Why bitwise operation (~0);
prints -1 ? In binary , not 0 should be 1 . why ?
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
.
Further reading:
~x
equals (-x)-1
"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...
~
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 = −1280 1 1 1 1 1 1 0 = 126
1 0 0 0 0 0 0 1 = −1271 1 1 1 1 1 1 1 = −1
0 0 0 0 0 0 0 0 = 01 1 1 1 1 1 1 0 = −2
0 0 0 0 0 0 0 1 = 11 0 0 0 0 0 0 1 = −127
0 1 1 1 1 1 1 0 = 1261 0 0 0 0 0 0 0 = −128
0 1 1 1 1 1 1 1 = 127
You could imagine the first bit in a signed number to be -(2x -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!