我刚刚收到了一些帮助，解决了我在这个问题( Trouble with MIPS array )中遇到的一些 MIPS 代码问题，他们给我的代码工作正常。现在我尝试添加一个排序算法来对数组中存储的整数进行排序，但我得到的输出只是 0 后跟作为输入的字符串(正确的输出应该是日期后跟字符串)。这就是我正在编写的代码:
.data
num_array: .space 400 #array
str_array: .space 400
in_name: .asciiz "\nInsert name: "
in_date: .asciiz "\nInsert date (mmdd): "
appt: .asciiz "\nList: "
spaz: .asciiz " "
.text
main:
la $s0, num_array #load array in s0
la $s1, str_array
addi $t0, $zero, 0 #t0=0 counter
addi $s2, $zero, 0 #s1=0 array size counter
j Input
Input:
# prompt for date
li $v0, 4
la $a0, in_date
syscall
# read date
li $v0, 5
syscall
# store date in the num_array
sw $v0, 0($s0)
# increment counter and move to next position in the array
addi $t0, $t0, 1
addi $s2, $s2, 1
addi $s0, $s0, 4
# prompt for name
li $v0, 4
la $a0, in_name
syscall #ask name
# store name (max 4 bytes) to str_array ($s1)
move $a0, $s1
li $a1, 4
li $v0, 8
syscall #read name
# move to the beginnig of next string in $s1
# increment by 4 because length of each input word is 4
addi $s1, $s1, 4
beq $s2, 5, sort #if array size=10 go to print
j Input #start over until s1=10
sort:
addi $t1, $zero, 0
addi $t2, $zero, 0
addi $s3, $zero, 0 #swap flag set at 0
move $t1, $s0 #first element of array in t1
addi $t2, $s0, 4 #second element of array in t2
jal ric
j print
ric:
add $sp, $sp, -4 #stack pointer
sw $ra, 0($sp)
pro:
addi $t5, $t5, 1 #counter t0++
beq $t5, $s2, exit #if counter=array size go to exit
bgt $t1, $t2, swap #swap if t1<t2
j pro #go back
swap:
move $t3, $t1 #$t3=$t1
move $t1, $t2 #$t1=$t2
move $t2, $t3 #$t2=$t3=$t1
addi $s3, $s3, 1 #s3++ (swap flag)
j pro #go back
exit:
move $s0, $t1
addi $s0, $s0, 4
bgtz $s3, sort #if 2 elements were swapped start over
lw $ra, 0($sp) #load return address
add $sp, $sp, 4
jr $ra #exit
print:
# print "List:"
la $a0, appt
li $v0, 4
syscall #print list
addi $t0, $zero, 0 #t0=0 counter
la $s0, num_array # address of num_array
la $s1, str_array # address of str_array
res:
# get number from num_array and print it
lw $a0, 0($s0)
li $v0, 1
syscall
# move to the next element, increment by 4 because numbers take 1 word = 4 bytes
addi $s0, $s0, 4
# print space
la $a0, 32
li $v0, 11
syscall
# get string from str_array and print it
la $a0, 0($s1)
li $v0, 4
syscall
# print space
la $a0, 32
li $v0, 11
syscall
# move to the next element, increment by 4 because strs have 4 bytes length (in your case)
addi $s1, $s1, 4
# increment loop counter
addi $t0, $t0, 1
blt $t0, $s2, res # $s2 is size of the array
end:
li $v0, 10
syscall #the end
有人可以指出我做错了什么吗？另外，关于如何将排序后的数字后跟相应的字符串作为输出，有什么建议吗？ (因此，第一个作为输入给出的日期应该遵循作为输入给出的第一个日期，即使它们已排序)
编辑:我按照星号的建议更改了代码并尝试交换数字。现在输出不只是 0，而是放入相同的输入，但未排序。我做错了什么？
最佳答案
一种方法是对数字数组进行排序并跟踪字符串数组中的相应条目(当您在数字数组中的值周围进行随机播放时，在字符串数组中执行相同的操作)。我建议您开始实现一个简单的排序算法并首先对数字进行排序。例如，考虑实现冒泡排序。为此，您可以考虑实现 swap 过程来交换 2 个值。
编辑 这是他的 mips 版本的程序。它实现了冒泡排序过程。我在文件顶部对数字和字符串数组进行了硬编码(因此您必须进行一些调整才能使其与您的代码一起使用)。这个想法如下:当我交换数字数组中的数字时，我也会交换字符串数组中相应的字符串。
.data
array: .word 3,2,1,4
str_array: .ascii "ccc\0bbb\0aaa\0ddd\0"
n: .word 4
.text
j main
str_swap:
# swap two strings
# $a0 - beginning of the first str
# $a1 - beginning of the second str
# Assume that str length is 3
li $t6, 0 # loop counter
li $t7, 3 # str length
str_loop:
bgt $t6, $t7, str_loop_end
lb $t8, 0($a0)
lb $t9, 0($a1)
sb $t8, 0($a1)
sb $t9, 0($a0)
addi $a0, $a0, 1
addi $a1, $a1, 1
addi $t6, $t6, 1
b str_loop
str_loop_end:
jr $ra
swap:
# $a0 - address of the first num
# $a1 - address of the second num
# $a2 - index of the first number in the array
addi $sp, $sp, -4
sw $ra, 0($sp)
lw $s0, 0($a0)
lw $s1, 0($a1)
sw $s0, 0($a1)
sw $s1, 0($a0)
la $a0, str_array
sll $a2, $a2, 2 # $a2 = $a2 * 4
add $a0, $a0, $a2 # address of the first str
addi $a1, $a0, 4 # address of the next string
jal str_swap
lw $ra, 0($sp)
addi $sp, $sp, 4
jr $ra
sort:
# sort in non-decreasing order
# $a0 - address of the array
# $a1 - length of the array
addi $sp, $sp, -12
sw $ra, 0($sp)
sw $a0, 4($sp)
sw $a1, 8($sp)
li $t0, 0 # outer index
move $t2, $a1
subi $t2, $t2, 1 # upper bound = length - 1
outer:
bgt $t0, $t2, outer_end
li $t1, 0 # inner index
lw $a0, 4($sp)
inner:
bge $t1, $t2, inner_end
lw $t3, 0($a0) # $t3 = a[i]
lw $t4, 4($a0) # $t4 = a[i+1]
move $t5, $a0 # save $a0 in $t5
bgt $t3, $t4, swap_elements # if a[i] > a[i+1]
b inner_continue
swap_elements:
addi $a1, $a0, 4
move $a2, $t1
jal swap
inner_continue:
addi $t1, $t1, 1 # increment inner counter
move $a0, $t5
addi $a0, $a0, 4 # move to the next position in the array
b inner
inner_end:
addi $t0, $t0, 1 # increment outer loop
b outer
outer_end:
lw $ra, 0($sp) # restore $ra
addi $sp, $sp, 12
jr $ra
main:
la $a0, array
lw $a1, n
jal sort # bubble sort the array
addi $t0, $zero, 0 #t0=0 counter
la $s0, array # address of num_array
la $s1, str_array # address of str_array
lw $s2, n
res:
# get number from num_array and print it
lw $a0, 0($s0)
li $v0, 1
syscall
# move to the next element, increment by 4 because numbers take 1 word = 4 bytes
addi $s0, $s0, 4
# print space
la $a0, 32
li $v0, 11
syscall
# get string from str_array and print it
la $a0, 0($s1)
li $v0, 4
syscall
# print space
la $a0, 32
li $v0, 11
syscall
# move to the next element, increment by 4 because strs have 4 bytes length (in your case)
addi $s1, $s1, 4
# increment loop counter
addi $t0, $t0, 1
blt $t0, $s2, res # $s2 is size of the array
end:
li $v0, 10
syscall #the end
因此给出这个输入:
array: .word 3,2,1,4
str_array: .ascii "ccc\0bbb\0aaa\0ddd\0"
n: .word 4
产生以下输出:
1 aaa 2 bbb 3 ccc 4 ddd
-- program is finished running --
关于arrays - 在 MIPS 中对整数进行排序，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20528668/