MIPS assembly addition program

Question

I am new to assembly and have made a simple addition program to sum two integers read from the keyboard. The program outputs correctly, but I want to know if there is a way to streamline my code. It seems a bit cumbersome for such a simple program and I may have instructions that are unnecessary.

# Author: Evan Bechtol
# Description: This program prompts the user to enter 2 integers and computes their sum.
#---------------------------------------------------------------------------------------#
 .data
A: .word # Store the number 4 as an integer in var1 # $t0 is used
B: .word # Store the number 2 as an integer in var2 # $t1 is used
S: .word # Store the sum of A and B # $t2 is used
Prompt1: .asciiz "Please enter first number: "
Prompt2: .asciiz "Please enter second number: "
Result: .asciiz "The sum of A and B is: "
 .text
main:
 #--------------------------------------------------------#
 #Display first prompt
 li $v0, 4 # Load instruction "print string"
 la $a0, Prompt1 # Load prompt into $a0
 syscall
 #Read first integer
 li $v0, 5 # Read 1st integer
 la $t0, A # $t0 = A
 syscall
 #Store first integer into memory
 move $t0, $v0 # Move contents in $v0 to $t0
 sw $t0, A # A = value at $t0 
 #--------------------------------------------------------#
 #Display second prompt
 li $v0, 4 # Load instruction "print string"
 la $a0, Prompt2 # Load prompt into $a0
 syscall
 #Read second integer
 li $v0, 5 # Read 1st integer
 la $t1, B # $t0 = A
 syscall
 #Store second integer into memory
 move $t1, $v0 # Move contents in $v0 to $t0
 sw $t1, B # A = value at $t0 
 #--------------------------------------------------------#
 #Add the two variables
 la $t2, S # $t2 = S 
 add $t2, $t0, $t1 # $t2 = $t0 + $t1
 sw $t2, S # S = value at $t2
 #Display the Result prompt
 la $a0, Result # Loads Output label to be printed
 li $v0, 4 # Sysycall to print string
 syscall
 #Display the sum
 lw $a0, S # $a0 = value at S
 li $v0, 1 # Syscall to print integer
 syscall
 #Exit the program
 li $v0, 10 # Load exit code to $v0
 syscall

Answer 1

• The comments are misleading:

  #Read second integer
  li $v0, 5 # Read 1st integer
  la $t1, B # $t0 = A
  

  umm... are we reading second or 1st? Bottomline is, do not overcomment the code.

• syscall 5 leaves a value in $v0. The contents of $t0 (or $t1) is irrelevant during the syscall. Set them up when you need them:

  li $v0, 5
  syscall
  la $t0, A
  move $t0, $v0
  

• You store data to memory just to load them back. This is very anti-assembly. Generally you want to use registers as much as possible, and avoid memory as much as possible:

  li $v0, 5
  syscall
  move $t0, $v0
  ...
  li $v0, 5
  syscall
  # At this moment you have first integer in $t0, and the second in $v0.
  # Just add them together. No memory access is necessary.
  

  Consult your documentation on which registers are guaranteed to survive a syscall (I suspect, all of them besides $v0).

• Nothing to simplify reading and printing.