3 bits can hold up to a maximum number of 7 (4 + 2 + 1). I'm trying to calculate this using a bitwise operation.
3 is 0b011 ~3 is 0b100 Doing a bitwise OR I would expect 0b111 (i.e. 7). Instead I get
int result = (~3) | 3; printf("%i\n", result); -1
What am I doing wrong?
You are doing everything right: N | ~N results in a number with binary representation consisting of all ones. Such number is interpreted as -1 in two's compliment representation of negative numbers.
~3 is 111111..111100, so the last three bits are indeed 0b100.
3 is an integer literal of type int. The ~ operator isn't well defined for signed integers.
How many bits wide is an int? You seem to think it's three bits wide. Certainly not correct! Guess again. What is ~0u? Try printf("%u\n", ~0u);. What about ~1u? ... and ~2u? Do you notice a pattern?
Note the u suffix, which tells the compiler that it's an unsigned literal. You can't work with signed integer types with the ~ operator... Well, you can, but you might run into trap representations and negative zeros, according to 6.2.6.2 of n1570.pdf. Using a trap representation is undefined behaviour. That might work on your system, but only by coincidence. Do you want to rely upon coincidence?
Similarly, I suggest using the %u directive to print unsigned values, as %d would produce undefined behaviour according to 7.21.6.1p29 of n1570.pdf.
When you do ~3 you are inverting the bits that make up 3 - so you turn 0000 0000 0000 0000 0000 0000 0000 0011 into 1111 1111 1111 1111 1111 1111 1111 1100. Since the high bit is set, this is interpreted as a negative number - all 1s is -1, one less than that is -2, one less -3 and so on. This number is the signed 32 bit integer for -4.
If you binary OR this with 3, you get all 1s (by definition) - which is the signed 32 bit integer for -1.
Your only problem is that you think you are working with 3 bit numbers, but you are actually working with 32 bit numbers.
After doing this in the code
int result = (~3) | 3; Add this line
result= result & 0x07 This will give you the answer that you expect.
#include <stdio.h> int main (){ unsigned d3 = 0b011; unsigned invd3 = ~d3; unsigned d4 = 0b100; unsigned result = d3 | invd3; printf("%X\n", result);//FFFFFFFF result = d3 | d4; printf("%X\n", result);//7 return 0; } 
