MIPS Fibonacci递归程序，我能正确理解吗？

Question

## fibonaci 
## Registers used:
## $a0 - initially n.
## $t0 - parameter n.
## $t1 - fib (n - 1).
## $t2 - fib (n - 2).
    .text
fib:
    bgt $a0, 1, fib_recurse # if n < 2, then just return a 1,
    li $v0, 1 # don't bother to build a stack frame.
    jr $ra

    # otherwise, set things up to handle

fib_recurse: # the recursive case:
    subu $sp, $sp, 32 # frame size = 32, just because...
    sw $ra, 28($sp) # preserve the Return Address.
    sw $fp, 24($sp) # preserve the Frame Pointer.
    addu $fp, $sp, 32 # move Frame Pointer to base of frame.

    move $t0, $a0 # get n from caller.
    # compute fib (n - 1):
    sw $t0, 20($sp) # preserve n.
    sub $a0, $t0, 1 # compute fib (n - 1)
    jal fib
    move $t1, $v0 # t1 = fib (n - 1)
    lw $t0, 20($sp) # restore n.

    # compute fib (n - 2):
    sw $t1, 16($sp) # preserve $t1.
    sub $a0, $t0, 2 # compute fib (n - 2)
    jal fib
    move $t2, $v0 # t2 = fib (n - 2)
    lw $t1, 16($sp) # restore $t1.

    add $v0, $t1, $t2 # $v0 = fib (n - 1) + fib (n - 2)

    lw $ra, 28($sp) # restore Return Address.
    lw $fp, 24($sp) # restore Frame Pointer.
    addu $sp, $sp, 32 # restore Stack Pointer.
    jr $ra # return.

main:
    li $a0, 5 # fib (5) is what we are trying to find.
    jal fib
    move $a0, $v0

    li $v0, 1
    syscall

    li $v0, 10
    syscall

这是我的问题：当在main中调用jal fib时，函数操作是否知道在a0上运行，如果我在a1，a2中调用fib之前放了一些值...这会影响什么吗？

我也在添加输出消息时遇到问题：＆＃34;输入n来计算fib（n）：＆＃34;然后当我收到＆＃34; n＆＃34;并把它放到$ a0，发现fib需要相当多的时间并且不断返回1或0 ..很奇怪。