Marking Scheme. Examination Paper. Module: Microprocessors (630313)

Transcription

1 Philadelphia University Faculty of Engineering Marking Scheme Examination Paper Department of CE Module: Microprocessors (630313) Final Exam First Semester Date: 30/01/2018 Section 1 Weighting 40% of the module total Lecturer: Coordinator: Internal Examiner: Dr. Qadri Hamarsheh Dr. Qadri Hamarsheh Dr. Nasser Halasa

2 Marking Scheme Microprocessors (630313) The presented exam questions are organized to overcome course material, the exam contains 5 questions; all questions are compulsory requested to be answered. Thus, the student is permitted to answer any question out of the existing ones in this section. Marking Assignments The following scheme shows the marks assignments for each question. They show also the steps for which a student can get marks along the related procedure he/she achieves. Question 1This question is attributed with 10 marks if answered properly The answer for this question as the following: 1) The first processor that includes Virtual Memory in the Intel microprocessor family was: 4004 c) Pentium Pro d) ) If SS = 90A3H, then the range of physical addresses for the stack segment is: 00000H 090A3H c) 090A3 190A2H 90A30H 9FA30H d) 90A30 A0A2FH 3) The special output bits in the adder/subtractor circuitry (like OF) are maintained in what register? EFLAGS c) EIP EBP d) ESP 4) Which directive is used when defining 64-bit IEEE long reals? REAL4 c) REAL64 REAL8 d) REAL 5) The instruction, CMP to compare source and destination operands it performs Addition c) Subtraction Division d) Multiplication 6) The instruction that loads effective address formed by destination operand into the specified source register is LES c) LDS LEA d) LAHF 7) What does the NEG instruction do? Two s Complement One s Complement 8) Given that BL contains B the effect of the following code OR BL, ADD BL, 2 is to clear BL c) store b in BL store in BL d) store d in BL 9) The instructions that are used to call a subroutine (procedure) from a main program and return to the main program after execution of called function are JMP,IRET c) CALL,JMP CALL,RET d) JMP,RET 10) Which of the following instructions changes the CS register: INT 21h c) INC CS MOV CS, DS d) CS cannot be changed

3 Question 2 This question is attributed with 7 marks if answered properly The answer for this question as the following: 1. TINY MODEL (.MODEL TINY): The model uses maximum of 64K bytes for Code and Data. 2. SMALL MODEL (.MODEL SMALL): The model uses maximum of 64K bytes for Code and 64K bytes for Data (Code<=64K and Data <=64K). This model is the most widely used memory model and is sufficient for all the programs to be used in this course. 3. MEDIUM MODEL, (.MODEL MEDIUM): The model uses maximum of 64K bytes for Data and Code can exceed 64K bytes (Code>64K and Data <=64K). 4. COMPACT MODEL, (.MODEL COMPACT): The model uses maximum of 64K bytes for Code and Data can exceed 64K bytes (Code<=64K and Data >64K). 5. LARGE MODEL, (.MODEL LARGE): Both Code and Data can exceed 64K bytes. However no single data set (i.e. array) can exceed 64K bytes (Code>64K and Data >64K). 6. HUGE MODEL, (.MODEL HUGE): Both Code and Data can exceed 64K bytes. Additionally, a single data set (i.e. array) can exceed 64K bytes (Code>64K and Data >64K). 7. FLAT MODEL, (.MODEL FLAT) Window NT Application Attributes of Memory Models 1. Direct addressing mode 2. Direct-offset addressing mode 3. Indirect addressing mode 4. Indexed addressing mode c) The interrupt vector table contains 256 four byte entries, containing the CS:IP interrupt vectors for each of the 256 possible interrupts. The table is used to locate the interrupt service routine addresses for each of those interrupts using the following equation:

4 Question 3 This question is attributed with 10 marks, if answered properly. The complete code for this question as the following: Question 3 i) ii) iii) iv) and ax,00ffh or ax, FF00h cmp dx, cx Jbe L1 MOV AX, Word ptr A1 Ñ Instruction Answer 1) MOV DS, BX V 2) test AL, X2 V 3) add eip, X1 I 4) MOV ES,DS I 5) sub si,di V 6) cmp 75,EAX I c) mov ax,valw ; ax=4567h MOV CL, TABLE[2] ; CL = 22 mov al,type Vector; al=4 mov eax,lengthof vec1 ; eax=6 mov ebx,lengthof vec2 ; ebx=12 mov ebx,sizeof vec2 ; ebx=48 Question 4 This question is attributed with 3 marks, if answered properly. The complete code for this question as the following: WhlLoop: mov ax, Avg cmp ax, Std jnae QuitLoop dec Avg inc Std jmp WhlLoop QuitLoop: exit Main endp mov cx, 33 mov si, OFFSET Name ; si = address of Name array Check: cmp BYTE PTR [si], 'A' ; if [si] == 'A' je HasAchar; then go to HasAchar inc si ; si = si + 1 dec cx jnz Check HasAchar. (10 marks) (4 marks) (1.5 marks) (1.5 marks)

5 Question 5 This question is attributed with 10 marks, if answered properly. The answer for this question as the following: TITLE (Array_Manipulation.asm) INCLUDE Irvine32.inc Swap PROTO, ; procedure prototype pvalx:ptr DWORD, pvaly:ptr DWORD.data array DWORD 16 DUP(?) msgbefore BYTE "Array before the swap: ",0 msgafter BYTE "Array after the swap: ",0 (1 mark).code main PROC call Randomize ; seed the random number generator mov esi, OFFSET array mov ecx, LENGTHOF array ; counter (1 mark) L1: call Random32 ; EAX = random number mov [esi], eax ; save random number in array add esi, TYPE array ; next entry in array loop L1 mov edx, OFFSET msgbefore ; message to display call WriteString call Crlf mov esi,offset array ; starting OFFSET mov ecx,lengthof array ; number of units mov ebx,type array ; doubleword format call DumpMem mov esi, OFFSET array mov ecx, LENGTHOF array ; number of elements in the array shr ecx, 1 ; divide counter by 2. (pair of elements) L2: INVOKE Swap, esi, ADDR [esi + 4] ;swap the pair add esi, TYPE array * 2 ;next pair loop L2 mov edx, OFFSET msgafter ; message to display call WriteString call Crlf mov esi,offset array ; starting OFFSET mov ecx,lengthof array ; number of units mov ebx,type array ; doubleword format call DumpMem exit main ENDP Swap PROC USES eax esi edi, pvalx:ptr DWORD, ; pointer to first integer pvaly:ptr DWORD ; pointer to second integer mov esi,pvalx ; get pointers mov edi,pvaly mov eax,[esi] ; get first integer xchg eax,[edi] ; exchange with second mov [esi],eax ; replace first integer ret Swap ENDP END main