MIPS printing strange empty square - mips

I am writing a code to read a NxN array using macros in MIPS, it has no problems reading, but when the function printString is called after a few uses, it starts printing a strange square. I found that commenting the line sw $v0, ($t9) in readArray will make the function print as expected, however the contents that were read won't be saved anymore (and the function will be useless).
My code is the following:
.data
.macro printString(%string)
.data
string: .asciiz %string
.text
li $a0, 0
li $v0, 0
la $a0, string
li $v0, 4
syscall
.end_macro
.macro printInt(%int)
.text
li $a0, 0
li $v0, 0
add $a0, $zero, %int
li $v0, 1
syscall
.end_macro
.macro alloc(%size)
.text
li $a0, 0
li $v0, 0
add $a0, $zero, %size
li $v0, 9
syscall
.end_macro
.macro end()
.text
li $v0, 10
syscall
.end_macro
.macro readInt()
.text
li $v0, 5
syscall
.end_macro
.macro readArray(%array, %size)
.data
array: %array
.text
rAMain:
la $t4, array
add $t5, $zero, %size
li $t6, 0
li $t7, 0
rALoop:
arrayIndex($t4, $t5, $t6, $t7)
move $t9, $v0
printString("Write a value:[")
printInt($t6)
printString("][")
printInt($t7)
printString("]: ")
readInt()
sw $v0, ($t9)
addi $t7, $t7, 1
blt $t7, $t5, rALoop
li $t7, 0
addi $t6, $t6, 1
blt $t6, $t5, rALoop
.end_macro
.macro arrayIndex(%array, %columnNumber, %i, %j)
.data
array: %array
.text
la $t0, array
add $t1, $zero, %i
add $t2, $zero, %j
add $t3, $zero, %columnNumber
mul $v0, $t1, $t3
add $v0, $v0, $t2
sll $v0, $v0, 2
add $v0, $t0, $v0
.end_macro
.text
main:
jal readN
alloc($s1)
move $s0, $v0
readArray($s0, $s1)
end()
readN: ######
printString("Write the value of N: ")
readInt()
move $s1, $v0
jr $ra
Picture of the error:

Usage of:
.data
array: %array
in your readArray macro does not allocate any storage when the corresponding parameter is a register — in fact, array: $t0 doesn't make any sense to me, but somehow MARS allocated zero bytes of storage after the array label without complaining of syntax error.
The data declarations .data ... array: are global declarations, which are performed at compile/assemble time, rather than dynamic allocations performed at runtime.  This is one reason I'm surprised that MARS accepts a register name as an argument here, but none-the-less, it surely isn't doing what you want.

Related

MIPS: how can I apply the integer which users input into arithmetic function?

This is my first time coding PCSPIM. I find that there is a little trouble with my code.
.data
user_input: .asciiz "\n\nEnter an Integer for the value of n: "
result_display: .asciiz "\nThe sum from 0 to n is "
Greeting: .asciiz "\n\nThank you!"
.text
main:
#user input
li $v0, 4
la $a0, user_input
syscall
#allow user input
li $v0, 5
syscall
#store the input value into t8
move $t8, $v0
#calculation
addi $s0, $zero, $t8
I wish to use the integer value ($t8) that users input into the #calculation section, but it ends up with error.
addi $t0, $zero, 0
loop1:
add $t0, $t0, $s0
addi $s0, $s0, -1
bne $s0, $zero, loop1
nop
nop
# Display the result
li $v0, 4
la $a0, result_display
syscall
# Print out the result
li $v0, 1
move $a0, $t0
syscall
# Greets the user
li $v0, 4
la $a0, Greeting
syscall
# Exit the program
li $v0, 10
syscall
Sorry for my broken English.
The error is in the way you are using the "addi" instruction. The instruction requires an immediate (number) value to be passed as the third operand and not an architectural register. If you update the "addi" instruction to "addu" the code should work.

MIPS : Printing n time a number in a file

I'm currently working on a project for school. I'm asked to generate a random list of number in a text file, and first of all I tried to print n times a number in a test file, 31 for example. However, no matter the value of n, I always get my test.txt file with "31 " printed in but only one time. Here's my code (I'm a beginner btw) :
.data
question: .asciiz "\nEnter a number : ?\n"
mynumber: .asciiz "31 "
file: .asciiz "test.txt"
.text
.globl __start
__start:
li $v0, 4
la $a0, question
syscall
li $v0, 5
syscall
move $t0 $v0
li $t1 0
Loop:
blt $t0 $t1 exit
addi $t1 $t1 1
jal open_file
jal fill_file
jal close_file
j Loop
open_file:
li $v0, 13
la $a0, file
li $a1, 1
li $a2, 0
syscall
jr $ra
fill_file:
move $a0, $v0
li $v0, 15
la $a1, mynumber
li $a2, 3
syscall
jr $ra
close_file:
li $v0, 16
syscall
jr $ra
exit:
li $v0, 10
Does somebody have a solution for my problem? Would appreciate ! (Sorry for my english, not a native speaker)

MIPS multiplication using addition

sorry, I try to multiply two integers but it doesn't work. I can't find where the problem is.Maybe because of register' names but I don't know how to correct it. I correct many times but it is not successful. Could anyone give me some points?
.data
prompt1: .asciiz "Please enter the first signed (decimal) integer: "
prompt2: .asciiz "Please enter the second signed (decimal) integer: "
result_msg: .asciiz "The result of these two 16-bit integers\' multiplication is: "
.text
.globl main
main:
li $v0, 4 #print prompt
la $a0, prompt1
syscall
li $v0, 5 #read multiplicand
syscall
move $s0, $v0
li $v0, 4 #print prompt
la $a0, prompt2
syscall
li $v0, 5 #read multiplier
syscall
move $s1, $v0
Mult: ori $t0,$zero,1 #mask
move $s3, $0 #initialize the result register
move $t1, $0
loop: beq $s1, $zero, end #if the multiplier is 0 then finished
and $t1, $t0, $s1 #mask
beq $t1, 1, mult_add
beq $t1, 0, shift
mult_add: addu $s3, $s3, $s0 #add to get product
shift:
sll $s0, $s0, 1 #shift multiplicand left
srl $s1, $s1, 1 #shift multiplier right
j loop
end:
jr $ra
result: #input the print_string
li $v0, 4
la $a0, result_msg
syscall
exit:
li $v0, 1 #input result
move $a0, $s3
syscall
li $v0, 10 #exit
syscall
Inspecting your code I see that you jump to label end when you are done multiplying.
The instruction at that label issues a jr $ra which "returns from a function", but I guess you just want to print the result and exit.
Therefore I'd suggest you remove that instruction so as to print the result and exit and maybe remove label result as it is not used anywhere in your code.

MIPS: Printing Out a Histogram

I'm writing a MIPS program (assembly language) that takes in 10 integers and prints out a histogram represented by asterisks.
E.g.:
User input of 1, 2, 3, 4
Output:
*
**
***
****
I have most of this code written already in MIPS. The problem I am running into is printing out the correct length of asterisks. As of now it is simply printing out the a histogram all of the same length; the FIRST user inputed integer.
# program functionality:
.data
menu: .asciiz "\n1. New Histogram\n2. Print Histogram\n3. Quit\n"
prompt: .asciiz "\nEnter 10 numbers between 0 and 50 (inclusive):\n"
prompt1: .asciiz "\nEnter a valid number:\n"
asterisk: .asciiz "*"
space: .asciiz "\n"
array: .word 0:10
.text
main:
do:
jal print_menu
li $v0, 5
syscall
beq $v0, 1, new
beq $v0, 2, print
beq $v0, 3, quit
j do # end do
new:
jal new_user
j do
print:
jal print_user
j do
j quit
print_menu:
la $a0, menu
li $v0, 4
syscall
jr $ra
new_user:
la $a0, prompt
li $v0, 4
syscall
enter_loop:
la $t0, array
li $t1, 10
enter_loop_2:
la $a0, prompt1
li $v0, 4
syscall
li $v0, 5
syscall
sw $v0, ($t0)
addi $t1, $t1, -1
beqz $t1, end_loop_2
addi $t0, $t0, 4
j enter_loop_2
end_loop_2:
jr $ra
print_user:
la $t0, array
li $t1, 10
pLoop:
la $a0, space
li $v0, 4
syscall
asterisk_fun:
li $v0, 1
lw $a0, ($t0)
syscall
counter:
la $a0, asterisk
li $v0, 4
syscall
addi $a0, $a0, -1
beqz $a0, asterisk_end
j counter
asterisk_end:
jr $ra
addi $t1, $t1, -1
beqz $t1, endpLoop
addi $t0, $t0, 4
j pLoop
endpLoop:
jr $ra
quit:
li $v0, 10
syscall
The problems is that you are overwriting register $a0 in counter with the address of the asterisk, and you also used $a0 to count the number of items in that bucket.
Easy solution is to use other register (e.g. $a1) to count the number of items:
That would be:
#... your code
asterisk_fun:
li $v0, 1
lw $a1, ($t0) # Load number in $a1
move $a0, $a1 # move to $a0 just to print it
syscall
la $a0, asterisk
counter:
li $v0, 4
syscall
addi $a1, $a1, -1 # we use $a1 to keep the counter
beqz $a1, asterisk_end
j counter
asterisk_end:
# ... more of your code

MIPS, Recursion

I trying to write a MIPS program that gets an unsigned integer as argument and returns the sum of all decimal digits in the integer recursively. For example if the argument is 75080 then the sum to be returned is 20 (7+5+0+8+0). Here is my code so far. Any help would be appreciated.
My way of thinking was to divide the number by 10 leaving me with the last integer in the number, add the reminder using mfhi.
.data
prompt: .asciiz "Enter a string of integer: "
output: .asciiz "\nThe total sum is: "
.text
.globl main
main:
la $a0, prompt
li $v0, 4
syscall
li $v0, 5
syscall
move $t2, $v0
la $a0, output
li $v0, 4
syscall
Loop:
div $t2, $t2, 10
mflo, $t1
mfhi, $t3
beqz $t1, Exit
add $t1, $t1, 0
b additive
additive:
add $t0, $t1, $t1
j Loop
Exit:
la $a0, output
li $v0, 4
syscall
la $v0, 10
syscall
What's this supposed to be doing? Adding 0 to the register won't change its value:
add $t1, $t1, 0
After dividing and copying to $t1 and $t3, the quotient is in $t1 and the remainder is in $t3. You're treating it the other way around when you add to the total.
This is actually going to give you $t0 = 2 * $t1: you're adding $t1 to itself and storing the result in $t0.
add $t0, $t1, $t1
You probably actually want:
add $t0, $t0, $t3
You're checking for $t1 == 0 before adding the remainder to the total, so the most significant digit will never get added. You don't really need a subroutine for adding to the total either. You can also use bnez Loop instead of beqz Exit -> b Loop. Lastly, you don't even need $t1, because the quotient is already in $t2.
Get rid of additive and replace Loop with this:
Loop:
div $t2, $t2, 10
mfhi, $t3
add $t0, $t0, $t3
bnez $t2, Loop
Your Exit is weird: you're printing the output string a second time instead of printing the integer.
Change it to this:
Exit:
move $a0, $t0
la $v0, 1
syscall
la $v0, 10
syscall
The approach is quite simple. You need recursive function, I named it SumDigits that will take the last digit and repeat procedure for all digits in the argument. After the recursive call returns you'll add digit to the previous result. The code is commented for easier understanding. The code follows:
.text
#calculates sum of digits recursively
SumDigits:
sub $sp, $sp, 12 #alloocate 12B on stack
sw $ra 0($sp) #save return address
sw $a0, 4($sp) #save argument
beq $a0, $0, exit_sumdigits #when there is no more digits return 0
rem $t0, $a0, 10 #get last digit
sw $t0, 8($sp) #save it on stack
div $a0, $a0, 10 #divide argument by 10
jal SumDigits #repeat procedure
lw $t0, 8($sp) #read digit from stack
add $v0, $v0, $t0 #add digit to previous result
lw $ra, 0($sp) #load return address
addi $sp, $sp, 12 #free stack
jr $ra #return
exit_sumdigits:
li $v0, 0 #there are no more digits, return 0
lw $ra, 0($sp) #load return address
addi $sp, $sp, 12 #free stack
jr $ra #return
main:
li $a0, 75080 #load number in $a0
jal SumDigits #call SumDigits
move $a0, $v0 #set a0 = result of SumDigits
li $v0, 1 #set $v0 for print int system call
syscall
li $v0, 10 #set $v0 for exit system call
syscall
.data