Mips recursion sum of digits -

so understand logic behind how calculate sum of digits. wondering whether recursive function correct. assume user entered number or wants digit sum , stored in $v0.

sum2:  li $s0, 0  move $a1, $v0  li $s1, 0   li $s2, 10   # adjusting stack frame size store 3 words   addi $sp, $sp,-12   # these sum value, return address, , number   sw $s1, 0($sp)   sw $ra, 4($sp)   sw $a1, 8($sp)  loop2: bne $s0, $a1, sum3   move $a0, $s1   li $v0, 1   syscall   lw $s1, 0($sp)   lw $ra, 4($sp)   lw $a1, 8($sp)   addi $sp, $sp 12   jr $ra sum3:    div $t0, $a1, $s2   mfhi $t0   add $s1, $s1, $t0   div $t1, $a1, $s2   mflo $a1   j loop2   

if logic isn't clear first checking see if number doesn't equal 0 , if doesn't modulus of number divided 10 , add sum 0 , divide user input number 10 , continue call function till number becomes zero. had 1 quick last question. in mips recursive function or iterative function execute faster?

your implementation corresponds this:

int sum_digits(int n) {     int res = 0;     while (n != 0) {         res += n % 10;         n /= 10;     }     return res; } 

actually you're printing result rather returning it, that's beside point. it's iterative function.

a recursive implementation this:

int sum_digits(int n) {     return (n >= 10) ? ((n % 10) + sum_digits(n / 10)) : (n % 10); } 

i'll leave translate mips assembly know algorithm looks like.