[PPT] - Fundamentals of Programming (C) Group 7 Lecturer: Mahdi Soltani PowerPoint Presentation

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Lecturer: Mahdi Soltani

Sharif University of Technology

Department of Computer Engineering

Fundamentals of Programming (C)

Group 7

Lecture 3 Number Systems

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Outline

Numeral Systems
Computer Data Storage Units
Numeral Systems Conversion
Calculations in Number Systems
Signed Integer Representation
Fractional and Real Numbers
ASCII Codes

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Numeral Systems

Decimal number system (base 10)
Binary number system (base 2)

– Computers are built using digital circuits – Inputs and outputs can have only two values: 0 and 1

1 or True (high voltage)
0 or false (low voltage)

– Writing out a binary number such as 1001001101 is tedious, and prone to errors

Octal and hex are a convenient way to represent binary numbers, as

used by computers

– Octal number system (base 8) – Hexadecimal number system (base 16)

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Numeral Systems

Hexadecimal Octal Binary Decimal 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 A (decimal value of 10) B (decimal value of 11) C (decimal value of 12) D (decimal value of 13) E (decimal value of 14) F (decimal value of 15)

Base B : 0 ≤ digit ≤ B -1

Base 10 : 0 ≤ digit ≤ 9 (10 -1)

Base 2 : 0 ≤ digit ≤ 1 (2-1) Base 8 : 0 ≤ digit ≤ 7 (8 -1) Base 16 : 0 ≤ digit ≤ 15 (16 -1)

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Computer Data Storage Units

Bit

– Each bit can only have a binary digit value: 1 or 0 – basic capacity of information in computer – A single bit must represent one of two states: 21=2

How many state can encode by N bit?

1 OR

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Binary Encoding

1 1 1 1 2 1 1 3 1 2 1 1 1 2 1 1 3 1 4 1 1 5 1 1 6 1 1 1 7

1 2 3 1 1 1 2 1 1 3 1 4 1 1 5 1 1 6 1 1 1 7 1 8 1 1 9 1 1 10 1 1 1 11 1 1 12 1 1 1 13 1 1 1 14 1 1 1 1 15

1 1

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Computer Data Storage Units

Byte: A sequence of eight bits or binary digits

– 28 = 256 (0..255) – smallest addressable memory unit

1 2 3 4 5 6 7 1 1 1 2 1 1 3 … … … … … … … … …

Bit Order One Bit Memory Address 1 2 3 …

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Kilo byte: 2 10 b = 1024 b ~ 103 b
2 11 = 21 x 210 = 2 KB = 2048 b
2 16 = 26 x 210 = 64 KB = 65536 b
Mega byte: 2 20 b = 1024 × 1024 b
2 21 = 21 x 220 = 2 M
Giga byte: 2 30
….

Computer Data Storage Units

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Numeral Systems Conversion

 Convert from Base-B to Base-10:

(A) B = (?) 10
(4173)10 = (4173)10
(11001011.0101)2 = (?)10
(0756)8 = (?)10
(3b2)16 = (?)10

 Convert from Base-10 to Base-B:

(N) 10 = (?) B
(4173) 10 = (?)2
(494) 10 = (?)8
(946) 10 = (?)16

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert from Base-B to Base-10

1. Define bit order
Example : Base-2 to Base-10

... 3

2
1
1

2 3 4 5 6 7 .. .

4

3

2
1
1

2 3 4 5 6 7 1 1 . 1 1 1 1 1

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert from Base-B to Base-10

2. Calculate Position Weight

– B bit order

Example : Base-2 to Base-10

2-4 2-3 2-2 2 1

20

21 22 23 24 25 26 27

4

3

2
1
1

2 3 4 5 6 7 1 1 . 1 1 1 1 1

B = 2 … Position Weight Decimal Point

102 101 100 10-1 10-2 100s 10s 1s 1/10s 1/100s 9 8 7 . 5 6

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert from Base-B to Base-10

3. Multiplies the value of each digit by the value
f its position weight

0.0625 0.125 0.25 0.5 1 2 4 8 16 32 64 128

4

3

2
1
1

2 3 4 5 6 7 1 1 . 1 1 1 1 1 0.0625 * 1 0.125 * 0.25 * 1 0.5 * 1 * 1 2 * 1 4 * 8 * 1 16 * 32 * 64 * 1 128 * 1 0.0625 0.25 . 1 2 8 64 128

*

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert from Base-B to Base-10

4. Adds the results of each section

0.0625 0.125 0.25 0.5 1 2 4 8 16 32 64 128

4

3

2
1
1

2 3 4 5 6 7 1 1 . 1 1 1 1 1 0.0625 * 1 0.125 * 0.25 * 1 0.5* 1 * 1 2 * 1 4 * 8 * 1 16 * 32 * 64 * 1 128 * 1 0.0625 0.25 . 1 2 8 64 128

203 + 0.3125 203.3125

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert from Base-B to Base-10

Examples:

(a n-1 a n-2 … a 0 . a -1 … a -m ) B = (N)10 N = (a n-1 * B n-1) + (a n-2 * B n-2) + … + (a 0 * B 0) + (a -1 * B -1) + … + (a -m * B –m)

(4173)10 = (4 * 103) + (1 * 102) + (7*101) + (3*100) = (4173)10
(0756)8 = (0 * 83) + (7 * 82) + (5*81) + (6*80) = (494)10
(3b2)16 = (3 * 162) + (11*161) + (2*160) = (946)10
(2E6.A3)16 = (2 * 162) + (14*161) + (6*160) +

(10 * (1 / 16)) + (3 * (1 / (16 * 16))) = (?)10

16 -1 16 -2

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert from Base-10 to Base-B

(N)10 = ( a n-1 a n-2 … a 0 . a -1 … a -m ) B

Integer part Fraction part

1. Convert integer part to Base-B

– Consecutive divisions

2. Convert fraction part to Base-B

– Consecutive multiplication

25.125 25 0.125

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert Integer Part to Base-B

Repeat until the quotient reaches 0
Write the reminders in the reverse order

– Last to first

Examples:

2 25 2 12 24 2 6 12 1 2 3 6 2 1 2 1 1

( 25 )10 = (11001) 2

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert Integer Part to Base-B

Examples:

8

494 8 61 488 8 7 56 6 5 7

(494)10 = (756)8

16 946 16 59 944 16 3 48 2 11 3

(946)10 = (3B2)16

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert Fraction Part to Base-B

Do
multiply fraction part by B (the result)
drop the integer part of the result (new fraction)
While
(result = 0) OR (reach to specific precision)
the integral parts from top to bottom are

arranged from left to right after the decimal point

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert Fraction Part to Base-B

Example:
0.125 * 2 = 0.25
0.25 * 2 = 0.50 (0.125)10 = (0.001)2
0.50 * 2 = 1.00
0.00 * 2 = 0.00

1.00

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Convert Fraction Part to Base-B

Example:

– 0.6 * 2 = 1.2 – 0.2 * 2 = 0.4 (0.6)10 = (0.1001)2 – 0.4 * 2 = 0.8 – 0.8 * 2 = 1.6 – 0.6 * 2 = …

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Conversion binary and octal

Binary to Octal
23 = 8
Each digit in octal format: 3 digits in binary format
If the number of digits is not dividable by 3, add

additional zeros:

43 = 043 = 000043 = 043.0 = 043.000
(10011.1101) 2 = (010011.110100) 2 = (23.64) 8

2 3 6 4

a3 a2 a1 a0

.

a-1 a-2

Octal 1 2 3 4 5 6 7 Binary 000 001 010 011 100 101 110 111

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Conversion binary and octal

Octal to Binary
Substitute each digit with 3 binary digits
(5)8 = (101)2
(1)8 = (001)2
(51)8 = (101 001)2
(23.61) 8 = (010 011.110 001) 2

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Conversion binary and Hexadecimal

Binary to Hexadecimal
24 = 16
Each digit in octal format: 4 digits in binary format
If the number of digits is not dividable by 4, add

additional zeros:

(1111101.0110) 16 = (01111101.0110) 16 = (7D.6) 2

7 D 6

a3 a2 a1 a0

.

a-1 a-2

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Conversion binary and Hexadecimal

Hexadecimal to Binary
Substitute each digit with 4 binary digits
(F25.03) 16 = (1111 0010 0101.0000 0011) 2

1111 0010 0101 . 0000 0011

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Conversion

Octal binary Hexadecimal
(345) 8 = (E5) 16
(345) 8 = (011100101) 2 = (011100101) 2 = (E5) 16

3 4 5 E 5

(3FA5) 16 = (0011111110100101) 2 =

(0011111110100101) 2 = (037645) 8 = (37645) 8

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Calculations in Numeral Systems

Addition
Binary

1 1 1 1 carried digits (13)10 1 1 0 1 (23)10 1 0 1 1 1 (36)10 1 0 0 1 0 1 + 1 → 1 0 1 + 1 + 1 → 1 1 1 1 1 1 1 1 0 1 1 (1 + 1 ) 10 → (2) 10 = (10) 2

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Calculations in Numeral Systems

Addition
Hexadecimal
Octal

4 5 6 7 8 4 B D A 1 1 7 8 4 B D A 1 B 5 E (8 + D ) 16 → (8 + 13) 10 = (21) 10 = (15) 16 1 1 1 1 1 7 7 7 1 4 7 6 1 0 1 2 (4 + 6 ) 10 → (10) 10 = (12) 8

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Calculations in Numeral Systems

Subtraction
Binary

2 0 2 Borrowed digits (13)10 1 1 0 1 (7)10 1 1 1 (6)10 0 1 1 (2) 10 = (10) 2 1 1 1 1 1 1 Borrowed digit

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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Calculations in Numeral Systems

Subtraction
Hexadecimal
Octal

16 + 0 = 16 16 + 3 = 19 2 3 5 + 16 =21 3 1 4 5 1 9 7 6 1 7 C F 3 6+ 8 = 14 4 6 7 3 7

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Department of Computer Engineering

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Signed Integer Representation

Negative numbers must be encode in binary

number systems

Well-known methods

– Sign-magnitude – One’s Complement – Two’s Complement

Which one is better?

– Calculation speed – Complexity of the computation circuit

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Department of Computer Engineering

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Sign-magnitude

One sign bit + magnitude bits

– Positive: s = 0 – Negative: s= 1 – Range = {(-127)10 .. (+127) 10} – Two ways to represent zero:

00000000 (+0)
10000000 (−0)

– Examples:

(+ 43)10 = 00101011
(- 43)10 = 10101011
How many positive and negative integers can be represented using

N bits?

– Positive: 2 N-1 - 1 – Negative: 2 N-1 - 1

1 2 3 4 5 6 7

s Magnitude Sign

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Number Systems – Lecture 3

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Department of Computer Engineering

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Sign-magnitude

Signed magnitude representation is easy for

people to understand, but it requires complicated computer hardware.

Another disadvantage of signed magnitude is

that it allows two different representations for zero: positive zero and negative zero.

For these reasons (among others) computers

systems employ complement systems for numeric value representation.

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Department of Computer Engineering

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One’s Complement

For example, using 8-bit one’s complement

representation:

+ 3 is: 00000011

3 is:

11111100

In one’s complement representation, as with signed

magnitude, negative values are indicated by a 1 in the high order bit.

Complement systems are useful because they

eliminate the need for subtraction. The difference of two values is found by adding the minuend to the complement of the subtrahend

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Number Systems – Lecture 3

Sharif University of Technology

Department of Computer Engineering

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One’s Complement

With one’s complement

addition, the carry bit is “carried around” and added to the sum.

– Example: Using one’s complement binary arithmetic, find the sum of 48 and - 19

We note that 19 in binary is

00010011,

so -19 in one’s complement is:

11101100.

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Department of Computer Engineering

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One’s Complement

Although the “end carry around” adds some

complexity, one’s complement is simpler to implement than signed magnitude.

But it still has the disadvantage of having two

different representations for zero: positive zero and negative zero.

Two’s complement solves this problem.
Two’s complement is the radix complement of

the binary numbering system; the radix complement of a non-zero number N in base r with d digits is rd – N.

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Department of Computer Engineering

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Two’s Complement

Negative numbers:

1. Invert all the bits through the number

~(0) = 1 ~(1) = 0

2. Add one

Example:
+1 = 00000001
- 1 = ?
~(00000001) → 11111110
11111110 + 1 → 11111111
Only one zero (00000000)
Range = {127 .. −128}
1

1 1 1 1 1 1 1 1

2

1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 Negative two's complement (11111101)

(00000011) = -3

~(11111101) + 1

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Department of Computer Engineering

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Two’s Complement

– Example: Using one’s complement binary arithmetic, find the sum of 48 and - 19.

With two’s complement arithmetic, all we do is add
ur two binary numbers. Just discard any carries

emitting from the high order bit.

We note that 19 in binary is:

00010011,

so -19 using one’s complement is:

11101100,

and -19 using two’s complement is: 11101101.

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Department of Computer Engineering

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When we use any finite number of bits to

represent a number, we always run the risk of the result of our calculations becoming too large

r too small to be stored in the computer.
While we can’t always prevent overflow, we can

always detect overflow.

In complement arithmetic, an overflow condition

is easy to detect.

Two’s Complement

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Department of Computer Engineering

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Example:

– Using two’s complement binary arithmetic, find the sum of 107 and 46.

We see that the nonzero carry

from the seventh bit overflows into the sign bit, giving us the erroneous result: 107 + 46 = -103.

But overflow into the sign bit does not always mean that we have an error.

Two’s Complement

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Department of Computer Engineering

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Example:

– Using two’s complement binary arithmetic, find the sum of 23 and

9.

– We see that there is carry into the sign bit and carry out. The final result is correct: 23 + (-9) = 14.

Rule for detecting signed two’s complement overflow: When the “carry in” and the “carry out” of the sign bit differ,

verflow has occurred. If the carry into the sign bit equals the

carry out of the sign bit, no overflow has occurred.

Two’s Complement

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Department of Computer Engineering

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The signed magnitude, one’s complement,

and two’s complement representation that we have just presented deal with signed integer values only.

Without modification, these formats are not

useful in scientific or business applications that deal with real number values.

Floating-point representation solves this

problem.

Floating-Point representation

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Department of Computer Engineering

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If we are clever programmers, we can perform

floating-point calculations using any integer format.

This is called floating-point emulation, because

floating point values aren’t stored as such; we just create programs that make it seem as if floating- point values are being used.

Most of today’s computers are equipped with

specialized hardware that performs floating-point arithmetic with no special programming required.

Floating-Point representation

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Department of Computer Engineering

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Floating-point numbers allow an arbitrary

number of decimal places to the right of the decimal point.

– For example: 0.5  0.25 = 0.125

They are often expressed in scientific notation.

– For example: 0.125 = 1.25  10-1 5,000,000 = 5.0  106

Floating-Point representation

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Department of Computer Engineering

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Computers use a form of scientific notation for

floating-point representation

Numbers written in scientific notation have three

components:

Floating-Point representation

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Department of Computer Engineering

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Computer representation of a floating-point

number consists of three fixed-size fields:

This is the standard arrangement of these fields.

Note: Although “significand” and “mantissa” do not technically mean the same thing, many people use these terms interchangeably. We use the term “significand” to refer to the fractional part of a floating point number.

Floating-Point representation

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Department of Computer Engineering

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We introduce a hypothetical “Simple Model” to

explain the concepts

In this model:

– A floating-point number is 14 bits in length – The exponent field is 5 bits – The significand field is 8 bits

Floating-Point representation

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Department of Computer Engineering

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Example:

– Express 3210 in the simplified 14-bit floating-point model.

We know that 32 is 25. So in (binary) scientific

notation 32 = 1.0 x 25 = 0.1 x 26. – In a moment, we’ll explain why we prefer the second notation versus the first.

Using this information, we put 110 (= 610) in the

exponent field and 1 in the significand as shown.

Floating-Point representation

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Department of Computer Engineering

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The IEEE has established a standard for

floating-point numbers

The IEEE-754 single precision floating point

standard uses an 8-bit exponent (with a bias of 127) and a 23-bit significand.

The IEEE-754 double precision standard uses

an 11-bit exponent (with a bias of 1023) and a 52-bit significand.

Floating-Point representation

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Department of Computer Engineering

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In both the IEEE single-precision and double-

precision floating-point standard, the significant has an implied 1 to the LEFT of the radix point.

– The format for a significand using the IEEE format is: 1.xxx… – For example, 4.5 = .1001 x 23 in IEEE format is 4.5 = 1.001 x

22. The 1 is implied, which means is does not need to be

listed in the significand (the significand would include only 001).

Floating-Point representation

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Department of Computer Engineering

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Example: Express -3.75 as a floating point number using

IEEE single precision.

First, let’s normalize according to IEEE rules:

– 3.75 = -11.112 = -1.111 x 21 – The bias is 127, so we add 127 + 1 = 128 (this is our exponent) – The first 1 in the significand is implied, so we have: – Since we have an implied 1 in the significand, this equates to

(1).1112 x 2 (128 – 127) = -1.1112 x 21 = -11.112 = -3.75.

(implied)

Floating-Point representation

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Department of Computer Engineering

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ASCII Codes

American Standard Code for Information Interchange

– First decision: 7 bits for representation – Final decision: 8 bits for representation

256 characters
ASCII (“P”) = (50) 16

ASCII (“=”) = (3D) 16

F E D C B A 9 8 7 6 5 4 3 2 1 Hexadecimal SI SO CR FF VT LF TAB BS BEL ACK ENQ EOT ETX STX SOH NUL US RS GS FS ESC SUB EM CAN ETB SYN NAK DC4 DC3 DC2 DC1 DLE 1 / .

,

+ * ( ) ' & % $ # " ! 2 ? < = > ; : 9 8 7 6 5 4 3 2 1 3 O N M L K J I H G F E D C B A @ 4 _ ^ [ \ ] Z Y X W V U T S R Q P 5

n

m l k j i h g f e d c b a ` 6 ~ { | } z y x w v u t s r q p 7

row number column number

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Department of Computer Engineering

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Summary

Numeral Systems

– Decimal, Binary, Octal, Hexadecimal

Computer Data Storage Units

– Bit, Byte, Kilo byte, Giga byte,

Numeral Systems Conversion

– Convert between different bases

Calculations in Number Systems

– Addition and subtraction

Signed Integer Representation

– Sing-magnitude: one sign bit + magnitude bits – Two’s complement : (-N) = ~(N) + 1

Fractional and Real Numbers
ASCII Codes