C和MIP中的分区功能

Question

我试图编写一些模拟C函数的MIPS代码，但是似乎我遇到了一个无法克服它的障碍。

int partition(int f, int l) {
  int pivot = v[l];
  int i = f;

  for (int j = f; j < l; j++) 
if(v[j] < pivot) 
  swap(i++,j);

  swap(i, l);
  return (i);
}

那就是C中的功能，这是我到目前为止用MIPS组装语言写的内容：

partition:
    
    addi $sp, $sp, -8               #creates space for 2 words
    sw $ra, 0($sp)                  #stores ra in stack 
    sw $a1, 4($sp)                  #stores a1 in stack
    la $s0, v                       #stores ad. of v[0] in s0
    
    sll $t0, $a1, 2                 #stores 4 * l in t0
    add $t0, $t0, $s0               #stores ad. of v[l] in t0
    lw $s1, 0($t0)                  #loads v[l] in s1, s1 is pivot
    add $t1, $a0, $zero             #loads f in t1, t1 is 'i'
    add $t2, $a0, $zero             #loads f in t2, t2 is 'j'
    
for1:   slt $t3, $t2, $a1           #checks if j < l
        beq $t3, $zero, exit
        sll $t4, $t2, 2             #t4 stores 4 * j
        add $t4, $t4, $s0           #t4 stores ad. of v[j]
        lw $t5, 0($t4)              #t5 stores value of v[j]
        slt $t3, $t5, $s1           #checks if v[j] < pivot
        beq $t3, $zero, bfor        #jumps to next repetition of the loop
        add $a0, $t1, $zero         #a0 is i
        add $a1, $t2, $zero         #a1 is j
        jal swap                    #call swap
        addi $t1, $t1, 1            # i++
        j bfor                      #continue loop
        
bfor:   lw $a1, 4($sp)              #restores a1
        addi $t2, $t2, 1            # j++
        j for1                      #continue loop
        


exit:    add $a0, $t1, $zero        #a0 is i    
         lw $a1, 4($sp)             #restores initial a1
         jal swap                   #call swap
         add $v0, $t1, $zero        #return i   
         lw $ra, 0($sp)             #restore initial ra
         addi $sp, $sp, 8


    
    

    jr      $ra

据说F和L分别存储在$ A0中和$ a1中，已创建了交换函数，并且向量V在内存。我不明白我的错误在哪里，欢迎任何帮助。提前致谢！

Answer 1

    partition:
        addi $sp, $sp, -16              #adjust stack for 4 items
        sw $ra, 0($sp)                  #store ra in stack
        sw $s0, 4($sp)                  #store s0
        sw $a0, 8($sp)                  #store a0
        sw $a1, 12($sp)                 #store a1
        
        la $s0, v                       #load address of v0 in s0
        
        sll $t0, $a1, 2                 #t0 stores 4 * l
        add $t0, $t0, $s0               #t0 stores ad. of v[l]
        lw $t1, 0($t0)                  #t1 contains v[l] (t1 pivot)
        add $t2, $a0, $zero             #t2 is f (t2 i)
        add $t3, $a0, $zero             #t3 is f (t3 j)
        
for1:   slt $t4, $t3, $a1               #sets 1 to t4 if j < l
        beq $t4, $zero, exit            
        sll $t5, $t3, 2                 #t5 is 4 * j
        add $t5, $t5, $s0               #t5 stores ad. of v[j]
        lw $t6, 0($t5)                  #t6 has the value of v[j]
        
        slt $t4, $t6, $t1               #sets 1 to t4 if v[j] < pivot
        beq $t4, $zero, bfor
        
        add $a0, $t2, $zero             #a0 is i
        add $a1, $t3, $zero             #a1 is j
        
        addi $sp, $sp, -12              #adjust stack for 3 items
        sw $t1, 0($sp)
        sw $t2, 4($sp)
        sw $t3, 8($sp)                  #store pivot, i, j in stack
        jal swap                        #call swap
        
        lw $t1, 0($sp)
        lw $t2, 4($sp)
        lw $t3, 8($sp)                  #restores pivot, i, j before the call
        addi $sp, $sp, 12               #return items to stack
        
        lw $a1, 12($sp)                 #restore initial a1
        
        addi $t2, $t2, 1                #i++
        j bfor                          
        
bfor:   addi $t3, $t3, 1                #j++
        j for1                          #continue loop        
        
        
exit:   add $a0, $t2, $zero             #a0 is i
        addi $sp, $sp, -4               #adjust stack for 1 item
        sw $t2, 0($sp)                  #store i
        jal swap                        #calls swap
        lw $v0, 0($sp)                  #returns i
        addi $sp, $sp, 4                #returns item to stack
        
        lw $ra, 0($sp)                  #restore initial ra in stack
        lw $s0, 4($sp)                  #restore initial s0
        lw $a0, 8($sp)                  #restore initial a0
        lw $a1, 12($sp)                 #restore initial a1
        addi $sp, $sp, 16               #return items to stack

        jr      $ra

我找到了解决方案。谢谢那些回答的人。

重要的更改是寄存器中的值 $ t1 ， $ t2 和 $ t3 在 swap ，因为交换对它们进行了修改。