MIPS Assembly code to find all the prime numbers below an inputted number

问题

Below I have posted my MIPS code. The java code I am basing it off on is...

Java Code...

for (int i=2;i<n;i++){
  p = 0;
  for (int j=2;j<i;j++){
  if (i % j == 0)
    p = 1;
  }
  if (p = 0) System.out.println(i);
}

I have added the line "beq $t3, 1, L4" to skip to L4 when p is set to 1 to save time. But when I added this line of code, the program outputs nothing. Before I added this line, it would print all the integers from 2~n.

MIPS Code...

# The number is read through the keyboard 
.text
.globl main

main:
# Display message to user for a number
li $v0, 4
la $a0, prompt1
syscall

# read keyboard into $v0 (number x is upper bound number to find primes)
li $v0, 5 
syscall

# move the number from $v0 to $t0
move $t0, $v0 # $t0 = n

# store 2 in $t1 and $t2
li $t1, 2 # i
li $t2, 2 # j

L3: # for (int i=2; i<n; i++)
# store 0 in $t3
li $t3, 0 # p = 0;

L2: # for (int j=2; j<i; j++)
# do div of two numbers
div $t2, $t1

# store the remainder in $t4
mfhi $t4

# branch if remainder is not 0 to L1
bne $t4, 0, L1 # if (i % j == 0)

# set $t3 as 1
li $t3, 1 # p = 1

# if p=1 break to next i
beq $t3, 1, L4

L1: # if (i % j == 0)
# add 1 to t2
addi $t2, $t2, 1 # j++

# repeat code while j < i
ble $t2, $t1, L2

# print integer function call 1 
# put the answer into $a0
li $v0, 1 
move $a0, $t1
syscall # System.out.println(i)
#print comma
li $v0, 4
la $a0, comma
syscall

L4:
# add 1 to t1
addi $t1, $t1, 1 # i++

# repeat code while i < n
ble $t1, $t0, L3 # for (int i=2; i<n; i++)

.data
prompt1:
 .asciiz "Enter a number "
comma:
 .asciiz ","

I think the error occurs because my assembly logic is not accounting for the j < i portion of the for loop. Am I on the right track?

回答1:

There are two problems:

When you increment i you forget to set j back to 2. L3 should be moved 1 line up.
ble in MIPS is less than or equal to, so your code is actually checking j <= i, not j < i. This causes your code to check i % j when i = j, which always has a remainder 0 and will register as not prime. Changing 'ble' to 'blt' should fix this. Same goes for the i < n check.

Also some additional constructive criticism: You set p=1 then immediately do a check to see if p = 1.

li $t3, 1 # p = 1
beq $t3, 1, L4 # if p=1 break to next i

You can take out the redundancy of having p, removing p altogether and replacing these two lines with an unconditional branch right to L4

b L4

来源：https://stackoverflow.com/questions/40246994/mips-assembly-code-to-find-all-the-prime-numbers-below-an-inputted-number

标签

assembly

mips

primes