MIPS匯編范例

1、mips匯編語言之實(shí)現(xiàn)swap函數(shù) 收藏 此程序用來交換兩個整數(shù)已在pcspim下編譯通過# programed by stevie zou # purpose:to swap two values # 10-15-2008 # # text segment # .text .globl mainmain: la $t0, number #讀取兩個整數(shù)并放入寄存器$t1,$t2 lw $t1, 0($t0) lw $t2, 4($t0) li $v0, 4 #打印msg1 la $a0, msg1 syscall li $v0, 1 #打印轉(zhuǎn)換前$t1中的值 move $a0, $t1 sys

2、call li $v0, 4 #打印msg2 la $a0, msg2 syscall li $v0, 1 #打印轉(zhuǎn)換前$t2中的值 move $a0, $t2 syscall move $t3, $t1 #關(guān)鍵部分，在寄存器間move數(shù)據(jù) move $t1, $t2 move $t2, $t3 li $v0, 4 #打印msg3 la $a0, msg3 syscall li $v0, 1 #打印轉(zhuǎn)換后$t1中的值 move $a0, $t1 syscall li $v0, 4 #打印換行符 /n la $a0, msg4 syscall li $v0, 1 #打印轉(zhuǎn)換后$t2中的值 mov

3、e $a0, $t2 syscall# data segment # .datanumber: .word 12,34msg1: .asciiz the first number is:nmsg2: .asciiz nthe second number is:nmsg3: .asciiz nnow they are swapped as:nmsg4: .ascii n# end of file程序運(yùn)行結(jié)果為：本文來自CSDN博客，轉(zhuǎn)載請標(biāo)明出處：mips匯編簡單實(shí)例一個小計(jì)算器 收藏 其實(shí)開始的時(shí)候一直在看 mips的指令格式，看了、忘了，沒什么效果?，F(xiàn)在通過這個例子，簡單訓(xùn)練一下： mips

4、匯編中：常用寄存器的使用；系統(tǒng)調(diào)用： 1(print_int),4(print_str),5(read_int),10(exit)；算術(shù)運(yùn)算： add,sub,mul,div；邏輯運(yùn)算及跳轉(zhuǎn)；上代碼：(匯編寫的少，代碼書寫、風(fēng)格還很差勁，感謝請大蝦指教)：view plaincopy to clipboardprint?#sample example a small calculater .text # text section .globl main # call main by SPIM main: la $t0, value # load address value into $t0 l

5、a $a0,msg0 li $v0,4 syscall #print please choose the operation: li $v0, 5 syscall sw $v0, 8($t0) la $a0,msg1 li $v0,4 syscall #print first num: li $v0, 5 syscall sw $v0, 0($t0) la $a0,msg2 li $v0,4 syscall #print second num: li $v0, 5 syscall sw $v0, 4($t0) #read the other num la $a0,newline li $v0,

6、4 syscall #print n lw $t1, 0($t0) # load the first num lw $t2, 4($t0) # load the second num lw $t3, 8($t0) # load the operation beq $t3,1,addOp # if + beq $t3,2,subOp # if - beq $t3,3,mulOp # if * beq $t3,4,divOp # if / addOp: add $t4, $t1, $t2 # $t1 + $t2 = $t4 sw $t4, 12($t0) # la $t5,addFlag j pr

7、intResult subOp: sub $t4, $t1, $t2 # $t1 - $t2 = $t4 sw $t4, 12($t0) la $t5,subFlag j printResult mulOp: mul $t4, $t1, $t2 # $t1 * $t2 = $t4 sw $t4, 12($t0) la $t5,mulFlag j printResult divOp: div $t4, $t1, $t2 # $t1 / $t2 = $t4 sw $t4, 12($t0) la $t5,divFlag j printResult printResult: lw $a0,0($t0)

8、 li $v0,1 syscall #read first number la $a0,0($t5) li $v0,4 syscall #print opflag lw $a0,4($t0) li $v0,1 syscall #print second number la $a0,equalStr li $v0,4 syscall #print = lw $a0,12($t0) li $v0,1 syscall # print sum result j exit exit: la $a0,newline li $v0,4 syscall #print n li $v0,10 syscall #

9、 exit # data section .data value: .word 0, 0, 0 ,0 ,0 # 0: first num ,4 : second num , 8 : operation , 12:result msg0 : .asciiz please choose the operation(14):ntt1 : +,addition ntt2 : -,subtracterntt3 : * multiplication ntt4 : /,divisionn msg1 : .asciiz first num: msg2 : .asciiz second num: addFlag

10、 : .asciiz + subFlag : .asciiz - mulFlag : .asciiz * divFlag : .asciiz / equalStr : .asciiz = newline : .asciiz n=n #sample example a small calculater.text # text section.globl main# call main by SPIMmain:la $t0, value# load address value into $t0la $a0,msg0 li $v0,4 syscall#print please choose the

11、operation:li$v0, 5syscallsw$v0, 8($t0)la $a0,msg1 li $v0,4 syscall#print first num:li$v0, 5syscallsw$v0, 0($t0)la $a0,msg2 li $v0,4 syscall#print second num:li$v0, 5syscallsw$v0, 4($t0)#read the other numla $a0,newline li $v0,4 syscall#print nlw $t1, 0($t0)# load the first numlw $t2, 4($t0)# load th

12、e second numlw $t3, 8($t0)# load the operationbeq $t3,1,addOp# if +beq $t3,2,subOp# if -beq $t3,3,mulOp# if *beq $t3,4,divOp# if /addOp:add $t4, $t1, $t2# $t1 + $t2 = $t4 sw $t4, 12($t0)# la $t5,addFlagj printResultsubOp:sub $t4, $t1, $t2# $t1 - $t2 = $t4 sw $t4, 12($t0)la $t5,subFlagj printResultmu

13、lOp:mul $t4, $t1, $t2# $t1 * $t2 = $t4 sw $t4, 12($t0)la $t5,mulFlagj printResultdivOp:div $t4, $t1, $t2# $t1 / $t2 = $t4 sw $t4, 12($t0)la $t5,divFlagj printResultprintResult:lw $a0,0($t0) li $v0,1 syscall#read first numberla $a0,0($t5) li $v0,4 syscall#print opflaglw $a0,4($t0) li $v0,1 syscall#pr

14、int second numberla $a0,equalStr li $v0,4 syscall#print = lw $a0,12($t0) li $v0,1 syscall# print sum resultj exitexit:la $a0,newline li $v0,4 syscall#print n li $v0,10 syscall# exit# data section.datavalue:.word 0, 0, 0 ,0 ,0 # 0: first num ,4 : second num , 8 : operation , 12:resultmsg0:.asciiz ple

15、ase choose the operation(14):ntt1 : +,addition ntt2 : -,subtracterntt3 : * multiplication ntt4 : /,divisionnmsg1:.asciiz first num:msg2:.asciiz second num:addFlag: .asciiz + subFlag: .asciiz - mulFlag: .asciiz * divFlag: .asciiz / equalStr: .asciiz = newline:.asciiz n=n 運(yùn)行效果截圖如下：2009/07/15/4346531.a

16、spx本文來自CSDN博客，轉(zhuǎn)載請標(biāo)明出處：高精度整數(shù)除法（MIPS匯編） 收藏 .datatitle: .asciiz tt高精度除法演示程序ntip1: .asciiz n請輸入被除數(shù)：tip2: .asciiz 請輸入除數(shù)：tip3: .asciiz 要輸出小數(shù)點(diǎn)后多少位：tip4: .asciiz n是否繼續(xù)？（y/n）err: .asciiz n除數(shù)不能為0，請重新輸入ndot: .asciiz .sign: .asciiz -buf: .byte 0:16 .text.globl _start_start:#輸出標(biāo)題 li $v0, 4 la $a0, title syscall

17、#提示輸入被除數(shù)并接收輸入 restart: li $v0, 4 la $a0, tip1 syscall li $v0, 5 syscall move $t0, $v0#提示輸入除數(shù)并接收輸入 input_again: li $v0, 4 la $a0, tip2 syscall li $v0, 5 syscall move $t1, $v0 #檢測除數(shù)是否為0,是則輸出錯誤信息并跳轉(zhuǎn)到除數(shù)輸入處,否則繼續(xù)執(zhí)行 bnez $t1, next li $v0, 4 la $a0, err syscall j input_again#提示輸入要求的小數(shù)位數(shù)并接收輸入next: li $v0, 4

18、la $a0, tip3 syscall li $v0, 5 syscall move $t2, $v0#兩數(shù)進(jìn)行除法運(yùn)算,把運(yùn)算結(jié)果保存到寄存器$t8和$t9 div $t0, $t1 move $s0, $zero mflo $t8 mfhi $t9#檢測商是否為0,是則繼續(xù)執(zhí)行下一步,否則跳轉(zhuǎn)到輸出整數(shù)結(jié)果處 bnez $t8, int_result #檢測被除數(shù)是否為非負(fù)數(shù),是則檢查除數(shù),否則置$s0為1,$s0用來記錄負(fù)數(shù)的個數(shù) bgez $t0, next_check addi $s0, $zero, 1#檢測除數(shù)是否為非負(fù)數(shù),是則輸出-號,否則$s0加1 next_check: bgez $t1, print_sign addi $s0, $s0, 1#輸出-號,如果$s0值不為1,即負(fù)數(shù)個數(shù)不為奇數(shù),則跳過此步 print_sign: bne $s0, 1, int_result li $v0, 4 la $a0, sign s