所以，我正在研究MIPS中的fibonacci，规则是我需要有一个解决问题的递归方法的序言。我的代码目前正在产生错误的输出，我无法识别应该编辑哪个部分。
.text
main: li $v0, 5
syscall
move $a0, $v0
move $v0, $zero
jal fibo
move $t2, $v0
li $v0, 1
move $a0, $t2
syscall
li $v0, 10
syscall
fibo: ####preamble#### #push from stack
subu $sp, $sp, 32
sw $ra, 0($sp) #store return address
sw $a0, 4($sp)
sw $fp, 8($sp)
sw $v0, 12($sp)
addu $fp, $sp, 32
####preamble####
zero: bne $a0, 0, one
move $v0, $a0
jr $ra
one: bne $a0, 1, fn1
move $v0, $a0
jr $ra
fn1: subi $a0, $a0, 1 #call for fibo(n-1)
jal fibo #recursive
lw $a0, 4($sp)
addi $a0, $a0, 1
subi $a0, $a0, 2 #call for fibo(n-2)
jal fibo
result: lw $ra, 0($sp) #load ra
lw $fp, 8($sp)
lw $t0, 12($sp)
add $v0, $t0, $v0
addu $sp, $sp, 32
addu $fp, $fp, 32
jr $ra
答案 0 :(得分：1)
好的，你有基本的结构和组件是正确的，但不幸的是，有一些错误。
我已经创建了两个版本的程序。一个评论详细说明了错误。并且，第二个版本清理，简化和工作
这里是带注释的版本[请原谅无偿的风格清理]：
.text
main:
li $v0,5
syscall
move $a0,$v0
# NOTE/BUG: doing this is unnecessary/wrong if fibo is correct
move $v0,$zero
jal fibo
move $t2,$v0
li $v0,1
move $a0,$t2
syscall
li $v0,10
syscall
####preamble#### #push from stack
fibo:
# NOTE/BUG: for simple functions like this, setting up fp is extra work
# NOTE/BUG: we want to have extra space in the frame but we don't need
# to push/pop for some
subu $sp,$sp,32
sw $ra,0($sp) # store return address
sw $a0,4($sp)
sw $fp,8($sp)
sw $v0,12($sp)
addu $fp,$sp,32
####preamble####
# NOTE/BUG: we must zero out t0 so that the add in result: is valid
# NOTE/BUG: a stack frame has already been established -- it must be popped
# we can _not_ just do a "jr" here
zero:
bne $a0,0,one
move $v0,$a0
jr $ra
# NOTE/BUG: a stack frame has already been established -- it must be popped
one:
bne $a0,1,fn1
move $v0,$a0
jr $ra
fn1:
subi $a0,$a0,1 # call for fibo(n-1)
jal fibo # recursive
# NOTE/BUG: we must save the result for fibo(n-1) in our stack frame
# NOTE/BUG: this is incorrect -- by doing it, we [effectively] do fibo(n-1)
# again (i.e.) (n+1)-2 --> (n-1) and _not_ (n-2) as we wish
# do one of these but _not_ both
lw $a0,4($sp)
addi $a0,$a0,1
subi $a0,$a0,2 # call for fibo(n-2)
jal fibo
# NOTE/BUG: fibo(n-1) must be added to fibo(n-2)
result:
lw $ra,0($sp) # load ra
lw $fp,8($sp)
# NOTE/BUG: this is misplaced because entering this block should already
# have v0 set correctly
lw $t0,12($sp)
add $v0,$t0,$v0
# NOTE/BUG: the correct way to restore $fp is using: lw $fp,8($sp)
addu $sp,$sp,32
addu $fp,$fp,32
jr $ra
这是工作版本：
.data
msg_ask: .asciiz "Enter n for fibonacci(n) (-1=stop): "
msg_fibo: .asciiz "fibonacci(n) is: "
msg_nl: .asciiz "\n"
.text
main:
# prompt user
la $a0,msg_ask
li $v0,4
syscall
# get number from user
li $v0,5
syscall
move $a0,$v0
bltz $a0,main_exit
jal fibo
move $t2,$v0 # save the result
# print message
la $a0,msg_fibo
li $v0,4
syscall
# print the result
li $v0,1
move $a0,$t2
syscall
# print message
la $a0,msg_nl
li $v0,4
syscall
j main
main_exit:
li $v0,10
syscall
# fibo -- recursive fibonacci
#
# RETURNS:
# v0 -- fibonacci(n)
#
# arguments:
# a0 -- the "n" for the Nth fibonacci number
#
# registers:
# t0 -- temporary
fibo:
# fibo(0) is 0 and fibo(1) is 1 -- no need to establish a stack frame
bgt $a0,1,fibo_full # do we need recursion? if yes, fly
move $v0,$a0 # no, just set return value
jr $ra # fast return
# establish stack frame
# we need an extra cell (to preserve the result of fibo(n-1))
fibo_full:
# this gives us a temp word at 0($sp)
subu $sp,$sp,12 # one more than we need
sw $ra,4($sp)
sw $a0,8($sp)
subi $a0,$a0,1 # call for fibo(n-1)
jal fibo # recursive
sw $v0,0($sp) # save result in our frame (in extra cell)
subi $a0,$a0,1 # call for fibo(n-2)
jal fibo # recursive
lw $t0,0($sp) # restore fibo(n-1) from our stack frame
add $v0,$t0,$v0 # result is: fibo(n-1) + fibo(n-2)
# restore from stack frame
lw $ra,4($sp)
lw $a0,8($sp)
addu $sp,$sp,12
jr $ra # return