UMBC CMSC 313 -- Boolean Operations Segment Previous | Next

Boolean Operations

Computer hardware often requires that several pieces of information be encoded into a single word or byte. Many programs also pack the information to save memory, disk space or communications time. There for bit operations consist of extracting or altering packed information, with either Boolean operatons and shifts.

Boolean Operations

There are four Boolean operations that are the same as in the bitwise logical operators in C: AND, OR, XOR (eXclusive OR), and NOT. These operations can be expressed in the form of truth tables, where false is zero and true is one:

AND Truth Table

A B A and B
0 0 0
0 1 0
1 0 0
1 1 1

Examples

        1011 1010
    AND 0000 1111
        0000 1010
  

        1111 1111
    AND 0000 1111
        0000 1111

OR Truth Table

A B A or B
0 0 0
0 1 1
1 0 1
1 1 1

Examples

        1011 1010
    OR  0000 1111
        1011 1010
  

        1111 1111
    OR  0000 1111
        0000 1111

XOR Truth Table

A B A xor B
0 0 0
0 1 1
1 0 1
1 1 0

Examples

        1011 1010
    XOR 0000 1111
        1011 0000
  

        1111 1111
    OR  0000 1111
        1111 0000

NOT Truth Table

A not A
0 1
1 0

Examples

   NOT 1011 1010
       0100 0101
  

   NOT 0000 1111
       1111 0000

Format

These instructions come in the form (No more than one operand is a memory reference):
and/or/xor reg/mem, reg/mem/constant
not reg/mem
The functionality is:
and dest, source ; dest := dest AND source
or dest, source ; dest := dest OR source
xor dest, source ; dest := dest XOR source
not dest ; dest := NOT dest
The not operation is a 1's complement.

The and operation is used to clear bits and the or operation is used to set bits. The xor operation is used to toggle bits to the opposite set. The source is known as the mask for the operation.

AND,OR, XOR set the zero and sign flags according to the result and the overflow and carry flags are cleared.

The most common usage is to AND with a mask and then do a jz to determine if any of the bits corresponding to the mask were a 1. It is so common that they built a special instruction for it: 

  test  reg/mem, reg/mem/constant

An Example

We have already talked about how the keyboard hardware does not know anything about ASCII code, it just knows about scan codes. It is simply a number (using the low seven bits in a byte). The high bit is used to show if the key was pressed (set to 0) or released (set to 1). Then the BIOS (BASIC Input-Output System) has to do its magic. In addition to keeping track of which key was pressed, the BIOS also keeps track of additional information which is stored at memory address 0417h. This is the keyboard status information:

Bit Mask Interpretation
7 10000000B Insert mode on
6 01000000B Caps lock on
5 00100000B Num lock on
4 00010000B Scroll lock on
3 00001000B Alt key down (either if more than one)
2 00000100B Ctrl key down (either if more than one)
1 00000010B Left shift key down
0 00000001B Right shift key down

If the state is on (Insert mode is set) then the corresponding bit in the keyboard status will be true (1). Forunately for us, the BIOS does all the work to keep the keyboard status up-to-date.

Suppose that we are writing a program where we will do one thing if a certain key is pressed, but do something slightly different if the combination of Alt and a certain key is press. We will want to set a flag based on the Alt key and then proceed with the processing. We could do something like: 

          .MODEL  small
          .STACK  100h 
ALTKEY    EQU    00001000B
          .DATA 
kbdStatus DB     ?
altFlag   DB     0             ; Initialized to off
          .CODE
example1  PROC
          mov    ax, @data
          mov    ds, ax
;
; Do something to get the keyboard status
          mov    ax, kbdStatus 
          and    ax, ALTKEY
          jne    noAlt
          mov    bx, 1         ; Set the flag because the Alt key was pressed
          mov    bx, altFlag
noAlt:    
          mov    al,0
          mov    ah, 4ch
          int    21h
example1  ENDP
          END     example1

Another Example

The directory information for a disk file constains the date of last modification in a single work in the following format:
year - 1900 bits 9 - 15 (leftmost seven bits)
month bits 5 - 8 (January is 1)
day bits 0 - 4 (rightmost five bits)

Now we can do the following: 

           .MODEL  small
           .STACK  100h
YEARMASK   EQU    1111111000000000B
MONTHMASK  EQU    0000000111100000B 
DAYMASK    EQU    0000000000011111B 

           .DATA
modDate    DW     ?        ; The year is still in bits 9 - 15 
funnyYear  DW     ?        ; The month is still in bits 5 - 8
funnyMonth DW     ?        ; The day is in bits 0 - 4 and
                           ;  all other bits are off 
OKDay      DW     ?

           .CODE   
example2   PROC    
           mov    ax, @data 
           mov    ds, ax 
; 
           ...             ; Do something to get the 
                           ; modification date 

           mov    ax, modDate 
           and    ax, YEARMASK 
           mov    funnyYear, ax 
           mov    ax, modDate
           and    ax, MONTHMASK
           mov    funnyMonth, ax
           mov    ax, modDate 
           and    ax, DAYMASK
           mov    ax, OKDay
                           ; At this point, the values 
                           ;   are not usualable yet 
                    
           mov    al,0
           mov    ah, 4ch
           int    21h
example2   ENDP	 
           END    example2

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©2004, Gary L. Burt