作業(yè)1—lab1指令集結構—計算機體系結構

1、Lab1 作業(yè)學號：12281166 姓名：崔雪瑩 Exercise 1: What will be 32-bit hex code for the above two examples.1. LW R2, 100(R3) 指令為讀指令，讀R3寄存器內容為首址，偏移地址為100的內容字，存到R2寄存器里，格式 LW rt, Imm(rs)。編碼為：100011 00011 00010 00000000 01100100Bit313029282726252423222120191817161514131211109876543210Field6-Bit Op Coders Fieldrt Fie

2、ld16-bit Immediate field100011000110001000000000011001002. SW R5, 100(R6) 指令為寫指令，將R5內容寫入以R6寄存器內容為首址，偏移地址為100的地址，格式 SW rt, Imm(rs)。編碼為：101011 00110 00101 00000000 01100100Bit313029282726252423222120191817161514131211109876543210Field6-Bit Op Coders Fieldrt Field16-bit Immediate field10101100110001010

3、000000001100100Exercise 2: Find 32-bit code for OR R7, R18, R12 and SUB R5, R4, R31 1. OR R7, R18, R12 位或操作，格式 OpCode rd, rs, rt編碼為：000000 10010 01100 00111 00000 100101Bit313029282726252423222120191817161514131211109876543210Field0 0 0 0 0 0rs Fieldrt Fieldrd fieldShift AmountFunction Field00000010

4、0100110000111000001001012. SUB R5, R4, R31 減法，格式 OpCode rd, rs, rt編碼為：000000 00100 11111 00101 00000 100010Bit313029282726252423222120191817161514131211109876543210Field0 0 0 0 0 0rs Fieldrt Fieldrd fieldShift AmountFunction Field00000000100111110010100000100010Exercise 3: Find 32-bit code for BNEQ

5、R7, R8, -5. 0000101 00111 01000 11111111 11111011 If the PC value for branch instruction is 124 what will be new value of PC if the branch is taken 124+4-5*4=108 .-5的補碼為 1111 1111 1111 1011 Bit313029282726252423222120191817161514131211109876543210Field000100 for BEQ 000101 for BNEQrs Fieldrt FieldBr

6、anch offset in Number of Instructions00010100111010001111111111111011Exercise 5: What is the 32-code for ADDI R7, R8, 600? 001000 01000 00111 0000 0010 0101 1000 . There is no SUBI, why? 因為減法可以用ADDI Rt, Rs, Imm（負數(shù)），實現(xiàn)減法。因為減法要考慮借位，加法考慮溢出，溢出很好實現(xiàn)，借位不易實現(xiàn)，所以可以用加一個負數(shù)實現(xiàn)。 Bit31302928272625242322212019181716

7、1514131211109876543210FieldOp Coders Fieldrt Field16-bit Immediate field00100001000001110000001001011000Homework:Write a program to sort an array using bubble sort. Home work should be submitted as rollno.s in the folder onIndus/common. It will be tested in the next lab.冒泡程序：（從小到大）C程void main()int a

8、ge=9,5,12,5,20,11,66,44,22,12;int x =0 ;for(int i=0 ; i<10 ; i+)if(agei>agei+1)for(int j=i;j>0;j-)if(agej>agej+1)x = agej;agej = agej+1;agej+1=x;Assembly Code（程序不能加載中文，所以實驗是去掉中文注釋）.globl main.dataage: .word 9,5,12,5,20,11,66,44,22,12.textmain:la $t2,age #t2為age0地址addi $t5,$zero,10 #數(shù)組長為1

9、0addi $t4,$zero,0 #循環(huán)loop1次數(shù)addi $t3,$zero,0 #循環(huán)loop2次數(shù)addi $t6,$zero,0#中間寄存器，用于交換loop1:lw $t0,0($t2)#t0存放ageiaddi $t2,$t2,4 #t2=t2+4 相當于i=i+1lw $t1,0($t2) #t1存放agei+1addi $t4,$t4,1 #loop1循環(huán)次數(shù)t4自加1bgt $t1,$t0,skip1 #如果agei+1大于agei，說明已經(jīng)i和i+1排好序#跳到skip1，否則說明agei+1需要排序add $t3,$t4,0#j=i，令子循環(huán)次數(shù)為已排好序的數(shù)的個數(shù)

10、loop2:la $t7,age #t7為age0地址sll $t6,$t3,2#t6=t3*4add $t8,$t7,$t6#t8為agej+1的地址lw $s0,0($t8)#s0存放agej+1lw $s1,-4($t8)#s1存放agejaddi$t3,$t3,-1# loop2循環(huán)次數(shù)t3自減1bgt $s0,$s1,skip1 #若agej+1大于agej，說明已經(jīng)以前的都排好序#退出子程序，跳skip1，否則說明繼續(xù)需要排序add $s3,$s1,0#交換agej+1和agejadd $s1,$s0,0add $s0,$s3,0sw $s0,0($t8)#寫入數(shù)據(jù)段sw $s1,-4($t8)skip2：bne $t3,$zero,loop2 #判斷j是否為0.#相等說明已經(jīng)比到最后一個數(shù)，退出loop2skip1:bne $t4,$t5,loop1 #不等，繼續(xù)循環(huán)，#否則說明已經(jīng)比到最后一個數(shù)，程序結束li $v0,10 #return，程序結束syscall加載后