汇编程序:递归?

时间:2013-03-11 01:37:01

标签: assembly recursion mips pcspim

我需要编写一个迭代10次的程序。每次更新一个值并将其打印到屏幕上。

我知道必须要创建堆栈并保存值,以便它可以迭代回到正确的部分继续执行程序。我试过很多东西,但我无法弄明白。这是我到目前为止的代码

# ############################################################### #
# Phase2.ASM                                                      #
#                                                                 #
# This program will recurse 10 times and show how much interest   #
# is made after 10        "months"                                        #
#                                                                 #
# ############################################################### #   

.data

PRINCIPAL:  .float  100.0   # principal = $100.00
INTEREST_RATE:  .float  0.012   # interest  = 1.2%

promptFirst:     .asciiz "Your starting Principal is $100.00: \n"
promptSecond:     .asciiz "Your interest rate is 1.2%: \n"
promptNow:          .asciiz "Interest Made After a Month:\n"
.text
.globl main

main:   



First:   
     # Prints the first prompt  
     li $v0, 4               # syscall number 4 will print string whose address is in $a0       
     la $a0, promptFirst     # "load address" of the string
     syscall                 # actually print the string   


Second:  
     # Prints the second prompt
     li $v0, 4               # syscall number 4 will print string whose     address is in $a0   
     la $a0, promptSecond    # "load address" of the string
     syscall                 # actually print the string    



jal CI


j EXIT



CI:




    la  $a0, PRINCIPAL  # load the address of the principal
    la  $a1, INTEREST_RATE  # load the address of the principal

    lwc1  $f2, ($a0)    # load the principal
    lwc1  $f4, ($a1)    # load the interest rate    
    mul.s $f12, $f4, $f2    # calculate the balance


    li $v0, 4            # syscall number 4 will print string whose address is in $a0   
    la $a0, promptNow    # "load address" of the string
    syscall              # actually print the string
    li  $v0, 2           # system call #2   
    syscall

jr $ra




EXIT:    
jr $ra


# END OF THE LINES ###############################################

到目前为止我的当前输出:

您的首席校长是$ 100.00:

您的利率为1.2%:

一个月后的兴趣:

1.20000005

非常感谢帮助。我在组装编程方面真的很糟糕。

PS:通过递归完成赋值

修改!新代码

# ############################################################### #
# Phase2.ASM                                                      #
#                                                                 #
# This program will recurse 10 times and show how much interest   #
# is made after 10     "months"                                       #
#                                                                 #
# ############################################################### #   

.data

PRINCIPAL:  .float  100.0   # principal = $100.00
INTEREST_RATE:  .float  1.012   # interest  = 1.2%

promptFirst:     .asciiz "Your starting Principal is $100.00: \n"
promptSecond:     .asciiz "Your interest rate is 1.2%: \n"
promptNow:          .asciiz "\nYour Balance After A Month:\n"
.text
.globl main

main:   



First:   
     # Prints the first prompt  
     li $v0, 4               # syscall number 4 will print string whose address is in $a0       
     la $a0, promptFirst     # "load address" of the string
     syscall                 # actually print the string   


Second:  
     # Prints the second prompt
     li $v0, 4               # syscall number 4 will print string whose address is in $a0   
     la $a0, promptSecond    # "load address" of the string
     syscall                 # actually print the string    


li $t1, 0
jal CI

ENDCI:
j EXIT



CI:



    add $t1, $t1, 1 
    la  $a0, PRINCIPAL      # load the address of the principal
    la  $a1, INTEREST_RATE  # load the address of the principal

    lwc1  $f2, ($a0)        # load the principal
    lwc1  $f4, ($a1)        # load the interest rate    
    mul.s $f12, $f4, $f2    # calculate the balance


    li $v0, 4               # syscall number 4 will print string whose address is in $a0   
    la $a0, promptNow       # "load address" of the string
    syscall                 # actually print the string
    li  $v0, 2              # system call #2    
    syscall

    beq $t1, 10, ENDCI
    jal CI
jr $ra




EXIT:    
jr $ra


# END OF THE LINES ###############################################

新产品:

我们的首席校长是100.00美元: 你的利率是1.2%:

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

一个月后的平衡:

101.19999695

所以我得到代码迭代10次。我需要更新金额,以便显示上个月+添加的利息

2 个答案:

答案 0 :(得分:1)

您需要在每次更新后存储当前余额,以便下次调用不会继续使用原始值。

即。像这样的东西:

lwc1  $f2, ($a0)        # load the principal
lwc1  $f4, ($a1)        # load the interest rate    
mul.s $f12, $f4, $f2    # calculate the balance
swc1  $f12, ($a0)

在每次额外调用CI之前,您还需要保存当前的返回地址,然后在返回之前将其恢复:

addi $sp,$sp,-4     # push the current return address
sw   $ra,($sp)      # ...
beq  $t1, 10, CIRET
jal  CI
CIRET:
lw   $ra,($sp)      # pop the saved return address
addi $sp,$sp,4      # ....
jr   $ra

答案 1 :(得分:0)

问题不是要求递归 - 一个简单的循环就可以了。程序集中的循环只是一个条件性的跳转代码(或无条件的跳转与前面的条件跳转相结合)。

条件是基于计数器的,计数器将从0变为9,或从9变为0,组装时通常更容易根据与零的比较进行条件跳转。