binary no

i.e 1111 1111 1111 1111 1111 1111 1111 1011->1's complement the most significant bit represents sign of the no(either - or +) if it is 1 then sign is '-' if it is 0 then sign is '+' as per this our result is a negative number, in java the negative numbers are stored in 2's complement form, the acquired result we have to convert into 2's complement( first perform 1's complement and just add 1 to 1's complement). all the one will become zeros,except most significant bit 1(which is our sign representation of the number,that means for remaining 31 bits 1111 1111 1111 1111 1111 1111 1111 1011 (acquired result of ~ operator) 1000 0000 0000 0000 0000 0000 0000 0100 (1's complement)

1 (2's complement)

1000 0000 0000 0000 0000 0000 0000 0101 now the result is -5 check out this link for the video <[Bit wise operators in java] https://youtu.be/w4pJ4cGWe9Y

Answer 2

Javascript tilde (~) coerces a given value to the one's complement--all bits are inverted. That's all tilde does. It's not sign opinionated. It neither adds nor subtracts any quantity.

0 -> 1
1 -> 0
...in every bit position [0...integer nbr of bits - 1]

On standard desktop processors using high-level languages like JavaScript, BASE10 signed arithmetic is the most common, but keep in mind, it's not the only kind. Bits at the CPU level are subject to interpretation based on a number of factors. At the 'code' level, in this case JavaScript, they are interpreted as a 32-bit signed integer by definition (let's leave floats out of this). Think of it as quantum, those 32-bits represent many possible values all at once. It depends entirely on the converting lens you view them through.

JavaScript Tilde operation (1's complement)

BASE2 lens
~0001 -> 1110  - end result of ~ bitwise operation

BASE10 Signed lens (typical JS implementation)
~1  -> -2 

BASE10 Unsigned lens 
~1  -> 14 

All of the above are true at the same time.

Answer 3

I know the answer for this question is posted a long back, but I wanted to share my answer for the same.

For finding the one’s complement of a number, first find its binary equivalent. Here, decimal number 2 is represented as 0000 0010 in binary form. Now taking its one’s complement by inverting (flipping all 1’s into 0’s and all 0’s into 1’s) all the digits of its binary representation, which will result in:

0000 0010 → 1111 1101

This is the one’s complement of the decimal number 2. And since the first bit, i.e., the sign bit is 1 in the binary number, it means that the sign is negative for the number it stored. (here, the number referred to is not 2 but the one’s complement of 2).

Now, since the numbers are stored as 2’s complement (taking the one’s complement of a number plus one), so to display this binary number, 1111 1101, into decimal, first we need to find its 2’s complement, which will be:

1111 1101 → 0000 0010 + 1 → 0000 0011

This is the 2’s complement. The decimal representation of the binary number, 0000 0011, is 3. And, since the sign bit was one as mentioned above, so the resulting answer is -3.

Hint: If you read this procedure carefully, then you would have observed that the result for the one’s complement operator is actually, the number (operand - on which this operator is applied) plus one with a negative sign. You can try this with other numbers too.

Answer 4

Basically action is a complement not a negation .

Here x= ~x produce results -(x+1) always .

x = ~2

-(2+1)

-3

Answer 5

tl;dr ~ flips the bits. As a result the sign changes. ~2 is a negative number (0b..101). To output a negative number ruby prints -, then two's complement of ~2: -(~~2 + 1) == -(2 + 1) == 3. Positive numbers are output as is.

There's an internal value, and its string representation. For positive integers, they basically coincide:

irb(main):001:0> '%i' % 2
=> "2"
irb(main):002:0> 2
=> 2

The latter being equivalent to:

irb(main):003:0> 2.to_s
"2"

~ flips the bits of the internal value. 2 is 0b010. ~2 is 0b..101. Two dots (..) represent an infinite number of 1's. Since the most significant bit (MSB) of the result is 1, the result is a negative number ((~2).negative? == true). To output a negative number ruby prints -, then two's complement of the internal value. Two's complement is calculated by flipping the bits, then adding 1. Two's complement of 0b..101 is 3. As such:

irb(main):005:0> '%b' % 2
=> "10"
irb(main):006:0> '%b' % ~2
=> "..101"
irb(main):007:0> ~2
=> -3

To sum it up, it flips the bits, which changes the sign. To output a negative number it prints -, then ~~2 + 1 (~~2 == 2).

The reason why ruby outputs negative numbers like so, is because it treats the stored value as a two's complement of the absolute value. In other words, what's stored is 0b..101. It's a negative number, and as such it's a two's complement of some value x. To find x, it does two's complement of 0b..101. Which is two's complement of two's complement of x. Which is x (e.g ~(~2 + 1) + 1 == 2).

In case you apply ~ to a negative number, it just flips the bits (which nevertheless changes the sign):

irb(main):008:0> '%b' % -3
=> "..101"
irb(main):009:0> '%b' % ~-3
=> "10"
irb(main):010:0> ~-3
=> 2

What is more confusing is that ~0xffffff00 != 0xff (or any other value with MSB equal to 1). Let's simplify it a bit: ~0xf0 != 0x0f. That's because it treats 0xf0 as a positive number. Which actually makes sense. So, ~0xf0 == 0x..f0f. The result is a negative number. Two's complement of 0x..f0f is 0xf1. So:

irb(main):011:0> '%x' % ~0xf0
=> "..f0f"
irb(main):012:0> (~0xf0).to_s(16)
=> "-f1"

In case you're not going to apply bitwise operators to the result, you can consider ~ as a -x - 1 operator:

irb(main):018:0> -2 - 1
=> -3
irb(main):019:0> --3 - 1
=> 2

But that is arguably of not much use.

An example Let's say you're given a 8-bit (for simplicity) netmask, and you want to calculate the number of 0's. You can calculate them by flipping the bits and calling bit_length (0x0f.bit_length == 4). But ~0xf0 == 0x..f0f, so we've got to cut off the unneeded bits:

irb(main):014:0> '%x' % (~0xf0 & 0xff)
=> "f"
irb(main):015:0> (~0xf0 & 0xff).bit_length
=> 4

Or you can use the XOR operator (^):

irb(main):016:0> i = 0xf0
irb(main):017:0> '%x' % i ^ ((1 << i.bit_length) - 1)
=> "f"

Answer 6

Simply ...........

As 2's complement of any number we can calculate by inverting all 1s to 0's and vice-versa than we add 1 to it..

Here N= ~N produce results -(N+1) always. Because system store data in form of 2's complement which means it stores ~N like this.

  ~N = -(~(~N)+1) =-(N+1). 

For example::

  N = 10  = 1010
  Than ~N  = 0101
  so ~(~N) = 1010
  so ~(~N) +1 = 1011 

Now point is from where Minus comes. My opinion is suppose we have 32 bit register which means 2^31 -1 bit involved in operation and to rest one bit which change in earlier computation(complement) stored as sign bit which is 1 usually. And we get result as ~10 = -11.

~(-11) =10 ;

The above is true if printf("%d",~0); we get result: -1;

But printf("%u",~0) than result: 4294967295 on 32 bit machine.