converting C code to MIPS problems with recursion and multiplication - c++

Question

hello everyone i need some help converting the following C code into MIPS:
main()
{
int i;
for (i=0;i<5;++i)
power(2,1)
factorial(i)
return 0;
}
int power(int base,int n)
{
int i,p;
p=0;
for(i=1;i<=n;++i)
p=p*base;
return p;
}
int factorial(int a)
{
if(a==1)
return 1;
else
{
a *=factorial(a-1);
return a;
}
}
i have only done the factorial part so far but i'm stuck at a*=factorial(a-1), i under stand that multi is used to multiply two registers together but how can it take the input of a recursive loop? also is it possible to multi to multiply with a constant instead of another register?
factorial:
addi $sp, $sp, -8 #adjust stack for 2 items
sw $ra, 4($sp) #save return address
sw $a0, 0($sp) #save argument
bne $a0, 1, Else #if !(a==1), go to else address
addi $v0, $zero, 1 #result is 1
addi $sp, $sp, 8 #pop 2 items from stack
jr $ra #return
Else: addi $a0, $a0, -1 #decrement s-1
jal factorial #recursive call
lw $a0, 0($sp) #restore original a
lw $ra, 4($sp) #and return address
addi $sp, $sp, 8 #pop two items from stack

Answer 1

Here is a code that I wrote, I think it will fit your purpose.
.text
factorial:
li $t0,13 # load 13 to check for overflow
blt $a0,$t0,no
li $v1,1
no:
bgtz $a0,find # if $a0>0 goto generic case
li $v0, 1 # base case, 0! = 1
jr $ra
find: sub $sp,8 # make room for $s0 and $ra
sw $s0,($sp) # store argument $s0=n
sw $ra,4($sp) # store return address
move $s0, $a0 # save argument
sub $a0, 1 # factorial(n-1)
jal factorial # v0 = (n-1)!
mul $v0,$s0,$v0 # n*(n-1)!
lw $s0,($sp) # restore $s0=n
lw $ra,4($sp) # restore $ra
add $sp,8 # reduce stack size
jr $ra # return
main:
la $a0,input
li $v0,4
syscall
li $v0,5
syscall
move $a0,$v0
jal factorial
move $t0,$v1
bnez $t0,overflow1
move $a0,$v0
li $v0,1
syscall
b end
overflow1:
la $a0,overflow
li $v0,4
syscall
end:
li $v0,10
syscall
.data
input:.asciiz"Input a number:"
overflow:.asciiz"There is an overflow!!"
It's complete with comments and explanations so there shouldn't be problems.