Question

我知道在c或java中，将升序排序更改为降序的快速修复方法是找到＆lt;或者＆gt;某处并扭转它。但是尽可能地尝试，我无法找到相应的（或等效的交换）来改变我编写的代码，而不是提升。（现在，它可以完美地按升序打印链表）。除了我所包含的内容之外我还有大约100个左右，因为他们没有进行排序，但是他们分配了值 - 所以这不是问题所在。我觉得我只是错过了一些明显的东西？这种变化会不会影响我的印刷方式或插入方式或两者兼而有之？任何帮助，将不胜感激。

loop_to_find_correct_spot:
    beqz $t0, if_current_node_next_equals_null
    l.s $f0, 0($t0)
    c.lt.s $f12, $f0
    bc1t if_current_node_does_not_equal_null

    or $t1, $t0, $zero
    lw $t0, 4($t0)

    j loop_to_find_correct_spot

if_current_node_next_equals_null:
    # link this node to the previous, $v0 = &(previous node)
    # copy address of the new node into the previous node
    sw $v0, 4($t1)

    # initalize the current node value to the passed in argument on the $f12 register
    swc1 $f12, 0($v0)

    # initalize the current node next value to null
    sw $zero, 4($v0)

    j end_insert_if_else

if_current_node_does_not_equal_null:
    # initalize the current node value to the passed in argument on the $f12 register
    swc1 $f12, 0($v0)

    sw $v0, 4($t1)
    sw $t0, 4($v0)

    j end_insert_if_else

end_insert_if_else:
    lw $ra, 0($sp)
    addi $sp, $sp, 4
    jr $ra

allocate_memory_for_node:
    addi $sp, $sp, -4
    sw $ra, 0($sp)

    # allocate 12 bytes in memory for the new node
    li $v0, 9
    li $a0, 8
    syscall

    lw $ra, 0($sp)
    addi $sp, $sp, 4
    jr $ra

print_linked_list:
    addi $sp, $sp, -4
    sw $ra, 0($sp)

    lw $s0, head # get a pointer to the head node
    lw $s0, 4($s0)

begin_loop_print_linked_list:
    # while (next != null)
    beqz $s0, end_loop_print_linked_list

    # get the value for the current node
    l.s $f12, 0($s0)

    # print out current item
    li $v0, 2
    syscall

    la $a0, end_line_prompt   #puts newlines in printed list
    li $v0, 4
    syscall

    #get the pointer to the next node
    lw $s0, 4($s0)
    b begin_loop_print_linked_list

end_loop_print_linked_list:
    lw $ra, 0($sp)
    addi $sp, $sp, 4
    jr $ra

在MIPS程序集中将升序排序更改为降序

0 个答案: