【发布时间】：2016-03-29 22:01:18
【问题描述】：
我正在尝试编写一个 MIPS 程序，它将罗马数字（作为 ascii 字符的字符串输入）转换为其十进制值。有一个 I/0 组件，提示用户输入，执行（有限的）有效性检查，然后输出原始字符串以及值。
为了将每个罗马数字字符转换为其十进制值，我使用了一个查找表（即，我在“罗马数字”数组中获取每个字符的索引，然后使用该索引值从“十进制值”数组。
罗马数字是从 L 读到 R，但我不能简单地将每个字符相加（即使总和纯递归），因为可能有这样的字符串：XIV，其中 X = 10，I = 1，并且V = 5，结果数 = 14（而不是 15）。我有适当的逻辑来输出这个结果，但是（不幸的是！），我是 MIPS 的新手，为了实现求和逻辑，我设置了一系列嵌套的子例程。
我知道对于不是通过“jal”指令到达以“知道”其返回地址的子例程，您应该将返回地址加载到堆栈中，并在子程序的末尾弹出它例程，然后是 jr $ra。然而， 我的代码中的一些数学逻辑依赖于分支，而不是“jal”指令。在所有这些混乱中的某个地方，我的堆栈指针不正确，并且我的代码终止得太快（即没有对字符串中的所有数字求和）。
任何帮助或指针将不胜感激。代码如下：
# This program runs in MIPS, and performs the following functions:
#
# (1) Reads input Roman Numerals as ASCII text characters
#
# (2) Calculates the equivalent decimal value
#
# (3) Writes the calculated value as Arabic numerals to the SPIM
# console, with appropriate commentary text
#
# (4) Tests by calculating the values of these Roman numerals:
# {I:1, V:5, X:10, L:50, C:100, D:500, M:1000}
#
# (5) Calculates and displays the results of the following Roman numerals:
# -> LXXIX: 79
# -> MMXVI: 2016
# -> MCMXLVIII: 1948
# -> MDCCCCVIIII: 1909
#
# NOTE: There is no check for "valid" Roman numerals (i.e., VIIII is not "valid" but will be computed as = 9)
# However, subtraction logic is used (i.e. numbers that decrease from L to R = addition; a smaller number to the left
# of a larger number represents subtraction, such that IX == VIIII == 9.
.data
# Lookup tables
#roman: .asciiz "IVXLCDMivxlcdm"
roman: .asciiz "IVXLCDM"
#values: .byte 1, 5, 10, 50, 100, 500, 1000, 1, 5, 10, 50, 100, 500, 1000
values: .byte 1, 5, 10, 50, 100, 500, 1000
# Input
input1: .word 4
input2: .asciiz "\n "
input3: .asciiz "\n "
input4: .asciiz "\n"
# Conversion
left: .word 0
sum: .word 0
counter: .word 0
temp: .word 0
# I/O and program control
convertAgain: .word 2
prompt1: .asciiz "\nPlease enter the Roman Numeral to convert in all uppercase OR all lowercase letters (i.e. VII or vii): "
outRecap: .asciiz "\nYou entered: "
outFinal: .asciiz "\nThe equivalent decimal value is: "
loopdriver: .asciiz "\nWould you like to convert another number? Enter 1 for YES and 2 for NO: "
stop: .asciiz "\n\nProgram complete.\n"
# Exception handling
error1: .asciiz "\nThe Roman numeral you have entered contains invalid character(s). Please try again.\n"
################################################################################
.text
.globl main
main:
#Prompt user to enter the Roman Numeral to convert
li $v0, 4 # print prompt1
la $a0, prompt1 # load address
syscall
# Load the word reserved for input, and read in the Roman Numeral string input by the user
la $a0, input1 # load contents of input1 (i.e. space allocated to save the input string) into $a0
la $a1, input1 # load the length allocated for input1 (i.e. 4 bytes or 32 bits) into $a1 to prevent overflow
li $v0, 8 # read string from user
syscall
# Go to sub-routines for lookup and conversion
jal init
# Display results for the user
# Load output string 1 to show the user their original input string
la $a0, outRecap # load contents of outRecap into $a0
li $v0, 4 # print string
syscall
# Load the string the user entered
la $a0, input1 # load contents of input1 into $a0
li $v0, 4 # print string
syscall
# Display the equivalent decimal value
la $a0, outFinal # load contents of outFinal (string) into $a0
li $v0, 4 # print string
syscall
# Print the sum
lw $a0, sum
li $v0,1
syscall
# Ask the user whether they want to convert another Roman Numeral (1 if YES; 2 if NO; default is NO)
la $a0, loopdriver
li $v0, 4
syscall
# Get user input from keyboard (1 = they want to convert another Roman Numeral; 2 = Exit; default = 2)
li $v0, 5 # read integer input
syscall
# Save the user's input and jump to "Exit"
sw $v0, convertAgain
lw $t0, convertAgain
bne $t0, 1, Exit # if the user enters a number != 1, go to Exit
j main # Go back to start if the user enters 1
######################################################################################################################################################
# Convert Roman Numeral string to integer decimal value
init:
sw $ra, 4($sp)
sw $a1, 0($sp)
addi $sp, $sp, -8 # move stack pointer
la $t2, input1 #load the address of the string (see by looking at starting index of each the length of each string
la $t3, roman
la $t4, values
# Loop through each char of the input string
loop1:
lb $a0, ($t2) # This gets the next byte of the string (from L to R); change the offset to get others (i.e. lb $a0, 3($t0) = I
beqz $a0, return # If the byte == 0, we have reached the end of the input string; jump to "return"
li $v0, 11 # print byte to console
syscall
# While string[i] != null:
jal index
addi $t2, $t2, 1 # Move to next character in input string
jal loop1 # iterate through loop1 again
# Find the index (i.e. offset) of the selected char in the "roman" lookup table
index:
lb $t5, ($t3) # load the first byte of romans
beqz $t5, invalid # if we reach the end of romans and the char isn't found, it's not a valid Roman numeral
sw $ra, 8($sp)
addi $sp, $sp, -4 # move stack pointer
beq $a0, $t5, getVal # once we find the index we want, where the romans value matches our char, need to look it up in values
jal iter # if we haven't found a match, jump to where we can increment $t3 and iterate through the loop again
# addi $t3, $t3, 1 # if we haven't found a match, increment $t3 and iterate through the loop again
# jal index # loop
# Use the index value to look up the char in the "values" array and get the decimal value
getVal:
la $t6, roman # Get the address of the array containing Roman numeral characters
la $t7, values
sub $t8, $t3, $t6 # Get the index value of the element that matches the byte we are working with [1:7]
add $t7, $t7, $t8
lb $t9, ($t7) # $t9 is the decimal value that corresponds to the letter
#added new
sw $t9, 12($sp)
addi $sp, $sp, -4 # move stack pointer
jal setup
addi $sp, $sp, 4 # reset the stack pointer
lw $ra, 8($sp) # fetch return address
jr $ra # go back to main so that results can be displayed
# This is the last part of the index loop
iter:
addi $t3, $t3, 1 # if we haven't found a match, increment $t3 and iterate through the loop again
jal index # loop
setup:
# Base case is that the string length == 1, and thus sum = the decimal equivalent of the only Roman numeral character
lw $s0, sum
beqz $s0, base # if sum == 0, this is our first pass through the loop, and we just want to add the first char's decimal value to sum and return to get the next char
jal calcSum
# Base case is our first pass through the loop, where sum == 0. We just want to add the value of the first number in the string to sum, and return to loop1 to get next char.
base: #lw $s0, sum
add $s0, $s0, $t9
sw $s0, sum
la $t3, roman
addi $sp, $sp, 4 # reset the stack pointer
lw $ra, 16($sp) # fetch return address
jr $ra # go back
#j return
# Here, we want to pop the previous letter's decimal value off the stack, compare it to current, process sum accordingly, and pop the current char onto the stack for the next iteration
calcSum: addi $sp, $sp, 8 #reset the stack pointer
lw $ra, 16($sp)
lw $t1, 12($sp) # pop the decimal value of the Roman numeral to the left of the current value in the original string
la $t3, roman
bge $t1, $t9, plus # If the decimal value of the previous char in the string is >= the current char, then we can add the current char to the sum
blt $t1, $t9, minus # if the decimal value of the previous char in the string is < the current char, we should add: current - (2*previous) to the sum
plus: lw $s0, sum
add $s0, $s0, $t9
la $t3, roman
addi $sp, $sp, 4 # reset the stack pointer
lw $ra, 16($sp) # fetch return address
jr $ra # go back
minus: lw $s0, sum
mul $t1, $t1, 2
sub $t9, $t9, $1
add $s0, $s0, $t9
la $t3, roman
addi $sp, $sp, 4 # reset the stack pointer
lw $ra, 16($sp) # fetch return address
jr $ra # go back to
# Go back to main
return:
addi $sp, $sp, 8 # reset the stack pointer
lw $a0, 0($sp) #
lw $ra, 4($sp) # fetch return address
jr $ra # go back to main so that results can be displayed
#Exception handling: User input string contains invalid character(s)
invalid:
li $v0, 4 # print string
la $a0, error1 # load address for error1 (a string)
syscall
j main
###############################################################################
#System Exit
Exit:
# Display exit message
li $v0, 4 # print string
la $a0, stop # load address for stop (a string)
syscall
li $v0, 10
syscall
################################################################################
【问题讨论】：
标签： assembly mips subroutine spim roman-numerals