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CDA 3100 Summer 2013

CDA 3100 Summer 2013. Special Thanks. Thanks to Dr. Zhenghao Zhang for letting me use his class slides and other materials as a base for this course. Course Information. About Me. Name: Britton Dennis Email: dennis@cs.fsu.edu Office: LOV 105A Office Hours: Thursday 9:30 AM – 12:30 PM

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CDA 3100 Summer 2013

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  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 • Email: dennis@cs.fsu.edu • Office: LOV 105A • Office Hours: Thursday 9:30 AM – 12:30 PM • Site: http://ww2.cs.fsu.edu/~dennis/cda3100_summer_2013 CDA3100

  5. 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 emails on the account on your University record CDA3100

  6. 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 CDA3100

  7. 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 CDA3100

  8. 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 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? • 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 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; }

  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; }

  16. Getting Started!

  17. We humans naturally use a particular numbering system Decimal Numbering System CDA3100

  18. 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 CDA3100

  19. 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. CDA3100

  20. Commonly Used Bases • Note that other bases are used as well including 12 and 60 • Which one is natural to computers? • Why? CDA3100

  21. 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 CDA3100

  22. Question • How many different numbers that can be represented by 4 bits?

  23. 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.

  24. 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? CDA3100

  25. Conversion Between Bases • From binary to decimal example CDA3100

  26. 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 101011two, in decimal, it is 25 + 23 + 21 + 20 = 43.

  27. 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.

  28. 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 = 16 + 2 + 1 = 1 * 24 + 0 * 23 + 0 * 22 + 1 * 21 + 1 * 20. • The binary representation is the binary coefficients. So 19ten in binary is 10011two.

  29. 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 * 101 + 9 * 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).

  30. 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.

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

  32. Addition in Binary • First, convert the numbers to binary forms. We are using 8 bits. • 39ten -> 001001112 • 57ten -> 001110012 • Second, add them. 00100111 00111001 01100000

  33. 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

  34. Subtraction in Binary • 57ten – 39ten = ?

  35. Subtraction in Binary 00111001 00100111 00010010

  36. 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.

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