汇编期末试卷08秋a-(含答案)090114v4

1、诚信考试 沉着应考 杜绝违纪浙江大学20082009学年 秋冬 学期汇编与接口课程期末考试试卷开课学院： 计算机学院 ，考试形式： 开 卷，允许带_1本书_入场考试时间： 09 年 1 月 11 日,所需时间： 120 分钟，任课教师 _考生姓名： _学号： 专业： _题序一二三四五六七总 分得分评卷人请把第15题的答案写在本页的答题纸上，第15题的评分以答题纸上的答案为准。一（10 points）Mark T for correct instruction or description, mark F foe incorrect instruction or description.（每题1分）12345678910二(20 points) Please select your best choice for the following questions. （每题2分）12345678910三(10 points) The following program runs under 32-bit Visual C+ environment，Please choose the best

2、 answer. （每题5分）四(10 points) The following program is developed under 32-bit Visual C+ environment, each integer takes 32 bits. The integer array “weight” holds many element （每题2分）（1）（2）（3）（4）（5）五(10 points) In protected mode, a Pentium 4 descriptor describes a segment, and the content of the descriptor is: 3CH, C0H, A2H, 6CH, 40H, 00H, 00H, 2FH（每题2分）(1)(2)(3)(4)(5)一. （10 points）Mark T for correct instruction or description, mark F foe incorrect instruction or description.（ ）1. POP CS（ ）2. IMUL D

3、H（ ）3. INSB（ ）4. MOV AX，ESIEDI（ ）5. MOV AX，SIDI（ ）6. JMP WORD PTR BP（ ）7. MOV EBX, ESI（ ）8. IMUL 125（ ）9. ADD CX，EAX（ ）10. In Mode 2, the 8254 usually generates a continuous square-wave.【答案】：（每题1分）（F）1. POP CS（T）2. IMUL DH（T）3. INSB（T）4. MOV AX，ESIEDI（F）5. MOV AX，SIDI（T）6. JMP WORD PTR BP（F）7. MOV EBX, ESI（F）8. IMUL 125（T）9. ADD CX，EAX（F）10. In Mode 2, the 8254 usually generates a continuous square-wave.二. (20 points)Please select your best choice for the following questions. （每题2分）1. In real mo

4、de, according to the following memory data, the entry address of INT 13H should be (_).0000:0030 72 10 A7 00 7C 10 A7 00-4F 03 80 05 8A 03 80 050000:0040 AD 06 14 02 A4 F0 00 F0-37 05 14 02 11 6D 00 C0A：6D11:C000B：116D:00C0C：00C0:116DD：C000:6D11【答案】：D2. MOVEAX, 3A4C0002HMOVEBX, -2MULBL then AX= (_)A：-4B：508C：-304HD：H【答案】：B3. MOVEAX，1007MOV EDX，100DIVDL then AX= (_)A：070AH B：-532HC：10D：7【答案】：A4. In real mode , variable X is defined in data segment, its segment address is 2345H, its offset is 2222

5、H, then its 20-bit physical address will be _ .A：04567H B：45670HC：25672HD：23452H【答案】：C5. Which instruction will do “EBX - ECX + 8*EAX 400H” ? A：MOV ECX, ECX + 8*EAX 400H B：LEA ECX, ECX + 8*EAX 400HC：MOV EBX, OFFSET ECX + 8*EAX 400HD：ADD ECX , 8*EAX 400H【答案】： B6. CF and OF are always set to 0 after instruction.A：SUB B：ANDC：ADDD：MUL【答案】： B7. Suppose that DS=3200H, BX=0300H, SI=0240H, offset of myarray=0120H, determine the physical address accessed by instruction: “MOV myarray SI, ECX” , assuming r

6、eal mode operation.A：32240H B：32120HC：3560HD：32360H【答案】： D8. What is the purpose of interrupt vector type number 0?A：to detect and respond to divide eror. B：to report hardware errorC：to do a single-step trapD：to process double fault【答案】： A9. Suppose that the 8254s control word is B, which description is correct?A：it sets counter 2 to read/write least-significant byte only. B：it sets counter 1 to read/write least-significant byte only.C：it sets counter 2 to read/write least-significant byte first

7、, followed by the most significant byte.D：it sets counter 2 to read/write least-significant byte only.【答案】： C10. Suppose that the 8254s control word is B, which description is correct?A：it selects mode 2. B：it selects BCD counting.C：it selects mode 3 and binary counting.D：it selects counter 1 and binary counting.【答案】： C三. (10 points) The following program runs under 32-bit Visual C+ environment.#include stdafx.h#include stdio.hchar STRING512=27 hosts,120 disks and 3 printers#;int main(int argc,

8、char* argv) _asm MOV DH,0XOR ECX,ECXLEA EBX,STRINGCONT:MOV DL,EBXCMP DL,#JE EXITINC EBXCMP DL,30HJB CONTCMP DL,39HJA CONTINC ECXCMP DL,DHJBE CONTMOV DH,DLJMP CONTEXIT:return 0;Please choose the best answer.1. When this program runs to the label “EXIT”, the register CX= .A. 5B.6C. 7D.82. When this program runs to the label “EXIT”, the register DH= .A. 34HB.35HC.36HD.37H（每题5分）1.【答案】： B2.【答案】： D四. (10 points) The following program is developed under 32-bit Visual C+ environment, each integer takes 32 bits. The integer array “weight” holds many element, each element is the weight of an elephant; The integer “count” holds the total number of arr

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