Question

So I am busy writing a MIPS program that will take an input string, and then print all possible UNIQUE permutations of that string. (AKA if the word is LoOp, LoOp and LOop are the same).

In order to do this, I know I need to NOT have a newline character on the end of my input string, but I don’t know to make sure it’s not added. Here is what I have so far:

.data


newLine:
    .asciiz "\n"
promptUser:
    .asciiz "Enter a 20 letter or less word:\n"
word:
    .space 21

.text

main:

    la $a0, promptUser
    li $v0, 4       # Ask User for Input
    syscall

    la $a0, word
    li $a1,21       # Max number of characters 20
    li $v0,8
    syscall         # Prompting User

    la $a0,newLine      # Newline   
    li $v0, 4
    syscall

    la $a0, word        # Printing Word
    li $v0, 4
    syscall

The only time a ‘\n’ isn’t included is when the number of letters entered is exactly 20. Any suggestions??

FIX:

This works:

    li $s0,0        # Set index to 0
remove:
    lb $a3,word($s0)    # Load character at index
    addi $s0,$s0,1      # Increment index
    bnez $a3,remove     # Loop until the end of string is reached
    beq $a1,$s0,skip    # Do not remove \n when string = maxlength
    subiu $s0,$s0,2     # If above not true, Backtrack index to '\n'
    sb $0, word($s0)    # Add the terminating character in its place
skip:

Answer 1

You can parse the string upon returning from syscall 8 to remove the character:

# your code to prompt the user        

    addu  $a2, $zero, $zero     # i = 0
loop:
    lbu $a3, word($a2)          # word[i]
    addiu $a2, $a2, 1
    bnez $a3, loop       # Search the NUL char code

    beq $a1, $a2, skip   # Check whether the buffer was fully loaded
                         # Otherwise 'remove' the last character
    sb $0, word-2($a2)   # and put a terminating NUL instead
skip:

# your code continues here

Also note that you didn’t reserve enough space for the word. You should reserve 21 bytes with

word: .space(21)