1. Review of the numeration systems The hardware/software representation of the computer and the coverage of that representation by this course. What is the instruction and ISA? CISC/RISC tradeoff. Learning outcome by the end of this class the students should know the main differences and distinguish RISC solutions from the CISC solutions The Goal of this class

2. Why to use the assembly and why we need this course at all? Operating systems behavior cannot be understood without knowledge of computer structure and the assembly language. PCB, context switch, threads, stack, heap, dynamic linking Assembly language advantages versus High level languages The primary reason to program in assembly language is that the speed or size of a program is critically important. A hybrid approach, in which most of a program is written in a high- level language and time-critical sections are written in assembly language, builds on the strengths of both languages Cases when no high-level language or compiler is available on a particular computer Ability to exploit specialized instructions, for example, string copy or pattern-matching instructions. How tens of programs work on one or couple of processors of your PC? It’s organized by the Operating System.

3. During the course we have to go through all the levels learn each of them separately learn the translation from one to another Finally we’ll put together the learned material and get the final picture of computer organization. The abstraction of the different parts of computer hardware / software allows us to create different Levels of Representation / Interpretation. This makes easier to understand each part separately and get the whole picture putting together all components. Human writes the code

4. Course Organization, Evaluation Student learning will be evaluated on the basis of the following weighted components:  Home assignments 20% of final grade  Grade of midterm examinations, Quizzes50%  Grade of final examination 30%  Bonus scores:  Class attendance/participation5-10%  Bonus Projects (deadline of taking February 28)5- 10% For the students who pass the waiver examination (2016.01.31 – 9:00 – 11:00) the waiver examination grade could be passed as a final grade and student is free to attend the classes. Waiver examination questions: 1. Write Assembly language program (cycles, i/o, arrays). 2. Representation of data in computers. (instructions, characters, numbers) 3. Create a simple combinational logic circuit. Office Hours - Thursday 19:00, Saturday 12:00 Aidela Karamyan 093249954

5. Course Organization, Evaluation Handouts’ topics fully cover all examination questions. (No questions will be out of handout topics). However handouts are not enough for full understanding the topics included there. The students need either to attend the class or work on textbooks for full understanding. Class attendance is graded with the bonus scores. Assignments are based only on the material of the current and previous classes’ handouts and exercises. They contain personal identifiers to make them different for different students. Assignments’ soft deadline is about a week. For example if the deadline is on Sunday then the assignment should be sent to lecturer before the Sunday evening 23:59. Delayed assignments’ grades are lowered. Assignments’ hard deadline is about 2 weeks. Assignments are not accepted after the hard deadline.

6. The course material is on the Academic server ac.aua.am

10. Instruction Set Architecture Basic job of a CPU: execute lots of instructions. Instructions are the primitive operations that the CPU may execute. Different CPUs implement different sets of instructions. The set of instructions a particular CPU implements defines the Instruction Set Architecture (ISA). Examples: Intel x86 (Pentium 4), Apple/IBM/Motorola PowerPC, MIPS, Sun Sparc,...

11. Computer architecture

12. As the Memory was expensive – the programs should have been short: Early trend was to add more and more instructions to new CPUs to do elaborate operations In order to have shorter programs to fit into the small memory VAX architecture had an instruction to multiply polynomials. intel x86 had string or array operation instructions. Memory becomes cheaper – the programs could be longer: RISC philosophy – Reduced Instruction Set Computing. Keep the instruction set small and simple, makes it easier to build fast hardware. Let software do complicated operations by composing simpler ones. CISC and RISC approaches

13. Complex Instruction Set Computer memc = mema + memb z = (a + b)(c + d) arr2[1..n] = arr1[1..n] bool1 = str1 cmp str2 short long instr longer instr instr... Processor (very complex) memc = mema + memb z = (a + b)(c + d) bool1 = str1 cmp str2 Small and expensive Memory Short programs with powerful instructions Many different types of instructions Complex and powerful instructions

14. Reduced Instruction Set Computer bool1 = str1 cmp str2 short long instr longer instr instr... Processor (very simple but fast with possibility to make parallelism) Large and cheap Memory Many and Long programs with simple instructions Complex and powerful instructions Few types of instructions y = c + d x = a + b c = a + b memc = mema + memb z = (a + b)(c + d) arr2[1..n] = arr1[1..n] c = a + b Simple, uniform size instructions c = a + b z = x * y inst... Many different types of instructions z = x * y a = mema b = memb memc = c a = mema memc = c

15. MIPS – semiconductor company that built one of the first commercial RISC architectures. We will study the MIPS architecture in detail in this course. Why MIPS instead of Intel 80x86? MIPS is simple. MIPS Architecture

16. RISC, CISC architectures tradeoff

17. High and Low level languages Intel x86 MIPSSun Sparc Assembly Program C Program C++ Program Java Program C#,.net Program

18. Numeration systems Review Doing binary correct operations in the computer we are sure that the translated result to decimal also is correct.

19. Numeration systems

22. Binary system in computers

23. Bit patterns

24. Bit grouping

26. Conversion Example From Decimal to Binary Convert the decimal 54.4062510 to binary: