Question

我正在尝试用MIPS汇编语言编写一个实现恢复除法算法的程序。

我找到了我必须要完成的任务的非常好的例子，该线程已经过时且已关闭，我并不完全理解它。

http://forums.extremeoverclocking.com/showthread.php?p=2693342

我几乎没有问题：

所以到底应该怎么样好的代码？因为我明白第一篇文章中的代码是错误的吗？

什么是恢复分区的好算法？

” 你要划分两个数字：a / b = c，余数= d

注册A = a 登记册B = b 寄存器P =一组“连接”的两个寄存器（64位寄存器）

将双重对手（P，A）向左移一位
- 对无符号值使用零符号扩展
- 强制MSB（P）脱离上端
从P
如果结果是＆lt; 0 LSB（A）= 0 否则LSB（A）= 1
如果结果是否定的 P = P + b

重复n次（对于n位值）寄存器A = a / b的内容寄存器P的内容=余数（a / b）“

这是一个很好的算法吗？有人可以用更简单的英语写出来吗？因为我不明白。

此外：

来自第一篇文章的代码：

.data
    GETA:           .asciiz "Enter A "
    GETB:           .asciiz "Enter B "
    QMSG:           .asciiz "Q = "
    RMSG:           .asciiz "R = "
    NL:             .asciiz "\n"
    ERR:            .asciiz "Divide by zero error\n"
    left:           .word 1
    right:          .word 1
.text
.globl main
main:
    # Prompt for an integer A: divisor
    li $v0, 4                            
    la $a0, GETA                           
    syscall                                     
    # Read A from user
    li $v0, 5                              
    syscall                              

    move $t0, $v0                         
    # Prompt for an integer B: dividend
    li $v0, 4                             
    la $a0, GETB                         
    syscall     

    # Read B from user
    li $v0, 5                               
    syscall                               
    move $a1, $v0                          

# Initialize quotient register to zero
# Initialize left half of Divisor regisiter with divisor
# Initialize remainder register with the dividend (right aligned)
# 1. Remainder = Remainder - Divisor
#    Remainder >= 0 goto Step 2a.
#    Remainder < 0 goto Step 2b.
# 2a. Shift quotient register to the left, setting rightmost bit to 1
# 2b. Restore the original value bu Remainder = Remainder + Divisor
#     Shift quotient register to the left setting new LSB to zero
# 3. Shift divisor register to the right 1 bit
# < 33 repitions, goto step 1
# >= 33 repititions, print result 
#
# Register usage:
# $a0 = divisor
# $a1 = dividend
# $a2 = quotient
# $a3 = remainder
# $t4 = counter of 33 repetitions
# $t0 = divisor (temp)
initialize:
    li $a2, 0                               # $a2 = quotient = 0
    swl $t0, left
    lwl $a0, left                           # $a0 = left half of divisor
    swr $a1, right
    lwr $a3, right                          # $a3 = remainder = $a2
    li $t4, 33                              # $t4 = counter = 33

step_one:
    sub $a3, $a3, $a0
    bge $a3, $zero, step_two_a
    blt $a3, $zero, step_two_b

step_two_a:
    #slt $a2, $zero, $a2
    sll $a2, $a2, 1
    ori $a2, $a2, 1
    j step_three

step_two_b:

    add $a3, $a3, $a0
    sll $a2, $a2, 1

step_three:

    sra $a0, $a0, 1

step_four:

    addi $t4, $t4, -1
    bge $t4, $zero, step_one
    beq $t4, $zero, print_result

print_result:

    li $v0, 4                  
    la $a0, QMSG                     
    syscall                   

    move $a0, $a2                      
    li $v0, 1                         
    syscall                            

    li $v0, 4                          
    la $a0, NL                         
    syscall                           

    li $v0, 4                    
    la $a0, RMSG                     
    syscall                     

    move $a0, $a3                      
    li $v0, 1                          
    syscall                            

    li $v0, 4                          
    la $a0, NL                         
    syscall

    j exit_program                    

exit_program:
    li $v0, 10                       
    syscall

可能遗漏了像

代码：

 # shift the double register (P,A) one bit left
    slt $t2, $v0, $zero                  # store MSB of hi
    sll $v0, $v0, 1                      # shift hi left 1 bit
    slt $t3, $v1, $zero                  # store MSB of lo
    or $v0, $v0, $t2                     # move MSB of lo into LSB of hi
    sll $v1, $v1, 1                      # shift lo left 1 bit

应该怎么样好的代码??

恢复MIPS大会的分部

0 个答案: