Understanding Bitwise Operators and Literals

[TerriSchiavosGhost]

So here are the lines of code I need help understanding:

  unsigned s = 555;
  int i = (s >> 4) & ~(~0 << 3);

I think I understand most of this, but I'm not sure I understand why ~0 = -1.

How exactly is the literal 0 being stored? I initially assumed that 0 would be treated as a signed int and stored as 000...00000 with the left-most bit being the sign bit, but if that were the case then ~0 would have to be 111...11111, or -2^16, right?

So it seems that it must be stored as only 2 bits, 00, with the left-most bit being the sign bit, and ~0 being stored as 11, or -1, and then the computer allocates more space in the registers or something as it uses the left shift bitwise operator, to get 11000, or -8. This seems wrong to me, but it's the only way I've been able to make sense of this...

Anyway, tell me why I'm wrong.

[vlad from moscow]

It would be more correctly to write

unsigned s = 555;
int i = (s >> 4) & ~(~0u << 3);

0u is unsigned int. ~0u produces all ones, that is 11111111 11111111 11111111 11111111 if sizeof( unsigned int ) == 4.
~0u << 3 produces

11111111 11111111 11111111 11111000

~( ~0u << 3 )

will invert bits

00000000 00000000 00000000 00000111.

So the aim of the expression

(s >> 4) & ~(~0u << 3)

is to get the last three bits (the less significant) of s >> 4.

[TerriSchiavosGhost]

That definitely makes sense, but I still don't understand how the 0 is being treated if not cast as an unsigned int.

Basically, I don't understand why

 std::cout << ~0;
 std::cout << (~0 << 3);

yields -1 and -8.

[vlad from moscow]

0 is a integer literal of type int. So then you use ~0 you get

11111111 11111111 11111111 11111111

that is equal to -1 because it is a value of type int and the sign bit is set. After ~0 << 3 you get

11111111 11111111 11111111 11111000

Again it is a negative value because its sign bit is set. It is equal to -8. If you will add 8 that is represented as

00000000 00000000 00000000 00001000

you will get

11111111 11111111 11111111 11111000
+
00000000 00000000 00000000 00001000
==========================
00000000 00000000 00000000 00000000

that is as you see the result will be equal to zero.

[TerriSchiavosGhost]

ah ok I wasn't understanding how negative integers were represented.

Thanks.