1. CDA 3100 Summer 2013

2. Special Thanks
Thanks to Dr. Zhenghao Zhang for letting me use his class slides and other materials as a base for this course

3. Course Information

4. About Me Name: Britton Dennis
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About Me
Name: Britton Dennis
Office: LOV 105A
Office Hours: Thursday 9:30 AM – 12:30 PM
Site:
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CDA3100

5. 11/27/2018
Class Communication
This class will use class web site to post news, changes, and updates. So please check the class website regularly
Please also make sure that you check your s on the account on your University record
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CDA3100

6. Lecture Notes and Textbook
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Lecture Notes and Textbook
All the materials that you will be tested on will be covered in the lectures
All lectures will be posted on the website in both (Microsoft Office 2010 published) .pptx and in the more universally viewable .pdf
The textbook isn’t required, but it may be worth purchasing for review and for further detail explanations
In particular, the appendix will be useful for MIPS instruction references
The lectures will be based on the textbook and any handouts distributed in class
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CDA3100

7. Required Textbook The required textbook for this class is
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Required Textbook
The required textbook for this class is
“Computer Organization and Design”
The hardware/software interface
By David A. Patterson and John L. Hennessy
Fourth Edition
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CDA3100

8. Grades and Weighting In Class Exercises – 5% Homework – 40% Exams
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Grades and Weighting
In Class Exercises – 5%
5-7 exercises will be given during class throughout the semester
Only top 4 will be graded
Homework – 40%
7 problems sets will be given throughout the semester
There will be two windows to turn them in.
Up to the first is the real deadline
Up to the second, you will receive a 10% deduction.
After that, the assignment won’t be accepted
For accreditation reasons, the department requires you to pass a particular assignment to pass the course
Exams
Midterm – 20%
Final – 35%
Cumulative
About 10 Multiple Choice and 2-3 Short Response
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CDA3100

9. Motivations

10. What you will learn to answer (among other things)
How does the software instruct the hardware to perform the needed functions
What is going on in the processor
How a simple processor is designed

11. Why This Class Important?
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Why This Class Important?
If you want to create better computers
It introduces necessary concepts, components, and principles for a computer scientist
By understanding the existing systems, you may create better ones
If you want to build software with better performance
If you want to have a good choice of jobs
If you want to be a real computer scientist
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CDA3100

12. Required Background

13. Required Background
You will probably suffer if you do not have the required C/C++ programming background.
In this class, we will be doing assembly coding, which is more advanced than C/C++.
If you do not have a clear understanding at this moment of array and loop , it is recommended that you take this course at a later time, after getting more experience with C/C++ programming.

14. Array – What Happens?
#include <stdio.h> int main (void) { int A[5] = {16, 2, 77, 40, 12071}; A[A[1]] += 1; return 0; }
A[1] = 2, so A[2] <- 78.

15. Loop – What Happens?
#include <stdio.h> int main () { int A[5] = {16, 20, 77, 40, 12071}; int result = 0; int i = 0; while (result < A[i]) { result += A[i]; i++; } return 0;
result = 113, i=3

16. Getting Started!

17. Decimal Numbering System
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Decimal Numbering System
We humans naturally use a particular numbering system
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18. Decimal Numbering System
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Decimal Numbering System
For any nonnegative integer , its value is given by
Here d0 is the least significant digit and dn is the most significant digit
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CDA3100

19. General Numbering System – Base X
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General Numbering System – Base X
Besides 10, we can use other bases as well
In base X,
Then, the base X representation of this number is defined as dndn-1…d2d1d0.
The same number can have many representations on many bases. For 23 based 10, it is
23ten
10111two
17sixteen, often written as 0x17.
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20. Commonly Used Bases Which one is natural to computers? Why? Base
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Commonly Used Bases
Base
Common Name
Representation
Digits 10
Decimal
5023ten or 5023
0-9 2
Binary
two
0-1 8
Octal
11637eight
0-7 16
Hexadecimal
139Fhex or 0x139F
0-9, A-F
Note that other bases are used as well including 12 and 60
Which one is natural to computers?
Why?
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CDA3100

21. Meaning of a Number Representation
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Meaning of a Number Representation
When we specify a number, we need also to specify the base
For example, 10 presents a different quantity in a different base
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22. 11/27/2018
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23. Question
How many different numbers that can be represented by 4 bits?

24. Question How many different numbers that can be represented by 4 bits?
Always 16 (24), because there are this number of different combinations with 4 bits, regardless of the type of the number these 4 bits are representing.
Obviously, this also applies to other number of bits. With n bits, we can represent 2n different numbers. If the number is unsigned integer, it is from 0 to 2n-1.

25. Conversion between Representations
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Conversion between Representations
Now we can represent a quantity in different number representations
How can we convert a decimal number to binary?
How can we then convert a binary number to a decimal one?
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CDA3100

26. Conversion Between Bases
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Conversion Between Bases
From binary to decimal example 1 15 14 13 12 11 10 9 8 7 6 5 4 3 2
215
214
213
212
211
210 29 28 27 26 25 24 23 22 21 20
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CDA3100

27. Converting from binary to decimal
Converting from binary to decimal. This conversion is also based on the formula:
d = dn-12n-1 + dn-22n-2 +…+ d222 + d121 + d020
while remembering that the digits in the binary representation are the coefficients.
For example, given two, in decimal, it is
= 43.

28. Conversion Between Bases
Converting from decimal to binary:
Repeatedly divide it by 2, until the quotient is 0.
Going from top to bottom, write down the remainder from right to the left.
Example: 11.
Quotient
Remainder 5 1 2

29. Digging a little deeper
Why can a binary number be obtained by keeping on dividing by 2, and why should the last remainder be the first bit?
Note that
Any integer can be represented by the summation of the powers of 2:
d = dn-12n-1 + dn-22n-2 +…+ d222 + d121 + d020
For example, 19 = = 1 * * * * * 20.
The binary representation is the binary coefficients. So 19ten in binary is 10011two.

30. Digging a little deeper
In fact, any integer can be represented by the summation of the powers of some base, where the base is either 10, 2 or 16 in this course. For example, 19 = 1 * * 100. How do you get the 1 and 9? You divide 19 by 10 repeatedly until the quotient is 0, same as binary!
In fact, the dividing process is just an efficient way to get the coefficients.
How do you determine whether the last bit is 0 or 1? You can do it by checking whether the number is even or odd. Once this is determined, you go ahead to determine the next bit, by checking (d - d020)/2 is even or odd, and so on, until you don’t have to check any more (when the number is 0).

31. Conversion between Base 16 and Base 2
Extremely easy.
From base 2 to base 16: divide the digits in to groups of 4, then apply the table.
From base 16 to base 2: replace every digit by a 4-bit string according to the table.
Because 16 is 2 to the power of 4.

32. Addition in binary
39ten + 57ten = ?
How to do it in binary?

33. Addition in Binary
First, convert the numbers to binary forms. We are using 8 bits.
39ten ->
57ten ->
Second, add them.

34. Addition in binary The addition is bit by bit.
We will encounter at most 4 cases, where the leading bit of the result is the carry:
0+0+0=00
1+0+0=01
1+1+0=10
1+1+1=11

35. Subtraction in Binary
57ten – 39ten = ?

36. Subtraction in Binary

37. Subtraction in binary Do this digit by digit. No problem if 0 - 0 = 0,
1 - 0 = 1
1 – 1 = 0.
When encounter 0 - 1, set the result to be 1 first, then borrow 1 from the next more significant bit, just as in decimal.
Borrow means setting the borrowed bit to be 0 and the bits from the bit following the borrowed bit to the bit before the current bit to be 1.
Think about, for example, subtracting 349 from 5003 (both based 10). The last digit is first set to be 4, and you will be basically subtracting 34 from 499 from now on.