Fibonacci recursion function in Mips - recursion

Question

I am trying to pratice how recursion works in Mips. So I tried to write a fibonacci function fib.
At first I had addi $a0, $a0, n to write a general solution, but I thought that if I want to check my results in Qtspim , maybe i need to add a real number as an argument. I do not want a full answer , if the thought behind the code is wrong, but some help in order to run it in Qtspim, and find my mistakes (logic mistakes) on my own
This is my code:
.globl main
.text
main:
addi $a0, $a0, 4
fib:
addi $sp, $sp, -8
sw $ra, 4($sp)
sw $a0, 0($sp)
slti $t0, $a0, 1
beq $t0, 1, L2 #if n<1
beq $a0, 1, L2 # if n=1
beq $t0, 0, L1 # if n>1
#what to do when n<=1
L2:
addi $v0, $v0, 1
jr $ra
#what to do when n>1
L1:
addi $a0, $a0, -1
jal fib
lw $a0, 0($sp)
lw $ra, 4($sp)
addi $sp, $sp, 8
lw $t1, 0($v0)
add $v0, $t1, $v0
jr $ra
li $v0,10
syscall
I get an error message as such:
Bad address in data/stack read: 0x00000000

Answer 1

.text
main:
addi $a0, $a0, 4
#### you've place fib inline inside main,
#### you should have all of main here, and "call" fib using jal instruction
#### and the syscall for exit goes up here as well
fib:
addi $sp, $sp, -8 # you will want one more word of stack space
sw $ra, 4($sp)
sw $a0, 0($sp)
slti $t0, $a0, 1 # i would have use 2 here instead of 1
beq $t0, 1, L2 #if n<1 # this is ok, but better to use bne $t0, $0, L2
beq $a0, 1, L2 # if n=1 # and then this would not be needed
beq $t0, 0, L1 # if n>1 # this doesn't need to be conditional, if the program reaches here then L1 is the thing to do next.
#what to do when n<=1
L2:
addi $v0, $v0, 1 # here you want to return just 1 not v0+1
jr $ra
#what to do when n>1
L1:
addi $a0, $a0, -1
jal fib # fib(n-1), good
lw $a0, 0($sp) # this reloads $a0 the original n
lw $ra, 4($sp)
addi $sp, $sp, 8
lw $t1, 0($v0) # after a call to fib $v0 holds fib(n)
# an integer value but you're treating it like a pointer and dereferencing it
add $v0, $t1, $v0 # here doing fib(n-1) + n
# you want fib(n-1) + fib(n-2) instead
# so you're missing a fib(n-2)
jr $ra
li $v0,10 # this is part of main, so move it to where main is
syscall # realize that code located here is unreachable (aka dead)
# anything after an unconditional branch (here the jr $ra just above)
# and without a label is very suspicious as unreachable

Answer 2

update , some improved (i hope ) code
.globl main
.text
main:
addi $a0, $a0, 4
jal fib
li $v0, 10
syscall
fib:
addi $sp, $sp, -8
sw $ra, 4($sp)
sw $a0, 0($sp)
slti $t0, $a0, 2
beq $t0, 1, L2
#what to do when n<=1
L2:
addi $v0, $v0, 1
jr $ra
#what to do when n>1
L1:
addi $a0, $a0, -1 # a0 -> n-1
jal fib # fib(n-1)
lw $a0, 0($sp) #load word from memory adress 0($sp) to register $a0
lw $ra, 4($sp) #load word from memory adress 4($sp) to register $ra
addi $sp, $sp, 8
add $t1, $zero, $v1 # in register $t1 store current $v1
add $v0, $t2, $v0 # v0_ n+1 =v0 _n + v0_n-1
add $t2, $t1, $zero # in $t2 save the previous value of v1

Answer 3

.data
prompt: .ascii "Fibonacci Program\n"
.asciiz "Enter N value: "
results: .asciiz "\nFibonacci of N = "
n: .word 0
answer: .word 0
.text
.globl main
.ent main
main:
# Read n value from user
li $v0, 4 # print prompt string
la $a0, prompt
syscall
li $v0, 5 # read N (as integer)
syscall
sw $v0, n
# Call Fibonacci function.
lw $a0, n
jal fib
sw $v0, answer
# Display result
li $v0, 4 # print prompt string
la $a0, results
syscall
li $v0, 1 # print integer
lw $a0, answer
syscall
# Done, terminate program.
li $v0, 10 # terminate
syscall # system call
.end main
# Fibonacci function
# Recursive definition:
# = 0 if n = 0
# = 1 if n = 1
# = fib(n-1) + fib(n-2) if n > 2
# Arguments
# $a0 - n
# Returns
# $v0 set to fib(n)
.globl fib
.ent fib
fib:
subu $sp, $sp, 8
sw $ra, ($sp)
sw $s0, 4($sp)
move $v0, $a0 # check for base cases
ble $a0, 1, fibDone
move $s0, $a0 # get fib(n-1)
sub $a0, $a0, 1
jal fib
move $a0, $s0
sub $a0, $a0, 2 # set n-2
move $s0, $v0 # save fib(n-1)
jal fib # get fib(n-2)
add $v0, $s0, $v0 # fib(n-1)+fib(n-2)
fibDone:
lw $ra, ($sp)
lw $s0, 4($sp)
addu $sp, $sp, 8
jr $ra
.end fib