This is what most tutorials usually start with. After reading this you will be confused, it is normal, you will get this by usage. Return to this chapter whenever needed. So let's go.

You know computer uses "bits", which are variables that can contain one of two possible values, 0 or 1. When value of bit is 0, we say it is "clear", when it is 1, we say it is "set"

terms:Now how can we store number in such bits. It is similar to storing word in two bytes (chapter 4.2, reread it). One bit contains value 0 or 1, so number that consists from only one bit can contain only values 0 and 1. When we add another bit, we can still store 0 and 1 in first bit, but we have another bit which now can hold 2*(0 or 1). Then another bit holds 4*(0 or 1), then 8*(0 or 1) etc.

bit- variable containing 0 or 1

clear bit- bit containing 0

set bit- bit containing 1

Like i said before, byte consists of 8 bits. So it can hold value

```
1*(0 or 1)+2*(0 or 1)+4*(0 or 1)+8*(0 or 1)+16*(0 or 1)+32*(0
or 1)+64*(0 or 1)+128*(0 or 1)
```

which is value from 0 (when all bits
are 0) to `1+2+4+8+16+32+64+128`

= 255 (when all bits are
1). See it?It is similar for word, just with 16 bits instead of 8, check it yourself if you wish.

Now some terms: bit which holds 1*(0 or 1) is bit 0, next, which holds 2*(0 or 1) is bit 1, and so until bit 7 which holds 128*(0 or 1). So bits are enumerated from 0, not from 1 as you would maybe await. bit 0 is called "low order bit", highest bit (which holds greatest value) is "high order bit". For example high order bit of some byte value is bit 7, and high order bit of word value is bit 15.

terms:low-order bit, high-order bit

Number encoded this way (in bits) is called "binary number".

You was using numeric constants before probably without deeper realizing you are using them. Numeric constants were just numbers you wrote in source file which were assembled into binary file. Numeric constants are for example "0", "50", "-100", "123456".

You used them here:

These numbers were decimal: ones which are used by (?) normal people. Assembler then translated them to binary form. But sometimes you want to specify directly binary numbers. Of course you dont have to hand-translate to decimal, you can specify them directly binary. Here are some examples of binary numbers: 0, 101011, 1101011, 11111, etc. To differ them from decimal numbers, every binary number must end with "b" character, so "0b", "101011b", "1101011b", "11111b" etc. Here first binary digit (first bit, first 0 or 1) is high-order bit, last one is low-order bit. So if you write "1101", then bit 0 = 1, bit 1 = 0, bit 2 =1, bit 3 = 1.db 5 mov al,20 cmp ax,5 db 'Some string',0 org 256

Example table:

decimal | binary |
---|---|

0 | 0b |

1 | 1b |

2 | 10b |

3 | 11b |

4 | 100b |

5 | 101b |

6 | 110b |

7 | 111b |

10 | 1010b |

15 | 1111b |

16 | 10000b |

I could learn you way how to translate number between decimal and binary forms, but you wont need it now anyway, and all other tutorials are full of such informations.

Binary numeric constants are just another way to express some number. Writing "5" is same as writing "101b". For example this will work too:

org 100000000b mov ah,1001b mov dx,string int 21h int 20h string db 'Hello world writen using binary constants',0

`org 100000000b`

is same as `org 256`

, and `mov ah,1001b`

is same as `mov ah,9`

Now let's think about what we can do with bit (which can hold value 0 or 1).

First, we can "set" it (set it's value to 1) or "clear" it (set it's value to 0). Then we can "flip" it's value (from 0 to 1, from 1 to 0). And that is probably all. This operation is called "bit complement" too (0 is complement of 1, 1 is complement of 0)

(lack of english math terms here, somebody could help me)

Now what we can do with 2 bits? If you think of bits as about boolean values, which can be either true (1) or false (0). Now what operations can we make with boolean values? If you programmed before you probably know this.

First is

`and`

(like "a and b" where "a" and "b" are boolean values
:). When we have two boolean values, result of `and`

ing them is true
only when they are both true, otherwise result if false. (Table will be below)Then comes

`or`

. You know, result of `or`

ing two values is
true, when at least one of them is true. And last, least known, is `xor`

,
which means "exclusive or" (the previous one was "inclusive or", but everyone calls it
just "or"). Result of `xor`

ing is 1 when one operand is 1 and other is 0.
Here is the table:

A | B | A and B | A or B | A xor B |
---|---|---|---|---|

0 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 1 |

1 | 0 | 0 | 1 | 1 |

1 | 1 | 1 | 1 | 0 |

Now interesting (?) part: In late times, processors designers didnt like having lot of instructions on their processors. But as you see, we defined 3 operations for single bit and 3 for two bits. So they found (obvious) way to achieve operations on single bit by using operations on two bits. To remind: operations on single bit were setting it to 0, setting it to 1 and flipping it's value (0->1, 1->0). How?

First let's talk about clearing bit (setting it's value to 0). Note that result of

`and`

is 0 whenever at least one of operands is 0. So if we
`and`

any bit (0 or 1) with 0 we always get 0, when we `and`

with 1 bit will reamin unchanged. And this is what we wanted. It is similar with
setting bit (to 1). Result of `or`

ing is 1 when at least one of
operands is 1. So `or`

ing any bit with 1 will always produce 1,
`or`

ing with 0 will leave bit unchanged.How to flip bit? Result of

`xor`

ing is 1 when one of operands is 1,
other 0. So `xor`

ing any value with 1 will always produce value's
complement, `xor`

ing with 0 will leave bit unchanged. This one is not
so obvious, so better look at it in table.First of all. You know smallest register we have are 8 bit (byte) registers. Also smallest part of memory we can access is byte (8 bits). For this reason binary operation instruction we will use will operate on two 8 bit numbers instead on two bits. What will be result? Bit 0 of result will be result of operation between bit 0 of first argument and bit 0 of second argument. Bit 1 of result will be result of operation on bits 1 of argument etc. You 'll see it.

First operation will be "and". Example:

first we loadmov al,00010001b mov bl,00000001b and al,bl

`al`

with 00010001b, so it's bits 0 and 4 contain 1,
other bits contain 0. Then we load `bl`

with 00000001b so it's bit 0
contains 1, other contain 0. Now when we `and`

`al`

with
`bl`

(this is how asm coder usually tells it), it works like some
`al = al and bl`

would eg. result of `and`

ing
`al`

with `bl`

is stored in `al`

.So what is the result (in

`al`

)? Bit 0 of `al`

contained
1 and was `and`

ed with 1. "1 and 1" is 1, so bit 0 of `al`

will be 1. Bits 1 to 2 and 5 to 7 would be "0 and 0" which is 0. Bit 3 would
contain "0 and 1" which is 0 too. Bit 4 will contain "1 and 0" which is 0 again. So result would be 00000001b.
Better way to write previus code would be
(i usedmov al,00010001b and al,00001001b

`bl`

in previous example only to reference second number
easier in text).Now

`or`

ing:
result will be 00011001b, due tomov al,00010001b or al,00001001b

`or`

description (one subchapter
upwards). And

`xor`

ing:
result will be 00011000b. Bitsmov al,00010001b xor al,00001001b

`xor`

ed with 0 will remain, bits
`xor`

ed will 1 will be flipped (to their complement).
instructions:These instructions take same arguments asand, or, xor

`mov`

eg. first argument
can be memory variable or register, second can be memory variable, register or
constant. Both arguments must be of same size, only one of arguments can be
memory variable.If you was programming before, you probably already know about boolean variables (rarely called logical). They can hold two values, TRUE or FALSE. You see that they can be finely stored in bit, 1 for TRUE, 0 for FALSE.

term:Problem here is, that smallest data directly accessible is byte (8 bits). You know, you can access byte register or byte memory variable, not bit. This is really so, there are no instruction which just access one bit. (Of course there are, you just don't need to know about them now :).boolean variable

But when you work with boolean variable you want to access only one bit, not all 8 bits or more. There are some tricks to do this:

Use only one bit of whole byte and leave other bits cleared. Thus if you want to see if bit is 0, you just check if whole byte is equal to 0. If it is not, then our bit is 1. Example:

wherecmp [byte_boolean_varaible],0 je byte_boolean_variable_is_false jnz byte_boolean_varaible_is_true

`byte_boolean_variable`

is byte varaible with only one bit
used. If such variable is 0 then value is FALSE, otherwise value is TRUE.`byte_boolean_variable_is_***`

are labels used for branching as
shown in previous chapter. By the way better "more assembly" way to do same as
previous is:
beacause in first versioncmp [byte_boolean_varaible],0 je byte_boolean_variable_is_false byte_boolean_varaible_is_true: <here value is TRUE> byte_boolean_varaible_is_false: <here value is FALSE>

`jnz`

conditonal jump would be always
taken, because instruction is executed only when `je`

wasn't taken.
If you dont understand read previous chapter.But using this way there are 7 unused bits, and that is waste of space. (Not for single variable, but surely for array of such variables). It is clear we can "pack" 8 boolean varaibles into single byte (8 bits). Problem is only setting and reading it.

First, we'll set all 8 bits (boolean variables) using

`mov`

instruction.
this would set all variables to zero (clear them). If we want set some of them to one, we just set bits in which they are stored.mov [eight_booleans],00000000b

this will set variables in bits 2 and 4, and leave all ather clear.mov [eight_booleans],00010100b

First, how to clear one bit and leave all other unmodified? We solved this before, we can do this with "and"ing:

This will clear bit 3 (and [eight_booleans],11110111b

`and`

ed with 0 so result will be 0), and leave
all other bits unchanged (`and`

ed with 1 so will stay unchanged).
This will clear bits 3 and 5:
it should be clear if you comprehended chapter 6.4.and [eight_booleans],11010111b

Now how to set one of variables:

This sets bit 3 to 1 (or [eight_booleans],00001000b

`or`

ing with 1 always gives 1) and leave other
unchanged (`or`

ing with 0 leaves unchanged).And, of course, using

`xor`

we can flip bit(s):
will flip bit 3 and leave others.xor [eight_booleans],00001000b

This was just to remind you, now let's go to checking value of bit. Checking value of bit is called "testing bit".

term:You often need to test value of some boolean variable and do something (jump somewhere) if it is (or isn't) TRUE. We did this with byte-sized boolean variable usingtesting bit

`cmp`

instruction, but it is impossible to use
`cmp`

for testing only one bit of byte. For this reason, there is
`test`

instruction. It takes same arguments like `mov`

,
`xor`

, `and`

, `cmp`

etc. It `and`

s
it's operands and if result of `and`

ing is 0 then it sets flags so
that if result is zero then `je`

will be taken, if result isn't zero
`je`

won't be taken (and `jnz`

would).
acts a little liketest arg1,arg2

but it doesn't modifyand arg1,arg2 cmp arg1,0

`arg1`

and you use `jz`

(jump if zero) and
`jnz`

(jump if not zero) conditional jumps. `jz`

jumps if result
of virtual `and`

ing (testing) is zero. Similary, `jnz`

jumps, if result is not zero (eg. at least one tested bit is non-zero)
`jz`

is same instruction as `je`

and `jnz`

is same as `jne`

, so using `jz`

is same as using `je`

would be in
and/cmp example
instruction:Some example of usingtest

`test`

all bits but third oftest [eight_booleans],00001000b jz bit_3_is_clear bit_3_is_set: <...> bit_3_is_clear: <...>

`eight_booleans`

are `and`

ed with
0 (but `eight_booleans`

stays unmodified), which means they are
cleared, only value of third will remain. Then result of operation will be zero
(and `je`

will taken) only if third bit is 0. If it is 1, result of
operation will be 00001000b, not 0 so `je`

won't be taken.Now some little more dificult example:

bits 3 and 5 oftest [eight_booleans],00101000b je bits_3_and_5_clear bits_3_and_5_not_both_clear: <...> bits_3_and_5_clear: <...>

`eight_booleans`

will remain, so result of operation
will be 0 (and `je`

will be taken) only when both these bits are 0.
If at least one of these bits is 1 result won't be 0 (it can be 00001000b or
00100000b or 00101000b) and `jz`

won't be taken. But testing two bits
at one time usually isn't used, at least not by beginners, i gave this example
just for better picture how `test`

works.