Assembly 汇编程序:递归?
我需要写一个程序,将迭代10次。每次它都会更新一个值并将其打印到屏幕上 我知道必须创建一个堆栈并保存值,这样它才能迭代回正确的部分继续程序。我试过很多东西,但我想不出来。这是到目前为止我的代码Assembly 汇编程序:递归?,assembly,recursion,mips,pcspim,Assembly,Recursion,Mips,Pcspim,我需要写一个程序,将迭代10次。每次它都会更新一个值并将其打印到屏幕上 我知道必须创建一个堆栈并保存值,这样它才能迭代回正确的部分继续程序。我试过很多东西,但我想不出来。这是到目前为止我的代码 # ############################################################### # # Phase2.ASM # #
# ############################################################### #
# 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:
您的起始本金为100.00美元:
您的利率是1.2%:
一个月后产生的利息:
1.20000005
任何帮助都将不胜感激。我真的很不擅长汇编编程
PS:赋值必须通过递归完成
编辑!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 ###############################################
新产出:
我们的起始本金为100.00美元:
您的利率是1.2%:
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
您一个月后的余额:
101.19999695
所以我让代码迭代10次。我需要更新金额,使其显示前一个月+增加的利息问题不需要递归-一个简单的循环就可以了。程序集中的循环只是代码的条件跳转(或无条件跳转与条件提前跳转相结合)
条件是基于计数器的,计数器将从0变为9,或从9变为0。在汇编中,通常更容易基于与零的比较进行条件跳转。每次更新后,您需要存储当前余额,以便下次调用不会继续使用原始值 也就是说,类似这样的事情:
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还原:
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
任务的一部分是,它必须通过recursionIv尝试一些lil的东西。没有什么真正值得一提的。我在课堂上有一些关于如何设置堆栈框架和框架指针的例子,但我不知道我在做什么。我知道它现在迭代了10次。但该值不会更新。我真的不知道如何使用浮点数,以及如何更新和保存它。非常感谢您的回复。帮了很多忙!真的很感激!