I am supposed to write a function in MIPS that determines if a number is prime. The function should pass the number to be judged as an argument (through $a0) and return a value to indicate whether the number is a prime or not(through $v0). Use this function in a program that determines and prints all the prime numbers between 1 and 1000. This is supposed to be a recursive function. I wrote it but im not sure if i wrote it recursively. #I/O console .data input_int: .asciiz "The prime numbers between 1 and 1000 are:\n" newline: .asciiz "\n" space: .asciiz " " .text .align 2 .globl main main: la $a0, input_int #print "The prime numbers between 1 and 1000 are:" li $v0,4 syscall li $a1,2 #the first prime is 2 li $s2,1000 #Stop searching for primes after 1000 li $t5,0 #zero initialized li $t2,6 #initialized to 6 li $s0,1 jal primeloop #jumps back primeloop: #Find primes beq $s2,0,exit sub $s2,$s2,1 #Subtract to keep track as in counter #Set $t1 to 2 li $t1,2 divide: div $a1,$t1 #Divides by 2 to get next prime mflo $t3 slt $v0,$t3,$t1 #if quotient less than divisor stop beq $v0,$s0,fdprime #determine if prime ($v0=1), if prime prints #If remainder is zero, it is a composite,not prime mfhi $t4 beq $t4,0,nprime #Try next divisor add $t1,$t1,1 j divide fdprime: li $v0,1 move $a0,$a1 syscall li $v0,4 la $a0,space syscall addi $t5,$t5,1 beq $t5,$t2,skip nprime: #Advances to the next number add $a1,$a1,1 jr $ra #goes back to find the primes skip: li $v0,4 la $a0,newline syscall li $t5,0 j nprime exit: oneohthree

I don't know MIPS, but a recursive program is one which calls itself. So in your situation you should ahve a counter that goes from 1 to 1000. the main function, after it determines wether the current number is prime or not, should do a check to see if the number is less that 1000. If so, the funtion should call itself (and increment the counter). If not, the function should end. Michael J

What I did was I made a counter that decrements from 1000 to 1. And in the divisor loop I go through the divisors. If the quotient is less than the divisor it goes it goes to another loop and increments the number and goes back to the divisor loop. I dont know if it is right oneohthree

Sounds recursive to me. Michael J