好的，我正在尝试在ARM7汇编中实现选择排序算法。它从数组的最后一个元素开始，一直到第一个元素。它适用于第一个交换，但在继续之前退出。我和我的教授谈过，他说使用堆栈变量，以便lr不会被覆盖等。但我仍然不明白具体如何做到这一点。我已经寻找了方法来做到这一点，但似乎找不到任何我明白如何在我的代码中实现。
我的代码如下：（a1是指向长度为a2的字符数组的指针。两者都是通过C传入的）

sort2:
    stmdb    sp!, {v1-v8, lr}          @ Copy registers to stack
    mov      v1, a2                    @store number of elements in v1,v2 (a2 is passed in via C)
    mov      v2, a2
    sub      v1, v1,#1                 @subtract 1 from a2 to get number last element in array
    sub      v2, v1,#1                 @subtract 1 from v1 to get element just below last element
    bl       loop2
    ldmia    sp!, {v1-v8, pc}

loop2:
    ldrb     v4, [a1,v1]               @store value of last element in array a1(passed in via C) into r4 
    ldrb     v5, [a1,v2]               @store value of next to last element in array a1(passed in via C) into r4 
    cmp      v4, v5
    blt      swap2                     @if last element is less than current element a[v2], swap them
    cmp      v2, #0                    @if v2 = 0, we cycled through one iteration
    subeq    v1, v1,#1                 @so now move on to a[v1-1] element in array                   
    moveq    v2, v1                    @if v2 = 0, move the value of v1 into v2
    sub      v2, v2,#1                 @subtract 1 from v2 to get element below a[v1]
    cmp      v1, #0                    @if v1 = 0, were done
    bne      loop2
    mov      pc, lr                    @return to sort2

swap2:
    mov      v8, lr                    @store return address in v8
    ldrb     v6, [a1,v2]               @store value of a[v2] into v6
    strb     v4, [a1,v2]               @store v4 into a[v2]
    strb     v6, [a1,v1]               @store value of v6 into a[v1]
    mov      lr, v8                    @falls through and exits here instead of returning to loop2