bitwise not operator .

Question

Why bitwise operation (~0); prints -1 ? In binary , not 0 should be 1 . why ?

Answer 1

Because ~ is not binary inversion, it’s bitwise inversion. Binary inversion would be ! and can (in Java) only be applied to boolean values.

Answer 2

In standard binary encoding, 0 is all 0s, ~ is bitwise NOT. All 1s is (most often) -1 for signed integer types. So for a signed byte type:

0xFF = -1    // 1111 1111
0xFE = -2    // 1111 1110
...
0xF0 = -128  // 1000 0000
0x7F = 127   // 0111 1111
0x7E = 126   // 0111 1110
...
0x01 = 1     // 0000 0001
0x00 = 0     // 0000 0000

Answer 3

It's binary inversion, and in second complement -1 is binary inversion of 0.

Answer 4

What you are actually saying is ~0x00000000 and that results in 0xFFFFFFFF. For a (signed) int in java, that means -1.

Answer 5

~ is a bitwise operator.

~0 = 1 which is -1 in 2's complement form  

http://en.wikipedia.org/wiki/Two's_complement

Some numbers in two's complement form and their bit-wise not ~ (just below them):

0 1 1 1 1 1 1 1 = 127
1 0 0 0 0 0 0 0 = −128

0 1 1 1 1 1 1 0 = 126
1 0 0 0 0 0 0 1 = −127

1 1 1 1 1 1 1 1 = −1
0 0 0 0 0 0 0 0 = 0

1 1 1 1 1 1 1 0 = −2
0 0 0 0 0 0 0 1 = 1

1 0 0 0 0 0 0 1 = −127
0 1 1 1 1 1 1 0 = 126

1 0 0 0 0 0 0 0 = −128
0 1 1 1 1 1 1 1 = 127

Answer 6

0 here is not a bit. It is a byte (at least; or more) - 00000000. Using bitwise or we will have 11111111. It is -1 as signed integer...

Answer 7

For 32 bit signed integer

~00000000000000000000000000000000=11111111111111111111111111111111 (which is -1)

Answer 8

You could imagine the first bit in a signed number to be -(2^{x -1}) where x is the number of bits.

So, given an 8-bit number, the value of each bit (in left to right order) is:

-128 64 32 16 8 4 2 1

Now, in binary, 0 is obviously all 0s:

    -128 64 32 16 8 4 2 1
0      0  0  0  0 0 0 0 0 = 0

And when you do the bitwise not ~ each of these 0s becomes a 1:

     -128 64 32 16 8 4 2 1
~0      1  1  1  1 1 1 1 1
 =   -128+64+32+16+8+4+2+1 == -1

This is also helpful in understanding overflow:

     -128 64 32 16 8 4 2 1
126     0  1  1  1 1 1 1 0  =  126
 +1     0  1  1  1 1 1 1 1  =  127
 +1     1  0  0  0 0 0 0 0  = -128  overflow!

Answer 9

You are actually quite close.

In binary , not 0 should be 1

Yes, this is absolutely correct when we're talking about one bit.

HOWEVER, an int whose value is 0 is actually 32 bits of all zeroes! ~ inverts all 32 zeroes to 32 ones.

System.out.println(Integer.toBinaryString(~0));
// prints "11111111111111111111111111111111"

This is the two's complement representation of -1.

Similarly:

System.out.println(Integer.toBinaryString(~1));
// prints "11111111111111111111111111111110"

That is, for a 32-bit unsigned int in two's complement representation, ~1 == -2.

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Further reading:

• Two's complement
  • This is the system used by Java (among others) to represent signed numerical value in bits
• JLS 15.15.5 Bitwise complement operator ~
  • "note that, in all cases, ~x equals (-x)-1"