if (4 && 3) 
   printf("hura!\n");

 if (4 & 3)
   printf("hura hura\n");

 0 = false
 nenulove = true

 4 && 3   => 1

  100
& 011
=====
  000   = 0

----------------------------------

 printf("%d\n", 4 || 3);    // vypise 1
 printf("%d\n", 4 | 3);     // vypise 7

  100
| 011
======
  111   => 7

!4  => 0
~4  => 

~(00000100) => 11111011  = 251

 11111111  => 255
100000000

 x + ~x = 255

  2^N

 1 * 2 * 2 * 2 * 2 

2 * 1010111011  => 10101110110

10010101100 / 2  => 1001010110


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

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

  x AND 0 => vynulujem bit
  x AND 1 => zachovam bit

  0010101101101
& 0000111011100
=================
  0000101001100

==================================

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

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

 x OR 0 => zachova
 x OR 1 => nastavi na 1

  011011101011011
| 100110101111100
====================
  111111101111111

=================================

 A  B   A XOR B
================
 0  0      0    
 0  1      1 
 1  0      1
 1  1      0

 17 XOR 42  =>  59
 59 XOR 42  =>  17

 010001
 101010 
========
 111011  => 59

 0 XOR 0  => 0
 1 XOR 0  => 1

 0 XOR 1  => 1  
 1 XOR 1  => 0

cvicenie:

 mam x a chcem sa dozvediet hodnotu 3. bitu, pricom vysledok chcem
 v podobe 0 alebo 1.

 x = 0101111110
   &       1000
=================
     0000001000

 (x & (1 << 3)) >> 3
 (x >> 3) & 1

 1 << 0  = 1
 1 << 1  = 2
 1 << 2  = 4
 1 << 3  = 8

 chytim 1 a posuniem ju 3-krat vlavo
 vysledok zadnujem s x
 vysledok posuniem 3-krat vpravo

 mam x a chcem sa dozvediet hodnotu i. bitu, pricom vysledok chcem
 v podobe 0 alebo 1.
 
 (x >> i) & 1
=============
  x = 0101110101
  y = x >> 1
  z = z << 1
=============

dalsie cvicenie:
 
  mam x a chcem ziskat rovnaku hodnotu ale taku, kde na i. bite je 1:

  x | (1 << i)
 
  mam x a chcem ziskat rovnaku hodnotu ale taku, kde na i. bite je 0:
 
  x & ~(1 << i)   

  11111111101111