Convert the C code to MIPS.

recursion$1.$c:

#include

int recursion$($int m$)\{$

if$($m $==-1)$

return $3$;

else if$($m $<=-2)\{$

if $($m $<-2)$

return $2$;

else

return $1$;

$\}$

else

return recursion$($m $-3)+$ m $+$ recursion$($m $-2)$;

$\}$

int main$()\{$

int x;

printf$("$Please enter a number: $")$;

scanf$("\%$d$",$ &x$)$;

printf$("\%$d$",$ recursion$($x$))$;

return $0$;

$\}$

Attempted MIPS code $($always returns $4$ no matter what input$)$:

$.$data

prompt: $.$asciiz "Please enter a number: $"$

result: $.$asciiz "The result is: $"$

$.$text

main:

li $v$0,4$ # syscall for print string

la $a$0,$ prompt

syscall

li $v$0,5$ # syscall for read int

syscall

move $a$0,$ $v$0$ # move read integer to $a$0$ for recursion

jal recursion # Call recursion with input x

li $v$0,4$ # syscall for print string

la $a$0,$ result

syscall

move $a$0,$ $v$0$ # Move result to $a$0$ for printing

li $v$0,1$ # syscall for print int

syscall

li $v$0,10$ # syscall to exit

syscall

recursion:

addi $sp$,$ $sp$,-16$ # stack frame for $4$ variables

sw $ra$,12($$sp$)$ # save return address

sw $a$0,8($$sp$)$ # save original m

# Base cases

li $v$0,3$ # return value for m $==-1$

beq $a$0,-1,$ finish

li $v$0,2$ # return value for m $==-2$

beq $a$0,-2,$ finish

li $v$0,1$ # return value for m $<-2$

blt $a$0,-2,$ finish

# Recursive case for m $>-1$

addi $a$0,$ $a$0,-3$

jal recursion

sw $v$0,4($$sp$)$ # Store result of recursion$($m $-3)$

lw $a$0,8($$sp$)$ # Restore original m

addi $a$0,$ $a$0,-2$

jal recursion # Recurse with m $-2$

lw $t$0,8($$sp$)$ # Load original m

lw $t$1,4($$sp$)$ # Load result of recursion$($m $-3)$

add $v$0,$ $v$0,$ $t$1$ # Add results of recursion$($m $-2)$ and recursion$($m $-3)$

add $v$0,$ $v$0,$ $t$0$ # Add m to sum

finish:

lw $ra$,12($$sp$)$ # Restore return address

addi $sp$,$ $sp$,16$ # Restore stack pointer

jr $ra