我尝试使用链接列表在MIPS中实现合并排序算法。我基本上试图翻译此代码：http://www.geeksforgeeks.org/merge-sort-for-linked-list/ 但是，我有一些问题。排序列表不完整，就像它丢失了整个递归分支一样，例如，这是要排序的链接列表： 4 - ＆gt; 3 - ＆gt; 1 - ＆gt; 2 输出是：1 - ＆gt; 2 - ＆gt; 4.当我尝试将列表写入文件时出现错误（0x004003fc处的运行时异常：地址超出范围0x00000000）。我猜这是因为我打算打印4个号码，但我在列表中只有3个号码。
我认为我需要帮助的地方在于MergeSort功能，因为我完全不了解它实际上在指针方面做了什么。这就是我认为它的作用：
它接收指向列表头部的指针作为参数。在函数内部，它创建了＆＃34; head＆＃34;指针，它指向headRef的VALUE（在这种情况下，＆amp; head在main上传递，所以现在＆＃34; head＆＃34;在函数内部有head的地址在main中）。然后它声明了两个指针a和b（在MIPS中，这就像做a = 0，b = 0？我可以＆＃39; t＆＃34;声明＆＃34;寄存器）。然后它检查列表是否包含1或0个元素，如果是，则返回。否则，它会使用FrontBackSplit（head，＆amp; a，＆amp; b）将列表分成两半，但这是棘手的...这个函数不会返回任何内容。相反，它修改了指针＆＃34; a＆＃34;和＆＃34; b＆＃34;，使它们分别指向列表的前半部分和后半部分。在MIPS中，我认为这是两个返回值$ v0和$ v1。然后它从左边开始递归地对列表进行排序，但是再一次，它不会返回任何内容......它会修改指针＆＃34; a＆＃34;和＆＃34; b＆＃34;。最后，它将指针的值更改为列表的头部，使其指向新的排序列表。
由于有很多** ptr和＆amp; ptr值，我对在堆栈上保存的内容感到困惑。关于这个功能，这就是我到目前为止所做的：
# void mergesort(node** headRef)
# input $a0: head of the list
mergesort:
move $t0, $s3 # node* head = headRef
li $t1, 0 # node* a
li $t2, 0 # node* b
# if(head == NULL || head->next == NULL)
beqz $t0, mergesort_return
lw $t3, node_next($t0) # $t3 = head->next
beqz $t3, mergesort_return
move $a0, $t0
move $a1, $t1
move $a2, $t2
# save head and $ra
addi $sp, $sp, -8
sw $t0, 0($sp)
sw $ra, 4($sp)
jal frontbacksplit
# restore head and $ra
lw $t0, 0($sp)
lw $ra, 4($sp)
addi $sp, $sp, 8
# save output results
move $t1, $v0 # a now points to the first half + 1 (if odd) of the list (frontRef)
move $t2, $v1 # b now points to the second half of the list (backRef)
# save head, a, b and $ra
addi $sp, $sp, -16
sw $t0, 0($sp)
sw $t1, 4($sp)
sw $t2, 8($sp)
sw $ra, 12($sp)
# mergesort(a)
move $a0, $t1
jal mergesort
# restore registers and $ra
lw $t0, 0($sp)
lw $t1, 4($sp)
lw $t2, 8($sp)
lw $ra, 12($sp)
addi $sp, $sp, 16
# save head, a, b and $ra
addi $sp, $sp, -16
sw $t0, 0($sp)
sw $t1, 4($sp)
sw $t2, 8($sp)
sw $ra, 12($sp)
# mergesort(b)
move $a0, $t2
jal mergesort
# restore registers and $ra
lw $t0, 0($sp)
lw $t1, 4($sp)
lw $t2, 8($sp)
lw $ra, 12($sp)
addi $sp, $sp, 16
move $a0, $t1
move $a1, $t2
# save head, a, b and $ra
addi $sp, $sp, -16
sw $t0, 0($sp)
sw $t1, 4($sp)
sw $t2, 8($sp)
sw $ra, 12($sp)
jal sortedmerge
# restore registers and $ra
lw $t0, 0($sp)
lw $t1, 4($sp)
lw $t2, 8($sp)
lw $ra, 12($sp)
addi $sp, $sp, 16
move $s3, $v0 # s3 is the saved head of the original list declared in main
mergesort_return:
jr $ra
答案 0 :(得分：0)
嗯......我做到了。这是一个家庭作业，并没有评分。我希望他们不要以为我复制了它。大声笑。
######### uses function frontbacksplit
######### WORKING
# mergesort
# input $a0: head of the list to merge
mergesort:
li $t0, 0 # head_one = NULL
li $t1, 0 # head_two = NULL
# base case
beqz $a0, mergesort_return
lw $t2, node_next($a0)
beqz $t2, mergesort_return
addi $sp, $sp, -8
sw $ra, 0($sp)
sw $a0, 4($sp)
# move $a0, $a0
move $a1, $t0
move $a2, $t1
jal frontbacksplit
move $t0, $v0
move $t1, $v1
lw $ra, 0($sp)
lw $a0, 4($sp)
addi $sp, $sp, 8
# mergesort(a)
addi $sp, $sp, -16
sw $ra, 0($sp)
sw $t0, 4($sp)
sw $t1, 8($sp)
sw $a0, 12($sp)
move $a0, $t0
jal mergesort # mergesort(a)
move $t9, $v0
lw $ra, 0($sp)
lw $t0, 4($sp)
lw $t1, 8($sp)
lw $a0, 12($sp)
addi $sp, $sp, 16
# mergesort(b)
addi $sp, $sp, -20
sw $ra, 0($sp)
sw $t0, 4($sp)
sw $t1, 8($sp)
sw $a0, 12($sp)
sw $t9, 16($sp)
move $a0, $t1
jal mergesort # mergesort(b)
move $t8, $v0
lw $ra, 0($sp)
lw $t0, 4($sp)
lw $t1, 8($sp)
lw $a0, 12($sp)
lw $t9, 16($sp)
addi $sp, $sp, 20
# t8 = a, t9 = b
addi $sp, $sp, -24
sw $ra, 0($sp)
sw $t0, 4($sp)
sw $t1, 8($sp)
sw $a0, 12($sp)
sw $t9, 16($sp)
sw $t8, 20($sp)
move $a0, $t8
move $a1, $t9
jal merge
lw $ra, 0($sp)
lw $t0, 4($sp)
lw $t1, 8($sp)
lw $a0, 12($sp)
lw $t9, 16($sp)
lw $t8, 20($sp)
addi $sp, $sp, 24
#move $v0, $v0
jr $ra
mergesort_return:
move $v0, $a0
jr $ra
######### WORKING
# input $a0: source node (list to split)
# input $a1: front ref (a) $s4
# input $a2: back ref (b) $s5
# output $v0: frontRef
# output $v1: backRef
frontbacksplit:
# $a0 = node* source, $a1 = node* frontRef, $a2 = node* backRef
move $t0, $a0 # $t0 = source
move $t1, $a1 # $t1 = frontRef
move $t2, $a2 # $t2 = backRef
li $t3, 0 # node* fast;
li $t4, 0 # node* slow;
# if(source == NULL)
beqz $t0, frontbacksplit_sorted
lw $t5, node_next($t0) # $t5 = source->next
# if(source->next == NULL)
beqz $t5, frontbacksplit_sorted
# else
move $t4, $t0 # slow = source
lw $t3, node_next($t0) # fast = source->next
frontbacksplit_while:
# while(fast != NULL)
beqz $t3, frontbacksplit_endwhile
lw $t3, node_next($t3) # fast = fast->next
# if(fast != NULL)
beqz $t3, frontbacksplit_while
lw $t4, node_next($t4) # slow = slow->next
lw $t3, node_next($t3) # fast = fast->next
j frontbacksplit_while
frontbacksplit_endwhile:
move $t1, $t0 # frontRef = source
lw $t2, node_next($t4) # backRef = slow->next
sw $zero, node_next($t4) # slow->next = NULL
move $v0, $t1
move $v1, $t2
j frontbacksplit_end
frontbacksplit_sorted:
move $v0, $t0 # frontRef = source
li $v1, 0 # backRef = NULL
frontbacksplit_end:
jr $ra
############ WORKING
# input $a0: head of the first list
# input $a1: head of the second list
# output $v0: pointer to the head of the merged-sorted list
merge:
beqz $a0, merge_return2
beqz $a1, merge_return1
li $t3, 0 # head_three = NULL
lw $t0, node_int($a0)
lw $t1, node_int($a1)
addi $sp, $sp, -16
sw $ra, 0($sp)
sw $a0, 4($sp)
sw $a1, 8($sp)
sw $t3, 12($sp)
bge $t0, $t1, merge_else
# if head_one->num < head_two->num
lw $a0, node_next($a0)
# $a1 already has head_two
jal merge
move $t0, $v0 # ptr = merge(head_one->next, head_two)
# restore values
lw $ra, 0($sp)
lw $a0, 4($sp)
lw $a1, 8($sp)
lw $t3, 12($sp)
move $t3, $a0
sw $t0, node_next($t3)
move $v0, $t3
addi $sp, $sp, 16
jr $ra
merge_else:
# $a0 already has head_one
lw $a1, node_next($a1)
jal merge
move $t0, $v0 # ptr = merge(head_one->next, head_two)
# restore values
lw $ra, 0($sp)
lw $a0, 4($sp)
lw $a1, 8($sp)
lw $t3, 12($sp)
move $t3, $a1
sw $t0, node_next($t3)
move $v0, $t3
addi $sp, $sp, 16
jr $ra
merge_return1:
move $v0, $a0
jr $ra
merge_return2:
move $v0, $a1
jr $ra
我知道它很乱，但它有效：P。