FCPS
Java
Packets
Computer Number
Systems
Based upon material developed by
Sally Bellacqua, Mary Johnson, and Janet Mulloy
Revised by Charles Brewer, August 2004
Revised by Shane Torbert, July 2006
Fairfax County Public Schools
Fairfax, Virginia
CN 1
1. INTRODUCTION
Throughout history, different civilizations have used various devices to facilitate
computations. The Babylonians, Chinese, and early Europeans used the abacus for
calculations. The abacus consists of a divided frame with sliding beads on the top
representing fives and sliding beads on the bottom representing ones. Today, devices
such as calculators and computers perform a similar function using a combination of
Random Access Memory (RAM), Read Only Memory (ROM), a Central Processing
Unit (CPU), and a variety of support chips and storage, display, and input devices.
2. BINARY NUMBER SYSTEM
Current computer chips have two states, ON or OFF, which conveniently can represent
0 and 1, the two digits in the binary number system. All instructions and data in
current computers are represented as a series of 0’s and 1’s. While high-level
languages shield the user from the need to input instructions and data in binary form,
knowledge of binary arithmetic is useful in understanding how computers operate and
their limitations. Since the binary representation of a decimal number can have 64 or
more digits, the hexadecimal (base 16) system is sometimes used as a convenient way
of expressing the binary number.
The table to the
right shows how
numbers are
represented in
different bases.
Base 10
(decimal)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Base 2
(binary)
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
10000
10001
10010
10011
Base16
(hexadecimal)
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
10
11
12
13
CN 2
A number in base b can be converted to its equivalent in base 10 by using place
values.
Place Value Formula
b8 b 7 b 6 b5 b 4 b3b 2 b1b 0 .b 1b 2 b 3b 4
Given 1394210, what do the individual digits represent?
13942 10 = 1(10)4 + 3(10)3 + 9(10)2 + 4(10)1 + 2(10)0
= 10,000 + 3000 + 900 + 40
+2
Likewise, the digits in the base 2 number 110101 represent powers of the base 2.
110101 2 = 1(2)5 + 1(2)4 + 0(2)3 + 1(2)2 + 0(2)1 + 1(2)0
= 1(32) + 1(16) + 0(8) + 1(4) + 0(2) + 1(1)
= 32 + 16
+0
+4
+0
+1
= 5310
Example:
Convert 13F16 to base ten.
13F16 = 1(16)2 + 3(16)1 + F(16)0
= 1(256) + 3(16) + 15(1)
= 256 + 48
+ 15
= 319 10
Exercise 1
1. How many digits are there in base 8 (octal)? ______ digits, from ____ through ____
2. Count to eighteen in octal. __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __
In problems 3 – 10, convert the base b number to its equivalent in base 10.
3.
159 16 = ________ 10
4.
01101100 2 = ________ 10
5.
3A 16 = _________ 10
6.
10110 2 = _________ 10
7.
DC 16 = __________ 10
8.
1010.101 2 = __________ 10
9.
5723 8 = _________ 10
10. 201 3 = ____________ 10
CN 3
2. ADDING IN DIFFERENT BASES
Adding in different bases follows the familiar addition rules, including when to carry the
number. You just need to pay attention to the base. For example, 7 8 + 38 has a carry
operation, namely, 78 + 38 = (1 eight and 2 left over) = 128
As another example, B16 + 616 = 1116
Addition in base 2 is so simple that you only have to memorize three rules:
02 + 0 2 = 0 2
and
1 2 + 02 = 12
and
12 + 12 = 102
Exercise 2.
Add in the indicated base. Leave your answers in that base.
1. 0010 1110 2
0011 1011 2
2. 342 8
517 8
3. 3A9 16
21C 16
4. 432 5
123 5
3. MOVING BETWEEN BINARY AND HEXADECIMAL
As previously noted, the representation of a decimal
number in binary form can require a significant number
of digits. For example, 65535 10 = 11111111111111112
These sixteen binary digits can be conveniently
expressed as four hexadecimal digits by grouping the
binary digits in groups of four and using the equivalent
values from the table at the right.
Example:
1111 1111 1111 1111 2 = F F F F 16.
Similarly, any hexadecimal number can be converted to
binary by replacing each hexadecimal digit with its four
binary digit equivalent.
Example:
3B9 16 = 0011 1011 1001 2
BINARY
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
HEX
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
CN 4
Exercise 3
Convert between binary and hexadecimal.
1.
01011001 2 = _____ 16
2.
54 16 = _____________ 2
3.
01000001 2 = _____ 16
4.
3E 16 = _____________ 2
5.
00110100.0110 2 = _____ 16
6.
6A.5 16 = _______________ 2
4. CONVERTING BASE 10 NUMBERS TO ANOTHER BASE
While we are primarily interested in being able to convert base 10 numbers to their
binary (or hexadecimal) equivalents, there is a convenient algorithm for the conversion
that works for all bases. The “Divide and Save” algorithm is simple to use and easy to
implement in a computer program.
The technique is to use successive integer division by the desired base until a zero
quotient is obtained. The remainders from each division are recorded from right to left
as the result.
Example:
Convert the base ten number 37 to binary.
0
R1
( one 25 digit )
21
R0
( zero 24 digit )
2 2
R0
( zero 23 digit )
2 4
R1
( one 22 digit )
29
R0
( zero 21 digit )
2 18
R1
( one 20 digit )
2 37
Therefore, 3710  1001012
CN 5
Example:
Convert the base ten number 111 to hexadecimal
0
R 6
16 6
R (15) E
16 111
Therefore 11110  6 E16
EXERCISE 4
Convert the following base ten numbers to the indicated base number.
1. 25 10 = ___________ 2
2. 73 10 = ____________ 16
3. 65 10 = ___________
4. 92 10 = ____________ 16
2
5. 32767 10 = _______________________ 2 = _______________________
16
6. 7 10 = ______________ 3
5. REPRESENTATION OF NEGATIVE INTEGERS
Different CPUs use 8, 16, or 32 bits to represent integers. If the integer is too small or
too big, the CPU can't handle it.
8 bits or 1 byte
16 bits or 2 bytes
32 bits or 4 bytes
Range
Minimum integer
Maximum integer
8
7
2 or 256
-2 or -128
27-1 or 127
216 or 65535
-215 or -32768
215 -1 or 32767
232
-231
231 -1
It is the programmer's responsibility to make sure that the code never goes outside
those upper and lower limits. Most high-level languages, including Java, do not give
error messages when the bytes "overflow"—they just keep
127
0111 1111
counting and suddenly the code is producing negative
...
...
numbers!
3
0000 0011
The representation of negative integers varies with the CPU.
The Intel family of CPUs uses the two’s-complement
method to represent negative integers. Using this method a
negative integer n is represented (in 8 bits) using the
relationship 256 - | n |, where | n | is the absolute value of n.
2
1
0
-1
-2
-3
...
-128
0000 0010
0000 0001
0000 0000
1111 1111
1111 1110
1111 1100
...
1000 0000
CN 6
For example,
-6 10 = (256 - | -6 |)10 = 250 10 = 1111 1010 2
-128 10 = (256 -| -128 |) 10 = 128 10 = 1000 00002
Notice that in the two’s-complement system, all positive integers begin with 0 and
negative integers begin with 1.
In the binary arithmetic used by the computer, it is a three-step algorithm to express a
negative number. For example, to represent –6 in binary:
First, change the absolute value of the number to binary
| - 6 10 | = 0000 0110
Second, form the one’s complement by interchanging 0s and 1s
1111 1001
Third, add 1
+ 1
to form the two’s complement
1111 1010
Example:
Convert -128 to its two-complement representation:
Step 1 – Change | -128 | to binary
Step 2 – Form the one’s complement
Step 3 -- Add 1
to form the two’s complement
1000 0000
0111 1111
+ 1
1000 0000
For convenience, 8-bit representation of signed integers will be used in the remainder
of this unit instead of the 16-bit, 32-bit, or 64-bit representations actually used.
The two’s-complement algorithm can of course be used to convert a signed binary
number to its signed integer equivalent.
Example:
What base ten integer does the signed binary number, 10111010, represent?
Since the leading bit is a 1, the number must be negative; therefore, the two’scomplement method must be used to find the absolute value.
Given signed binary number
Step 1 – Form one’s complement
Step 2 – Add 1 to
form the two’s complement
1011 1010
0100 0101
+ 1
0100 0110
Step 3 – Convert the two’s complement
to the base ten representation
0100 01102 = 7010
Step 4 – Insert the negative sign. Therefore, 1011 10102 = -7010.
CN 7
EXERCISE 5
1 – 4. Form the one’s and two’s complements of the following binary numbers.
BINARY NUMBER
1010 1011
ONE’S COMPLEMENT
TWO’S COMPLEMENT
0111 0000
0000 0001
0000 0000
5 – 8. Represent each negative integer in 8-bit two’s complement form.
DECIMAL
INTEGER
-1
ABSOLUTE VALUE
IN BINARY
ONE’S
COMPLEMENT
TWO’S
COMPLEMENT
-2
-30
-8
6. ADDING SIGNED BINARY NUMBERS
Since the leading bit indicates the sign, would it be possible to use a system where,
e.g. 410 = 0000 01002 and – 410 = 1000 01002? While such a system could be used, it
is not used because addition in that system is hard. In contrast, the two’s-complement
representation eases the performance of binary arithmetic, which is why it is used.
Addition becomes relatively simple to perform and as you may have guessed,
subtraction is merely the addition of two’s complement. Computers never actually
subtract; they just add the two's complement.
Example:
What is the answer to 10 10 + (– 10 10) ?
The 8-bit binary representation of 10 is
The 8-bit binary representation of –10
is the two’s-complement
Add the binary numbers and discard the
overflow digit, the resulting being 0000 0000 2
or 0 10.
0000 1010
0000 1010
1111 0101
+1
1111 0110
1111 0110
1 0000 0000
CN 8
Example:
Add these two 8-bit signed binary numbers and
then check your work by converting all three
numbers to base ten.
1110 1100
+ 1101 0110
Add – Add the bits, column by column. If
carrying creates an overflow, discard
the overflow bit.
1110 1100
+ 1101 0110
1 1100 0010
Check – Check the addition by converting the three numbers to base ten to. Since all
three are negative in this example, the two’s complement method will be used to find
their absolute values.
BINARY
(1’s comp + 1 = 2’s comp)
DECIMAL
1110 1100  0001 0011  1  0001 0100   20
1101 0110  0010 1001  1  0010 1010   42
1100 0010  0011 1101  1  0011 1110   62
Exercise 6
Add these 8-bit signed binary numbers. Check your work by converting all three
numbers to base ten. (The check is more work than the original addition problem.)
1. 0000 1001
1111 1100
2. 0001 1100
1011 0101
Represent these decimal numbers as 8-bit signed binary numbers. Add the binary
number and then convert each answer back to base ten as a check.
3.
-7
+ 9
4.
-17
+ -13
CN 9
6. REPRESENTATION OF REAL NUMBERS
Real numbers are represented using one of the standard formats, either float or
double. The Institute of Electrical and Electronic Engineers (IEEE) standard for 32-bit
or 4-byte representation is used for floats. The standard consists of 3 parts: 1 bit is
used for sign information, 8 bits for weighted exponent, and 23 bits for normalized
mantissa.
Example: Express 7.5 as a float.
First, convert the decimal to binary. 7.5 10 = 0111.12 = 1.111 2 * 2 2
Second, determine the weighted exponent by adding 127 to the power of 2 resulting in
129 or in binary form ( 1000 0010 ).
Third, create the normalized mantissa (fraction) by dropping the 1 to the left of the
binary point and adding zeroes to the right to give 23 digits.
The resulting float representation is:
0100 0001 0111 0000 0000 0000 0000 0000
Normalized Mantissa
Weighted Exponent
Sign
Special Values
If the exponent (E) is 255 and the fraction (F) is nonzero, the value is NaN for Not a
Number. If E=255 and F=0 the value is Infinity or -Infinity, depending on the sign bit.
If E=0 and F=0 the value if 0 or -0, depending on the sign bit. If E=0 and F is nonzero
the fraction is considered unnormalized (i.e., we assume a 0 in front of the binary point
instead of a 1).
The largest positive value is when E=254, so the exponent is E-127=127. This is on
the order of 2^127=(2^10)^12*2^7=(10^3)^12*128=10^36*10^2=10^38, approximately.
The smallest positive value is when E=0 and the fraction is unnormalized. If F is as
small as possible we get an exponent of -126 and a fraction of 2^-23. Multiplying this is
2^-149=(2^10)^-15=(10^3)^-15=10^-45, approximately. Positive values smaller than
this cannot be represented in a float variable. Underflow occurs, and the resulting
value stored is exactly equal to zero.
Float values are denser closer to zero, and generally the number of distinct floats
between consecutive powers of two is constant. This means that very large values
have very large gaps between them, and swamping is possible.
CN 10
Double
A double uses twice the bits of a float. It has 1 sign bit, 11 exponent bits, and 52
fraction bits.
If E=2047 and F is nonzero, the value is NaN for Not a Number. If E=2047 and F=0
the value is Infinity or -Infinity, depending on the sign bit.
If E=0 and F=0 the value if 0 or -0, depending on the sign bit. If E=0 and F is nonzero
the fraction is considered unnormalized.
Its largest value is 2^1023=(2^10)^102*2^3=(10^3)^102*10=10^307, approximately.
Its smallest positive value is 2^-1022*2^-52=10^-324, approximately.
Code Peculiarities in Java
double x = Math.sqrt(-1);
System.out.println( x +" " + (x == x) );
NaN
double y = 1.0 / 0.0;
System.out.println( y +" " + (y == y+ 1) );
Infinity
y *= -1.0;
System.out.println( y +" " + (y == y + 1) );
-Infinity
false
true
true
Evidently not-a-number is not equal to itself. It seems that infinity-plus-one equals one.
Notice that NaN, Infinity, and Negative Infinity are actually stored in double variables.
Precision
The fractional part of the scientific notation determines precision. A double’s fraction is
accurate to 2^-52=10^-16. This is why you see 16 digits after the decimal point when
you print out, say, a random number from Math.random.
Precision determines the spread of values that can be represented in a double, which
in turn determines when swamping occurs. One important value is given the special
name machine epsilon. This is the smallest number that can be added to 1 and yield a
result that is different from 1. For doubles in 32-bit compilers, the machine epsilon is
approximately 10^-16, or approximately 2^-52. That is, 1.0 + something less than
10^-16 is indistinguishable from 1.0.
Note that machine epsilon is not the smallest number that can be represented, which is
10^-324, or 2^-1074,
CN 11
7. MULTIPLICATION IN BINARY
The Arithmetic Logic Unit (ALU) in the CPU multiplies binary numbers using a “shift
left” and “shift right” logic. The following function uses a multiplying algorithm that
illustrates the operation of this shift logic.
int result (int x, int y)
{
int z = 0;
while ( y != 0 )
{
if (y % 2 == 1)
z = z + x;
x = 2 * x;
y=y/2;
}
return z;
}
// If y is odd add x to interim result (z)
// Shifts binary number left
// Shifts binary number right
// When y is 0, return interim result
A call to this function with x = 7 and y =13 would return 91 in this way:
Iteration
0
1
2
3
4
z
0
7
7
35
91
Decimal
x
7
14
28
56
112
y
13
6
3
1
0
z
0000 0000
0000 0111
0000 0111
0010 0011
0101 1011
Binary
x
0000 0111
0000 1110
0001 1100
0011 1000
0111 0000
y
0000 1101
0000 0110
0000 0011
0000 0001
0000 0000
8. REPRESENTATIONS OF CHARACTERS
Characters are normally represented on PCs using the 8-bit ASCII (American Standard
Code for Information Interchange). The ASCII code defines 128 characters from 0 to
127 using the seven of the eight bits in a byte. The extended ASCII set includes the
code from 128 to 255 and varies from machine to machine and font to font.
A standard for 16-bit representation of characters called Unicode has been created and
is used by the Java compiler. This larger data size provides for non-English
characters. A table of Unicode characters is on the next page. Note that the Unicode
for 'A' is '\u0041' (in hexadecimal) and the unicode for 'a' is '\u0061' (in hexadecimal).
What is the code for 'A' in decimal? _____ What is the code for 'a' in decimal? ______
Some mainframe computers use still another format for character representation called
EBCDIC (Extended Binary Coded Decimal Interchange Code).
CN 12
Unicode, expressed in hexadecimal
The Primitive Data Types in Java
Real
Integers
Keyword
Size
Intel Format
Typical Use
byte
8-bit
two's complement
short
16-bit
two's complement
int
32-bit
two's complement
for integers
elapsed time in milliseconds,
financial calculations,
large Fibonacci numbers
where space is a concern
long
64-bit
two's complement
float
32-bit
IEEE 754
double
64-bit
IEEE 754
scientific uses
char
16-bit
Unicode character
single letters or characters
boolean
1 bit
0, 1
true or false
CN 13
9. UNIT REVIEW EXERCISES
1.
The octal number system is based on what number?
___________
2.
The largest digit in base 16 is?
___________
3.
What is the base-ten value of the underlined bit in the
binary number 0 1 1 0 1 1 1 0?
___________
4.
Express the hexadecimal number FF in base ten.
___________
5.
Express the hexadecimal number 12A in binary.
___________
6.
Find the two’s complement of 1101 0111.
___________
7.
Add these hexadecimal numbers.
Leave your answer in hexadecimal.
8.
4F
+ 68
___________
1100 1111 is the signed 8-bit representation
of what base ten number?
___________
9.
Express the base ten number 37 in binary.
___________
10.
Express the binary 0101 1011.011 in base sixteen.
___________
11.
Express the base ten number 87 in base sixteen.
___________
12.
Does a base ten number greater than one require
more digits than the same number written in binary?
___________
13.
Perform the following addition of two
8-bit signed binary numbers. Check
your work by converting each binary
number into base ten.
14.
Convert the following base ten numbers
to binary and add. Check your result
by converting the answer back to
base ten
1111 0011
0000 1101
- 37
23