1. 1 Syllabus Instructor: Dr. Wesam Ashour Office: B231 Email: Washour@iugaza.eduWashour@iugaza.edu Office Hours: Saturday 10:00  11:00, 13:00  14:00 Sunday 11:00  13:00. Wednesday 10:00  11:00.

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3. 3 Syllabus Course: ECOM 3310 – Computer Architecture Contents: 1- Ch.1: Introduction. 2- Ch.2: Instructions: Language of the Computer. 3- Ch.3: Arithmetic for Computers. 4- Ch.4: Assessing and Understanding Performance. 5- Ch.5: The Processor: Datapath and Control. 6- Ch.6: Enhancing Performance with Pipelining. 7- Ch.7: Large and Fast: Exploiting Memory Hierarchy 8- Ch.8: Input/Output Systems We will cover as much as we can!

4. 4 Syllabus Grades: 1- 30% Mid Term Exam. 2- 50% Final Exam. 3- Teaching Assistants control %20. Midterm Exam: Tuesday, 16 th November 2010, 11:00am-12:30pm Final Exam: Saturday, 15 th January 2011, 11:00-14:00 Important Notice: Sometimes (& Suddenly) I will test you by giving Quizzes at the beginning of the lecture! - To solve it easily and to get the highest mark, you should always read the last two lectures before attending the class – Good Luck. Important Advice: From first day in the semester you should work very hard and you must activate your mind to think!

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42. 42 lw $t0, 1200($t1) add $t0, $s2, $t0 sw $t0, 1200($t1) What is the MIPS machine language code for these three instructions? 1200 (in base 10) = 0000 0100 1011 0000 (binary) $t0 register is number 8 $t1 register is number 9 $s2 register is number 18 Opcode -> lw is 35 (100011), sw is 43 (101011), R-type (000000) add func is 32 (100000)

43. 43 lw $t0, 1200($t1) add $t0, $s2, $t0 sw $t0, 1200($t1) What is the MIPS machine language code for these three instructions? 1200 (in base 10) = 0000 0100 1011 0000 (binary) $t0 register is number 8 $t1 register is number 9 $s2 register is number 18 Opcode -> lw is 35 (100011), sw is 43 (101011), R-type (000000) add func is 32 (100000) 100011 01001 01000 0000010010110000 000000 10010 01000 01000 00000 100000 101011 01001 01000 0000010010110000

44. 44 or $t0, $t1, $t2 ori $t0, $t1, 255 sll $t0, $t1, 4 and $t0, $t1, $t2

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49. 49 if (save[i] >= k) Code; Assume save[i] is stored in $t1 and k is stored in $t2

50. 50 slt $t0, $t1, $t2 bne $t0, $zero, Exit Code Exit: if (save[i] >= k) Code; Assume save[i] is stored in $t1 and k is stored in $t2

51. 51 if (save[i] < k) Code; Assume save[i] is stored in $t1 and k is stored in $t2

52. 52 slt $t0, $t1, $t2 beq $t0, $zero, Exit Code Exit: if (save[i] < k) Code; Assume save[i] is stored in $t1 and k is stored in $t2

53. 53 if (save[i] > k) Code; Assume save[i] is stored in $t1 and k is stored in $t2

54. 54 slt $t0, $t1, $t2 bne $t0, $zero, Exit beq $t1, $t2, Exit Code Exit: if (save[i] > k) Code; Assume save[i] is stored in $t1 and k is stored in $t2

55. 55 if (save[i] <= k) Code; Assume save[i] is stored in $t1 and k is stored in $t2 Try it by yourself

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