Why is ~0xF equal to 0xFFFFFFF0 on a 32-bit machine?

Question

Why is ~0xF equal to 0xFFFFFFF0?

Also, how is ~0xF && 0x01 = 1? Maybe I don't get 0x01 either.

Answer 1

Question 1

Why is ~0xF equal to 0xFFFFFFF0?

First, this means you run this on a 32-bit machine. That means 0xF is actually 0x0000000F in hexadecimal,

And that means 0xF is 0000 0000 0000 0000 0000 0000 0000 1111 in binary representation.

The ~ operator means the NOT operation. Tt changes every 0 to 1 and every 1 to 0 in the binary representation. That would make ~0xF to be: 1111 1111 1111 1111 1111 1111 1111 0000 in binary representation.

And that is actually 0xFFFFFFF0.

Note that if you do this on a 16-bit machine, the answer of ~0xF would be 0xFFF0.

Question 2

You wrote the wrong statement, it should be 0xF & 0x1. Note that 0x1 0x01, 0x001, and 0x0001 are all the same. So let’s change this hexdecimal number to binary representation:

0xF would be:

0000 0000 0000 0000 0000 0000 0000 1111

and 0x1 would be:

0000 0000 0000 0000 0000 0000 0000 0001

The & operation follows the following rules:

0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
1 & 1 = 1

So doing that to every bit, you get the result:

0000 0000 0000 0000 0000 0000 0000 0001

which is actually 0x1.

Additional

| means bitwise OR operation. It follows:

0 | 0 = 0
0 | 1 = 1
1 | 0 = 1
1 | 1 = 1

^ means bitwise XOR operation. It follows:

0 ^ 0 = 0
0 ^ 1 = 1
1 ^ 0 = 1
1 ^ 1 = 0

You can get more information here.

Answer 2

If you store 0xF in a 32-bit "int", all 32 bits flip, so ~0xF = 0xFFFFFFF0.

See this: http://teaching.idallen.com/cst8214/08w/notes/bit_operations.txt

They give a good explanation