BINARY NUMBER SYSTEM
The Binary Number Base Systems
Most modern computer systems (including the IBM PC) operate using binary logic. The
computer represents values using two voltage levels (usually 0V for logic 0 and either
+3.3 V or +5V for logic 1). With two levels we can represent exactly two different
values. These could be any two different values, but by convention we use the values
zero and one. These two values, coincidentally, correspond to the two digits used by the
binary number system.
Since there is a correspondence between the logic levels used by the computer and the
two digits used in the binary numbering system, it should come as no surprise that
computers employ the binary system. The binary number system works like the decimal
number system except the Binary Number System:
uses base 2
includes only the digits 0 and 1 (any other digit would make the number an
invalid binary number)
The weighted values for each position is determined as follows:
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 2^-1 2^-2
128 64
32
16
8
4
2
1
.5
.25
In the United States among other countries, every three decimal digits is separated with a
comma to make larger numbers easier to read. For example, 123,456,789 is much easier
to read and comprehend than 123456789. We will adopt a similar convention for binary
numbers. To make binary numbers more readable, we will add a space every four digits
starting from the least significant digit on the left of the decimal point. For example, the
binary value 1010111110110010 will be written 1010 1111 1011 0010.
Number Base Conversion
Binary to Decimal
It is very easy to convert from a binary number to a decimal number. Just like the decimal
system, we multiply each digit by its weighted position, and add each of the weighted
values together. For example, the binary value 1100 1010 represents:
1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 1*2^3 + 0*2^2 + 1*2^1 + 0*2^0 =
1 * 128 + 1 * 64 + 0 * 32 + 0 * 16 + 1 * 8 + 0 * 4 + 1 * 2 + 0 * 1 =
128 + 64 + 0 + 0 + 8 + 0 + 2 + 0 =
202
Decimal to Binary
To convert decimal to binary is slightly more difficult. There are two methods, that may
be used to convert from decimal to binary, repeated division by 2, and repeated
subtraction by the weighted position value.
Repeated Division By 2
For this method, divide the decimal number by 2, if the remainder is 0, on the side write
down a 0. If the remainder is 1, write down a 1. This process is continued by dividing the
quotient by 2 and dropping the previous remainder until the quotient is 0. When
performing the division, the remainders which will represent the binary equivalent of the
decimal number are written beginning at the least significant digit (right) and each new
digit is written to more significant digit (the left) of the previous digit. Consider the
number 2671.
Division Quotient Remainder Binary Number
2671 / 2
1335
1
1
1335 / 2
667
1
11
667 / 2
333
1
111
333 / 2
166
1
1111
166 / 2
83
0
0 1111
83 / 2
41
1
10 1111
41 / 2
20
1
110 1111
20 / 2
10
0
0110 1111
10 / 2
5
0
0 0110 1111
5/2
2
1
10 0110 1111
2/2
1
0
010 0110 1111
1/2
0
1
1010 0110 1111
The Subtraction Method
For this method, start with a weighted position value greater that the number.
If the number is greater than the weighted position for the digit, write down a 1
and subtract the weighted position value.
If the number is less than the weighted position for the digit, write down a 0 and
subtract 0.
This process is continued until the result is 0. When performing the subtraction, the digits
which will represent the binary equivalent of the decimal number are written beginning at
the most significant digit (the left) and each new digit is written to the next lesser
significant digit (on the right) of the previous digit. Consider the same number, 2671,
using a different method.
Weighted Value Subtraction
Remainder Binary Number
2^12 = 4096
2671 - 0
2671
2^11 = 2048
2671 - 2048 623
01
2^10 = 1024
623 - 0
623
0 10
2^9 = 512
623 - 512
111
0 101
2^8 = 256
111 - 0
111
0 1010
2^7 = 128
111 - 0
111
0 1010 0
2^6 = 64
111 - 64
47
0 1010 01
2^5 = 32
47 - 32
15
0 1010 011
2^4 = 16
15 - 0
15
0 1010 0110
2^3 = 8
15 - 8
7
0 1010 0110 1
2^2 = 4
7-4
3
0 1010 0110 11
2^1 = 2
3-2
1
0 1010 0110 111
2^0 = 1
1-1
0
0 1010 0110 1111
0
Binary Number Formats
We typically write binary numbers as a sequence of bits (bits is short for binary digits).
We have defined boundaries for these bits. These boundaries are:
Name
Size (bits) Example
Bit
1
1
Nibble
4
0101
Byte
8
0000 0101
Word
16
0000 0000 0000 0101
Double Word 32
0000 0000 0000 0000 0000 0000 0000 0101
In any number base, we may add as many leading zeroes as we wish without changing its
value. However, we normally add leading zeroes to adjust the binary number to a desired
size boundary. For example, we can represent the number five as:
Bit
101
Nibble 0101
Byte
0000 0101
Word
0000 0000 0000 0101
We'll number each bit as follows:
1. The rightmost bit in a binary number is bit position zero.
2. Each bit to the left is given the next successive bit number.
Bit zero is usually referred to as the LSB (least significant bit). The left-most bit is
typically called the MSB (most significant bit). We will refer to the intermediate bits by
their respective bit numbers.
The Bit
The smallest "unit" of data on a binary computer is a single bit. Since a single bit is
capable of representing only two different values (typically zero or one) you may get the
impression that there are a very small number of items you can represent with a single
bit. Not true! There are an infinite number of items you can represent with a single bit.
With a single bit, you can represent any two distinct items. Examples include zero or one,
true or false, on or off, male or female, and right or wrong. However, you are not limited
to representing binary data types (that is, those objects which have only two distinct
values).
To confuse things even more, different bits can represent different things. For example,
one bit might be used to represent the values zero and one, while an adjacent bit might be
used to represent the values true and false. How can you tell by looking at the bits? The
answer, of course, is that you can't. But this illustrates the whole idea behind computer
data structures: data is what you define it to be.
If you use a bit to represent a boolean (true/false) value then that bit (by your definition)
represents true or false. For the bit to have any true meaning, you must be consistent.
That is, if you're using a bit to represent true or false at one point in your program, you
shouldn't use the true/false value stored in that bit to represent red or blue later.
Since most items you will be trying to model require more than two different values,
single bit values aren't the most popular data type. However, since everything else
consists of groups of bits, bits will play an important role in your programs. Of course,
there are several data types that require two distinct values, so it would seem that bits are
important by themselves. however, you will soon see that individual bits are difficult to
manipulate, so we'll often use other data types to represent boolean values.
The Nibble
A nibble is a collection of bits on a 4-bit boundary. It wouldn't be a particularly
interesting data structure except for two items: BCD (binary coded decimal) numbers and
hexadecimal (base 16) numbers. It takes four bits to represent a single BCD or
hexadecimal digit.
With a nibble, we can represent up to 16 distinct values. In the case of hexadecimal
numbers, the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F are represented with
four bits. BCD uses ten different digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and requires four bits.
In fact, any sixteen distinct values can be represented with a nibble, but hexadecimal and
BCD digits are the primary items we can represent with a single nibble.
b3 b2 b1 b0
The Byte
Without question, the most important data structure used by the 80x86 microprocessor is
the byte. This is true since the ASCII code is a 7-bit non-weighted binary code that is
used on the byte boundary in most computers. A byte consists of eight bits and is the
smallest addressable datum (data item) in the microprocessor.
Main memory and I/O addresses in the PC are all byte addresses. This means that the
smallest item that can be individually accessed by an 80x86 program is an 8-bit value. To
access anything smaller requires that you read the byte containing the data and mask out
the unwanted bits.
The bits in a byte are numbered from bit zero (b0) through seven (b7) as follows:
b7 b6 b5 b4 b3 b2 b1 b0
Bit 0 is the low order bit or least significant bit, bit 7 is the high order bit or most
significant bit of the byte. We'll refer to all other bits by their number.
A byte also contains exactly two nibbles. Bits b0 through b3 comprise the low order
nibble, and bits b4 through b7 form the high order nibble. Since a byte contains exactly
two nibbles, byte values require two hexadecimal digits.
Since a byte contains eight bits, it can represent 2^8, or 256, different values. Generally,
we'll use a byte to represent:
1.
2.
3.
4.
unsigned numeric values in the range 0 => 255
signed numbers in the range -128 => +127
ASCII character codes
other special data types requiring no more than 256 different values. Many data
types have fewer than 256 items so eight bits is usually sufficient.
Since the PC is a byte addressable machine, it turns out to be more efficient to manipulate
a whole byte than an individual bit or nibble. For this reason, most programmers use a
whole byte to represent data types that require no more than 256 items, even if fewer than
eight bits would suffice. For example, we'll often represent the boolean values true and
false by 00000001 and 00000000 (respectively).
Probably the most important use for a byte is holding a character code. Characters typed
at the keyboard, displayed on the screen, and printed on the printer all have numeric
values. To allow it to communicate with the rest of the world, the IBM PC uses a variant
of the ASCII character set. There are 128 defined codes in the ASCII character set. IBM
uses the remaining 128 possible values for extended character codes including European
characters, graphic symbols, Greek letters, and math symbols.
The Word
NOTE:
The boundary for a Word is defined as either 16-bits or the size of the data bus for the
processor, and a Double Word is Two Words. Therefore, a Word and a Double Word is
not a fixed size but varies from system to system depending on the processor. However,
for our discussion, we will define a word as two bytes.
For the 8085 and 8086, a word is a group of 16 bits. We will number the bits in a word
starting from bit zero (b0) through fifteen (b15) as follows:
b15 b14 b13 b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1 b0
Like the byte, bit 0 is the LSB and bit 15 is the MSB. When referencing the other bits in a
word use their bit position number.
Notice that a word contains exactly two bytes. Bits b0 through b7 form the low order
byte, bits 8 through 15 form the high order byte. Naturally, a word may be further broken
down into four nibbles. Nibble zero is the low order nibble in the word and nibble three is
the high order nibble of the word. The other two nibbles are "nibble one" or "nibble two".
With 16 bits, you can represent 2^16 (65,536) different values. These could be the
unsigned numeric values in the range of 0 => 65,535, signed numeric values in the range
of -32,768 => +32,767, or any other data type with no more than 65,536 values. The three
major uses for words are
1. 16-bit integer data values
2. 16-bit memory addresses
3. any number system requiring 16 bits or less
The Double Word
A double word is exactly what its name implies, two words. Therefore, a double word
quantity is 32 bits. Naturally, this double word can be divided into a high order word and
a low order word, four bytes, or eight nibbles.
Double words can represent all kinds of different data. It may be
1.
2.
3.
4.
an unsigned double word in the range of 0 => 4,294,967,295,
a signed double word in the range -2,147,483,648 => 2,147,483,647,
a 32-bit floating point value
any data that requires 32 bits or less.
Octal Number System
The Octal Number Base System
Although this was once a popular number base, especially in the Digital Equipment
Corporation PDP/8 and other old computer systems, it is rarely used today. The Octal
system is based on the binary system with a 3-bit boundary. The Octal Number System:
uses base 8
includes only the digits 0 through 7 (any other digit would make the number an
invalid octal number)
The weighted values for each position is as follows:
8^5
8^4
8^3
8^2
8^1
8^0
32768
4096
512
64
8
1
Binary to Octal Conversion
It is easy to convert from an integer binary number to octal. This is accomplished by:
1. Break the binary number into 3-bit sections from the LSB to the MSB.
2. Convert the 3-bit binary number to its octal equivalent.
For example, the binary value 1010111110110010 will be written:
001 010 111 110 110 010
1
2
7
6
6
2
Octal to Binary Conversion
It is also easy to convert from an integer octal number to binary. This is accomplished by:
1. Convert the decimal number to its 3-bit binary equivalent.
2. Combine the 3-bit sections by removing the spaces.
For example, the octal value 127662 will be written:
1
2
7
6
6
2
001 010 111 110 110 010
This yields the binary number 001010111110110010 or 00 1010 1111 1011 0010 in our
more readable format.
Octal to Decimal Conversion
To convert from Octal to Decimal, multiply the value in each position by its Octal weight
and add each value. Using the value from the previous example, 127662Q, we would
expect to obtain the decimal value 44978.
1*8^5
2*8^4
7*8^3 6*8^2 6*8^1 2*8^0
1*32768 2*4096 7*512 6*64
6*8
2*1
32768
48
2
8192
3584
32768 + 8192 + 3584 + 384 + 48 + 2 = 44978
384
Hexadecimal Number System
The Hexadecimal Number Base System
A big problem with the binary system is verbosity. To represent the value 202 requires
eight binary digits.
The decimal version requires only three decimal digits and, thus, represents numbers
much more compactly than does the binary numbering system. This fact was not lost on
the engineers who designed binary computer systems.
When dealing with large values, binary numbers quickly become too unwieldy. The
hexadecimal (base 16) numbering system solves these problems. Hexadecimal numbers
offer the two features:


hex numbers are very compact
it is easy to convert from hex to binary and binary to hex.
Since we'll often need to enter hexadecimal numbers into the computer system, we'll need
a different mechanism for representing hexadecimal numbers since you cannot enter a
subscript to denote the radix of the associated value.
The Hexadecimal system is based on the binary system using a Nibble or 4-bit boundary.
In Assembly Language programming, most assemblers require the first digit of a
hexadecimal number to be 0, and we place an H at the end of the number to denote the
number base.
The hexadecimal number system:
uses base 16
includes only the digits 0 through 9 and the letters A, B, C, D, E, and F
In the Hexadecimal number system, the hex values greater than 9 carry the following
decimal value:
Binary
Octal Decimal Hex
0000B
00Q
00
00H
0001B
01Q
01
01H
0010B
02Q
02
02H
0011B
03Q
03
03H
0100B
04Q
04
04H
0101B
05Q
05
05H
0110B
06Q
06
06H
0111B
07Q
07
07H
1000B
10Q
08
08H
1001B
11Q
09
09H
1010B
12Q
10
0AH
1011B
13Q
11
0BH
1100B
14Q
12
0CH
1101B
15Q
13
0DH
1110B
16Q
14
0EH
1111B
17Q
15
0FH
1 0000B 20Q
16
10H
This table provides all the information you'll ever need to convert from one number base
into any other number base for the decimal values from 0 to 16.
To convert a hexadecimal number into a binary number, simply brake the binary number
into 4-bit groups beginning with the LSB and substitute the corresponding four bits in
binary for each hexadecimal digit in the number.
For example, to convert 0ABCDh into a binary value, simply convert each hexadecimal
digit according to the table above. The binary equivalent is:
0ABCDH = 0000 1010 1011 1100 1101
To convert a binary number into hexadecimal format is almost as easy. The first step is to
pad the binary number with leading zeros to make sure that the the binary number
contains multiples of four bits. For example, given the binary number 10 1100 1010, the
first step would be to add two bits in the MSB position so that it contains 12 bits. The
revised binary value is 0010 1100 1010.
The next step is to separate the binary value into groups of four bits, e.g., 0010 1100
1010. Finally, look up these binary values in the table above and substitute the
appropriate hexadecimal digits, e.g., 2CA.
The weighted values for each position is as follows:
16^3 16^2 16^1 16^0
4096 256
16
1
Binary to Hex Conversion
It is easy to convert from an integer binary number to hex. This is accomplished by:
1. Break the binary number into 4-bit sections from the LSB to the MSB.
2. Convert the 4-bit binary number to its Hex equivalent.
For example, the binary value 1010111110110010 will be written:
1010 1111 1011 0010
A
F
B
2
Hex to Binary Conversion
It is also easy to convert from an integer hex number to binary. This is accomplished by:
1. Convert the Hex number to its 4-bit binary equivalent.
2. Combine the 4-bit sections by removing the spaces.
For example, the hex value 0AFB2 will be written:
A
F
B
2
1010 1111 1011 0010
This yields the binary number 1010111110110010 or 1010 1111 1011 0010 in our more
readable format.
Hex to Decimal Conversion
To convert from Hex to Decimal, multiply the value in each position by its hex weight
and add each value. Using the value from the previous example, 0AFB2H, we would
expect to obtain the decimal value 44978.
A*16^3
F*16^2 B*16^1 2*16^0
10*4096 15*256
11*16
2*1
40960
176
2
3840
40960 + 3840 + 176 + 2 = 44978
Decimal to Hex Conversion
To convert decimal to hex is slightly more difficult. The typical method to convert from
decimal to hex is repeated division by 16. While we may also use repeated subtraction by
the weighted position value, it is more difficult for large decimal numbers.
Repeated Division By 16
For this method, divide the decimal number by 16, and write the remainder on the side as
the least significant digit. This process is continued by dividing the quotient by 16 and
writing the remainder until the quotient is 0. When performing the division, the
remainders which will represent the hex equivalent of the decimal number are written
beginning at the least significant digit (right) and each new digit is written to the next
more significant digit (the left) of the previous digit. Consider the number 44978.
Division
Quotient Remainder Hex Number
44978 / 16 2811
2
2
2811 / 16
175
11
B2
175 / 16
10
15
FB2
10 / 16
0
10
0AFB2
As you can see, we are back with the original number. That is what we should expect.
When you use hex numbers in an 8085 program, the Assembler usually requires the most
significant hex digit to be 0 even if this number of digits exceed the size of the register.
This is an Assembler requirement and your value will be assembled correctly.
BCD (Binary Coded Decimal) Number System
The BCD Number System
You should now be familiar with the Binary, Decimal and Hexadecimal Number System.
If we view single digit values for hex, the numbers 0 - F, they represent the values 0 - 15
in decimal, and occupy a nibble. Often, we wish to use a binary equivalent of the decimal
system. This system is called Binary Coded Decimal or BCD which also occupies a
nibble. In BCD, the binary patterns 1010 through 1111 do not represent valid BCD
numbers, and cannot be used.
Conversion from Decimal to BCD is straightforward. You merely assign each digit of the
decimal number to a byte and convert 0 through 9 to 0000 0000 through 0000 1001, but
you cannot perform the repeated division by 2 as you did to convert decimal to binary.
Let us see how this works. Determine the BCD value for the decimal number 5,319.
Since there are four digits in our decimal number, there are four bytes in our BCD
number. They are:
Thousands
Hundreds
Tens
Units
5
3
1
9
00000101 00000011 00000001 00001001
Since computer storage requires the minimum of 1 byte, you can see that the upper nibble
of each BCD number is wasted storage. BCD is still a weighted position number system
so you may perform mathematics, but we must use special techniques in order to obtain a
correct answer.