# Logical Operations

Before we leave the topic of binary and hex numbers, it's probably a good time to talk about logical operations. You are probably used to using logic in your code, maybe in a statement like `if ((conditionA) and (conditionB))`. With programs that deal with hardware, you often have to manipulate individual bits in numbers.

### AND operation

Here are all possible results of an AND operation between two one-bit numbers.

 x y result 0 0 0 0 1 0 1 0 0 1 1 1

The result of an AND is only one if both values are one. When you combine two numbers with the AND operation, each bit of one number is ANDed with the corresponding bit of the other number. The result is stored in the that bit of the destination. It may be easier just to look at the math:

 binary hex source x 0 1 1 0 1 0 1 1 \$6B source y 1 1 0 1 0 0 1 0 \$D2 x AND y 0 1 0 0 0 0 1 0 \$42

In C, the logical and operation is a single ampersand '&'.

### OR operation

The OR operation works similarly, the difference is that the result will be one if either x or y is a one.

 x y result 0 0 0 0 1 1 1 0 1 1 1 1

 binary hex source x 0 1 1 0 1 0 1 1 \$6B source y 1 1 0 1 0 0 1 0 \$D2 x OR y 1 1 1 1 1 0 1 1 \$FB

In C, the logical or operation is the vertical bar '|'.

### Why do we care?

With a lot of older processors but particularly with arcade games, the game will often want to work with only one bit of a number. You'll see a lot of code like this:

```   /* Example 1: read the control panel */
char *buttons_ptr = (char *)0x2043;
char buttons = *buttons_ptr;
if (buttons & 0x4)
HandleLeftButton();

/* Example 2: turn on one LED on the control panel */
char * LED_pointer = (char *) 0x2089;
char led = *LED_pointer;
led = led | 0x40; //set LED controlled by bit 6
*LED_pointer = led;

/* Example 3: turn off one LED */
char * LED_pointer = (char *) 0x2089;
char led = *LED_pointer;
led = led & 0xBF; //mask out bit 6
*LED_pointer = led;

```

In example 1, memory mapped location \$2043 is the location of the buttons on the control panel. This code reads and responds to the pushed button. (Of course, in Space Invaders this code will be in assembly language!)

In example 2, the game wants to light a LED which is located at bit 6 of memory mapped location \$2089. The code must read the value that is already there, modify only the one bit, and write it back out.

In example 3, the LED from example 2 needs to be turned off, so the code needs to clear bit 6 of \$2089. The way to do that is to AND the LED control byte with a value with only bit 6 clear. This will effect only bit 6, and leave the rest of the bits alone.

This is usually called a mask. A mask is usually written using C's not operator, the tilde '~'. So instead of writing `0xBF`, I'd write `~0x40`, which yields the same number but I'd not have to think about it as hard.

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