CODEWITHSUNDEEP
c
Maverick
Maverick

Reputation: 19

How bitwise or and not operator works?

#include<stdio.h>
void main()
{
  int a, b, c;
  a=5;
  b=8;
  c= ~(a|b);
  printf("%d",c);
}

The expected the output is -13 but the results shows -14. How is it - 14? .

Upvotes: 1

Views: 62

Answers (5)

Shawn
Shawn

Reputation: 721

It's because of 2's complement.

Using 8 bit signed values (it works the same with 32 bit signed, but this is easier to demonstrate):

8:        00001000
5:        00000101
8 | 5:    00001101
~(8 | 5): 11110010 = 242 (unsigned)

Quick way to do 2's complement when the MSb is high is to subtract 2^n (here n is 8, 2^8 is 256)

242-256=-14

Upvotes: 0

Jo&#227;o Neto
Jo&#227;o Neto

Reputation: 1840

Let's assume you use 8-bit integers:

  • a=5 is represented as 00000101
  • b=8 is represented as 00001000
  • a|b is 00001101

When you invert the bits, it will be 11110010, which in Two's complement is -14.

Upvotes: 0

dbush
dbush

Reputation: 223689

This is due to two's complement representation of negative numbers. Two's complement is performed by inverting all bits then adding one.

First, you have a|b (for simplicity's sake I'll show only the low order 8 bits):

  a  00000101  5
| b  00001000  8
 ------------
     00001101  13

Then the bitwise NOT:

   ~ 00001101  13
   ----------
     11110010  -14

Performing a bitwise NOT on a positive value does not give you its negative value, it gives you one less.

Upvotes: 0

Lundin
Lundin

Reputation: 213276

Because you have a 2's complement computer.

  • 5 | 8 gives binary 0101 | 1000 = 1101 = 13 dec.
  • Invert this with ~ and you get 1111....0010.
  • Representing that as a signed type, then in 1's complement this would have been -13 dec.
  • But for 2's complement we subtract 1 and get -14 dec.

For the same reason as ~0 gives 2's complement -1 and not -0.

Upvotes: 2

bruno
bruno

Reputation: 32586

you confuse the complement to one (~) and the complement to two / unary operator -

~(5|8) is -(5|8) - 1 so -13 - 1 so -14

Upvotes: 0

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