PowerPoint Presentation
Basic Concepts
COMSC 260
Assembly Language
Computer Organization
Outline
Welcome to COMSC 260
Assembly-, Machine-, and High-Level Languages
Assembly Language Programming Tools
Programmer’s View of a Computer System
Basic Computer Organization
2
Memory Devices
Random-Access Memory (RAM)
Usually called the main memory
It can be read and written to
It does not store information permanently (Volatile , when it is powered off, the stored information are gone)
Information stored in it can be accessed in any order at equal time periods (hence the name random access)
Information is accessed by an address that specifies the exact location of the piece of information in the RAM.
DRAM = Dynamic RAM
1-Transistor cell + trench capacitor
Dense but slow, must be refreshed
Typical choice for main memory
SRAM: Static RAM
6-Transistor cell, faster but less dense than DRAM
Typical choice for cache memory
Memory Devices
ROM (Read-Only-Memory)
A read-only-memory, non-volatile i.e. stores information permanently
Has random access of stored information
Used to store the information required to startup the computer
Many types: ROM, EPROM, EEPROM, and FLASH
FLASH memory can be erased electrically in blocks
Cache
A very fast type of RAM that is used to store information that is most frequently or recently used by the computer
Recent computers have 2-levels or more levels of cache; the first level is faster but smaller in size (usually called internal cache), and the second level is slower but larger in size (external cache).
Processor-Memory Performance Gap
1980 – No cache in microprocessor
1995 – Two-level cache on microprocessor
CPU: 55% per year
DRAM: 7% per year
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Processor-Memory
Performance Gap:
(grows 50% per year)
Performance
“Moore’s Law”
The Need for a Memory Hierarchy
Widening speed gap between CPU and main memory
Processor operation takes less than 1 ns
Main memory requires more than 50 ns to access
Each instruction involves at least one memory access
One memory access to fetch the instruction
Additional memory accesses for instructions involving memory data access
Memory bandwidth limits the instruction execution rate
Cache memory can help bridge the CPU-memory gap
Cache memory is small in size but fast
Typical Memory Hierarchy
Registers are at the top of the hierarchy
Typical size < 1 KB
Access time < 0.5 ns
Level 1 Cache (8 – 64 KB)
Access time: 0.5 – 1 ns
L2 Cache (512KB – 8MB)
Access time: 2 – 10 ns
Main Memory (1 – 2 GB)
Access time: 50 – 70 ns
Disk Storage (> 200 GB)
Access time: milliseconds
Microprocessor
Registers
L1 Cache
L2 Cache
Memory
Disk, Tape, etc
Memory Bus
I/O Bus
Faster
Bigger
Magnetic Disk Storage
Track 0
Track 1
Sector
Recording area
Spindle
Direction of rotation
Platter
Read/write head
Actuator
Arm
Track 2
Disk Access Time =
Seek Time +
Rotation Latency +
Transfer Time
Seek Time: head movement to the desired track (milliseconds)
Rotation Latency: disk rotation until desired sector arrives under the head
Transfer Time: to transfer data
Example on Disk Access Time
Given a magnetic disk with the following properties
Rotation speed = 7200 RPM (rotations per minute)
Average seek = 8 ms, Sector = 512 bytes, Track = 200 sectors
Calculate
Time of one rotation (in milliseconds)
Average time to access a block of 32 consecutive sectors
Answer
Rotations per second
Rotation time in milliseconds
Average rotational latency
Time to transfer 32 sectors
Average access time
= 7200/60 = 120 RPS
= 1000/120 = 8.33 ms
= time of half rotation = 4.17 ms
= (32/200) * 8.33 = 1.33 ms
= 8 + 4.17 + 1.33 = 13.5 ms
Data Representation
Chapter 2
Outline
Introduction
Numbering Systems
Binary & Hexadecimal Numbers
Base Conversions
Integer Storage Sizes
Binary and Hexadecimal Addition
Signed Integers and 2’s Complement Notation
Binary and Hexadecimal subtraction
Carry and Overflow
Character Storage
Introduction
Computers only deal with binary data (0s and 1s), hence all data manipulated by computers must be represented in binary format.
Machine instructions manipulate many different forms of data:
Numbers:
Integers: 33, +128, -2827
Real numbers: 1.33, +9.55609, -6.76E12, +4.33E-03
Alphanumeric characters (letters, numbers, signs, control characters): examples: A, a, c, 1 ,3, “, +, Ctrl, Shift, etc.
Images (still or moving): Usually represented by numbers representing the Red, Green and Blue (RGB) colors of each pixel in an image,
Sounds: Numbers representing sound amplitudes sampled at a certain rate (usually 20kHz).
So in general we have two major data types that need to be represented in computers; numbers and characters.
Numbering Systems
Numbering systems are characterized by their base number.
In general a numbering system with a base r will have r different digits (including the 0) in its number set. These digits will range from 0 to r-1
The most widely used numbering systems are listed in the table below:
Binary Numbers
Each digit (bit) is either 1 or 0
Each bit represents a power of 2
Every binary number is a sum of powers of 2
Converting Binary to Decimal
Weighted positional notation shows how to calculate the decimal value of each binary bit:
Decimal = (dn-1 2n-1) + (dn-2 2n-2) + … + (d1 21) + (d0 20)
d = binary digit
binary 10101001 = decimal 169:
(1 27) + (1 25) + (1 23) + (1 20) = 128+32+8+1=169
Convert Unsigned Decimal to Binary
Repeatedly divide the decimal integer by 2. Each remainder is a binary digit in the translated value:
37 = 100101
stop when quotient is zero
least significant bit
most significant bit
Another Procedure for Converting from Decimal to Binary
Start with a binary representation of all 0’s
Determine the highest possible power of two that is less or equal to the number.
Put a 1 in the bit position corresponding to the highest power of two found above.
Subtract the highest power of two found above from the number.
Repeat the process for the remaining number
Another Procedure for Converting from Decimal to Binary
Example: Converting 76d to Binary
The highest power of 2 less or equal to 76 is 64, hence the seventh (MSB) bit is 1
Subtracting 64 from 76 we get 12.
The highest power of 2 less or equal to 12 is 8, hence the fourth bit position is 1
We subtract 8 from 12 and get 4.
The highest power of 2 less or equal to 4 is 4, hence the third bit position is 1
Subtracting 4 from 4 yield a zero, hence all the left bits are set to 0 to yield the final answer
Hexadecimal Integers
Binary values are represented in hexadecimal.
Converting Binary to Hexadecimal
Each hexadecimal digit corresponds to 4 binary bits.
Example: Translate the binary integer 000101101010011110010100 to hexadecimal
Converting Hexadecimal to Binary
Each Hexadecimal digit can be replaced by its 4-bit binary number to form the binary equivalent.
Converting Hexadecimal to Decimal
Multiply each digit by its corresponding power of 16:
Decimal = (d3 163) + (d2 162) + (d1 161) + (d0 160)
d = hexadecimal digit
Examples:
Hex 1234 = (1 163) + (2 162) + (3 161) + (4 160) =
Decimal 4,660
Hex 3BA4 = (3 163) + (11 * 162) + (10 161) + (4 160) =
Decimal 15,268
Converting Decimal to Hexadecimal
Decimal 422 = 1A6 hexadecimal
stop when quotient is zero
least significant digit
most significant digit
Repeatedly divide the decimal integer by 16. Each remainder is a hex digit in the translated value:
Integer Storage Sizes
What is the largest unsigned integer that may be stored in 20 bits?
Standard sizes:
Binary Addition
Start with the least significant bit (rightmost bit)
Add each pair of bits
Include the carry in the addition, if present
0
0
0
0
0
1
1
1
0
0
0
0
0
1
0
0
+
0
0
0
0
1
0
1
1
1
(4)
(7)
(11)
carry:
0
1
2
3
4
bit position:
5
6
7
Hexadecimal Addition
Divide the sum of two digits by the number base (16). The quotient becomes the carry value, and the remainder is the sum digit.
36 28 28 6A
42 45 58 4B
78 6D 80 B5
1
1
21 / 16 = 1, remainder 5
Important skill: Programmers frequently add and subtract the addresses of variables and instructions.
Signed Integers
Several ways to represent a signed number
Sign-Magnitude
1’s complement
2’s complement
Divide the range of values into 2 equal parts
First part corresponds to the positive numbers (≥ 0)
Second part correspond to the negative numbers (< 0)
Focus will be on the 2's complement representation
Has many advantages over other representations
Used widely in processors to represent signed integers
Two's Complement Representation
8-bit Binary
value Unsigned
value Signed
value
00000000 0 0
00000001 1 +1
00000010 2 +2
. . . . . . . . .
01111110 126 +126
01111111 127 +127
10000000 128 -128
10000001 129 -127
. . . . . . . . .
11111110 254 -2
11111111 255 -1
Positive numbers
Signed value = Unsigned value
Negative numbers
Signed value = Unsigned value - 2n
n = number of bits
Negative weight for MSB
Another way to obtain the signed value is to assign a negative weight to most-significant bit
= -128 + 32 + 16 + 4 = -76
1
0
1
1
0
1
0
0
-128
64
32
16
8
4
2
1
Forming the Two's Complement
Sum of an integer and its 2's complement must be zero:
00100100 + 11011100 = 00000000 (8-bit sum) Ignore Carry
The easiest way to obtain the 2's complement of a binary number is by starting at the LSB, leaving all the 0s unchanged, look for the first occurrence of a 1. Leave this 1 unchanged and complement all the bits after it.
starting value 00100100 = +36
step1: reverse the bits (1's complement) 11011011
step 2: add 1 to the value from step 1 + 1
sum = 2's complement representation 11011100 = -36
Sign Bit
Highest bit indicates the sign. 1 = negative, 0 = positive
If highest digit of a hexadecimal is > 7, the value is negative
Examples: 8A and C5 are negative bytes
A21F and 9D03 are negative words
B1C42A00 is a negative double-word
Sign Extension
Step 1: Move the number into the lower-significant bits
Step 2: Fill all the remaining higher bits with the sign bit
This will ensure that both magnitude and sign are correct
Examples
Sign-Extend 10110011 to 16 bits
Sign-Extend 01100010 to 16 bits
Infinite 0s can be added to the left of a positive number
Infinite 1s can be added to the left of a negative number
10110011 = -77
11111111 10110011 = -77
01100010 = +98
00000000 01100010 = +98
Two’s Complement of a Hexadecimal
To form the two’s complement of a hexadecimal
Subtract each hexadecimal digit from 15
Add 1
Examples:
2’s complement of 6A3D = 95C3
2’s complement of 92F0 = 6D10
2’s complement of FFFF = 0001
No need to convert hexadecimal to binary
Two’s Complement of a Hexadecimal
Start at the least significant digit, leaving all the 0s unchanged, look for the first occurrence of a non-zero digit.
Subtract this digit from 16.
Then subtract all remaining digits from 15.
Examples:
2’s complement of 6A3D = 95C3
2’s complement of 92F0 = 6D10
2’s complement of FFFF = 0001
F F F 16
6 A 3 D
————–
9 5 C 3
F F 16
9 2 F 0
————–
6 D 1 0
Binary Subtraction
When subtracting A – B, convert B to its 2’s complement
Add A to (–B)
0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0
0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 (2’s complement)
0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 (same result)
Carry is ignored, because
Negative number is sign-extended with 1’s
You can imagine infinite 1’s to the left of a negative number
Adding the carry to the extended 1’s produces extended zeros
Practice: Subtract 00100101 from 01101001.
–
+
Hexadecimal Subtraction
When a borrow is required from the digit to the left, add 16 (decimal) to the current digit’s value
Last Carry is ignored
Practice: The address of var1 is 00400B20. The address of the next variable after var1 is 0040A06C. How many bytes are used by var1?
C675
A247
242E
-1
–
16 + 5 = 21
C675
5DB9 (2’s complement)
242E (same result)
1
+
1
Ranges of Signed Integers
The unsigned range is divided into two signed ranges for positive and negative numbers
Practice: What is the range of signed values that may be stored in 20 bits?
Carry and Overflow
Carry is important when …
Adding or subtracting unsigned integers
Indicates that the unsigned sum is out of range
Either < 0 or > maximum unsigned n-bit value
Overflow is important when …
Adding or subtracting signed integers
Indicates that the signed sum is out of range
Overflow occurs when
Adding two positive numbers and the sum is negative
Adding two negative numbers and the sum is positive
Can happen because of the fixed number of sum bits
0
1
0
0
0
0
0
0
0
1
0
0
1
1
1
1
+
1
0
0
0
1
1
1
1
79
64
143
(-113)
Carry = 0 Overflow = 1
1
1
0
0
1
1
1
0
1
1
1
0
1
1
0
1
0
+
0
1
1
1
0
1
1
1
218 (-38)
157 (-99)
119
Carry = 1 Overflow = 1
1
1
1
Carry and Overflow Examples
We can have carry without overflow and vice-versa
Four cases are possible
1
1
1
1
1
0
0
0
0
0
0
0
1
1
1
1
+
0
0
0
0
0
1
1
1
15
245 (-8)
7
Carry = 1 Overflow = 0
1
1
1
1
1
0
0
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1
0
0
0
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0
0
0
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1
1
1
+
0
0
0
1
0
1
1
1
15
8
23
Carry = 0 Overflow = 0
1
Character Storage
Character sets
Standard ASCII: 7-bit character codes (0 – 127)
Extended ASCII: 8-bit character codes (0 – 255)
Unicode: 16-bit character codes (0 – 65,535)
Unicode standard represents a universal character set
Defines codes for characters used in all major languages
Used in Windows-XP: each character is encoded as 16 bits
UTF-8: variable-length encoding used in HTML
Encodes all Unicode characters
Uses 1 byte for ASCII, but multiple bytes for other characters
Null-terminated String
Array of characters followed by a NULL character
ASCII Codes
Examples:
ASCII code for space character = 20 (hex) = 32 (decimal)
ASCII code for ‘A’ = 41 (hex) = 65 (decimal)
ASCII code for ‘a’ = 61 (hex) = 97 (decimal)
Control Characters
The first 32 characters of ASCII table are used for control
Control character codes = 00 to 1F (hex)
Examples of Control Characters
Character 0 is the NULL character used to terminate a string
Character 9 is the Horizontal Tab (HT) character
Character 0A (hex) = 10 (decimal) is the Line Feed (LF)
Character 0D (hex) = 13 (decimal) is the Carriage Return (CR)
The LF and CR characters are used together
They advance the cursor to the beginning of next line
One control character appears at end of ASCII table
Character 7F (hex) is the Delete (DEL) character
Parity Bit
Data errors can occur during data transmission or storage/retrieval.
The 8th bit in the ASCII code is used for error checking.
This bit is usually referred to as the parity bit.
There are two ways for error checking:
Even Parity: Where the 8th bit is set such that the total number of 1s in the 8-bit code word is even.
Odd Parity: The 8th bit is set such that the total number of 1s in the 8-bit code word is odd.
1
1
1
1
1
1
1
1
2
7
2
6
2
5
2
4
2
3
2
2
2
1
2
0
M1023.swf
M1021.swf
byte
16
8
32
word
doubleword
64
quadword
1
1
1
1
0
1
1
0
0
0
0
0
1
0
1
0
sign bit
Negative
Positive
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