Unaligned memory reference when reading and storing multiple strings - arrays

I'm trying to read multiple strings from a file and store them all together. The way the input is designed, the following preconditions are guaranteed:
The number of string I have to read in is pre-determined
The length of all strings (including a newline and a NUL-byte) is the same
I have the following code, however I've been running into an unaligned memory word reference error the second time the loop is entered.
.data
.align 4
aLine:
.space 128
arr:
.space 1024
.text
//...
//...
build:
addi $sp,$sp,-24
sw $ra, 20($sp)
sw $s4, 16($sp)
sw $s3, 12($sp)
sw $s2, 8($sp)
sw $s1, 4($sp)
sw $s0, 0($sp)
li $t0,0 #incrementing value for index
move $s0,$a0 #$a0 is the number of strings
addi $s1,$a1,2 #$a1 is the length of each string, $s1 makes room for the newline character and the null
la $a2,arr
build_loop:
beq $t0,$s0,build_done
li $v0, READ_STRING #READ_STRING is set to 8 earlier in the program
la $a0,aLine
move $a1,$s1
syscall
mul $t1,$t0,$s1
add $t2,$t1,$a2
sw $v0,($t2)
addi $t0,$t0,1
jal build_loop
build_done:
lw $ra, 20($sp)
lw $s4, 16($sp)
lw $s3, 12($sp)
lw $s2, 8($sp)
lw $s1, 4($sp)
lw $s0, 0($sp)
addi $sp,$sp,24
jr $ra
I got this mostly from an online resource on creating an array of integers, tweaking the values to work with strings of a constant length.
What exactly am I missing here?

You are using jal build_loop, which is effectively recursion, not iteration.Unfortunately, jal build_loop stores the return address to a register, (load-store-architecture, remember?) which happens to be $ra (GPR 31). This clashes badly with the rest of build_loop, which neglects to save $ra on the stack.The result is, of course, that $ra is clobbered.

Related

Mips program infinite loop issue

I am currently trying to figure out why my mars program for creating an array is getting stuck in an infinite loop. I am trying to create an array using different functions and in the create array function the program is supposed to jump into a get number function and return the value to store it into the array but the counter seems to not be decrementing and it just keeps asking for an element for the array. Here is the code:
.data
str5: .asciiz "Please enter a number of elements for the array between 0 and 20: "
str6: .asciiz "Please enter an element: "
array: .word 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
.text
main:
begin:
li $v0, 4
la $a0, str5
syscall
jal readnum
add $a0, $v0, $0
jal verify
add $a0, $v0, $0
jal createarray
la $s1, array
jal printarray
li $v0, 10
syscall
printarray:
createarray:
add $t0, $ra, $0
add $s0, $a0, $0
add $t9, $0, $0
la $s1, array
loop:
beq $s0, 0, done
li $v0, 4
la $a0, str6
syscall
jal readnum
add $t1, $v0, $0
sw $t1, 0($s1)
addi $s1, $s1, 4
addi $s0, $s0, -1
addi $t9, $t9, 1
j loop
done:
add $ra, $t0, $0
jr $ra
verify:
add $s0, $a0, $0
bge $s0, 20, begin
ble $s0, 0, begin
add $v0, $s0, $0
jr $ra
readnum:
li $v0, 5
syscall
jr $ra
EDIT: After re-reading the code, what is causing the error is actually the following code:
jal printarray
You never did the following code in printarray:
jr $ra
So, it would go back down to the other functions.
Use the $sp (the proper way to handle the $ra when calling functions within functions)
The following segment is the cause of the problem:
syscall
jal readnum
add $t1, $v0, $0
Moreso specifically the jal call. Calling a function within a function requires some more steps to be taken place.
First, understand what the jal instruction is doing. The jal instruction, put simply, marks where you currently are and stores it in the $ra register. Then, it will jump to that function.
But there's a problem: you are already wishing for the $ra register to remember where you once were because you, in fact, called createarray!
So, you called createarray, making $ra store where you once were, then within that function you called readnum, making $ra store where you are in createarray. **But now, you have lost where you once were before called createarray. So, it will keep looping back to what $ra used to be, which is inside your createarray function.
Fortunately, the $sp register is just what you need
How to use the $sp register
To store where we are, we push to the stack:
addi $sp, $sp, -4 # by convention, we use negative numbers when pushing
sw $ra, ($sp)
To get where we once were after we called the function (which replaced our $ra in the first place), we pop the stack:
lw $ra, ($sp)
addi $sp, $sp, 4 # by convention, we use positive numbers when popping
So in the grand scope of things, here is what you do:
# push to the stack
addi $sp, $sp, -4
sw $ra, ($sp)
jal theFunctionWeWantToCallInOurFunction
# pop the stack, get back our $ra
lw $ra, ($sp)
addi $sp, $sp, 4
Let's apply this solution to your codebase:
.data
str5: .asciiz "Please enter a number of elements for the array between 0 and 20: "
str6: .asciiz "Please enter an element: "
array: .word 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
.text
main:
begin:
li $v0, 4
la $a0, str5
syscall
jal readnum
add $a0, $v0, $0
jal verify
add $a0, $v0, $0
jal createarray
la $s1, array
jal printarray # comment off this line and you won't have an infinite loop
li $v0, 10
syscall
printarray:
createarray:
add $t0, $ra, $0
add $s0, $a0, $0
add $t9, $0, $0
la $s1, array
loop:
beq $s0, 0, done
li $v0, 4
la $a0, str6
syscall
# store where we wish to come back to so that $ra can be overriden without losing data.
addi $sp, $sp, -4
sw $ra, ($sp)
# call the function. jal will replace our current $ra
jal readnum
# retrieve what our $ra once was
lw $ra, ($sp)
addi $sp, $sp, 4
add $t1, $v0, $0
sw $t1, 0($s1)
addi $s1, $s1, 4
addi $s0, $s0, -1
addi $t9, $t9, 1
j loop
done:
add $ra, $t0, $0
jr $ra
verify:
add $s0, $a0, $0
bge $s0, 20, begin
ble $s0, 0, begin
add $v0, $s0, $0
jr $ra
readnum:
li $v0, 5
syscall
jr $ra

MIPS: How to access an array full of zero with loop

I have an array like this:
r_clues: .word 0 : 512 # array full of zero
I do
la $s0, r_clues
lw $t1, 0($s1)
and I take the address of first 4 bytes and if i want to take, for example, 4th address i will do
lw $t1, 16($s1)
because its 4(address) * 4(bytes)
How i can access to this array with one loop and load word to a register for each 8 bytes ?
Calculate address of elements and load words.
la $s0, r_clues # the address
addiu $s2, $zero, 0 # offset
addiu $s3, $zero, 64 # number of loops
loop_begin:
addu $s1, $s0, $s2 # address = base + offset
lw $t1, 0($s1) # load the array
addiu $s2, $s2, 8 # proceed to the next element
addi $s3, $s3, -1 # substract the counter
bne $s3, $zero, loop_begin # if there are more elements to load, go to loop
nop # prevent next instruction from being executed before exiting the loop

Mips reading integer and store them in array then find the max

again
I'm writing a MIPS program that reads 5 integers and stores them in array. It then creates a new array whose values are the values of the initial array multiplied with their index. After that I need to find the max and min of the second array.
I'm new to MIPS, here is the code I wrote :
.data
Prompt: .asciiz "\n Enter 5 Integers :" #gets number of integers
op: .asciiz "\n Enter Option: \n 1-Find the mult \n 2- Find the max \n 3-Find the min \n 4-Exit"
invalidm: .asciiz "\n Bad Input:"
vec: .space 20
vec2: .space 20
.text
.globl __start
__start:
la $a0,Prompt
li $v0,4
syscall
#Reading integers and store theme in the array
options:
la $a0,op
li $v0,4
syscall
li $v0,5
syscall
blt $v0,1,invalid#
bgt $v0,4,invalid#
beq $v0, 1, multp#
beq $v0, 2, max#
beq $v0, 3, min#
beq $v0, 4, exitpro#
j options
multp:
#multiply every element in the array with its index and store them in the new array vec2
j options
max:
#find the max of the array vec
j options
min:
#find the min of the array vec
j options
invalid:
la $a0,invalidm
li $v0,4
syscall
j options
exitpro:
li $v0,10
syscall
I'd like to make a quick note on coding style: assembly is a bit diabolical. It's very tempting to say "oh, I do these 5 lines a lot, I should just jal here and reuse it". This makes sense at first, but tends to result in confusing spaghetti code that can't go 3 lines without jumping somewhere. As such, my code has a bit of code repetition, but nothing too bad.
I altered your initial conditions a little, the length of the array is hard coded, but it uses a stack-allocated array rather than one that's statically allocated at initialization in the .data header. Instead the array's length is allocated in the data section. This was just personal preference. It has the benefit that the code should work for any length > 0.
.data
PromptHead: .asciiz "\n Enter "
PromptTail: .asciiz " integers\n"
Minstr: .asciiz "Min: "
Maxstr: .asciiz "Max: "
newline: .asciiz "\n"
inputs: .word 5
.text
main:
# Output prompt for input
## "\n Enter "
la $a0, PromptHead
li $v0, 4
syscall
## "5"
lw $a0, inputs
li $v0, 1
syscall
## " integers\n"
la $a0, PromptTail
li $v0, 4
syscall
# Backup sp value before allocating array
move $fp, $sp
# calculate size of vector (numinputs * sizeof(int) = numinputs * 4 = numinputs << 2), store in $s1
lw $s0, inputs
sll $s1, $s0, 2
# dynamically grow stack to include array of ints
add $sp, $sp, $s1
# Loop initialization
# i = 0
move $s1, $zero
# while i < numInputs
ReadInput:
slt $s2, $s1, $s0
beq $s2, $zero, ReadInputDone
# Read integer
li $v0, 5
syscall
# Calculate array offset for this loop
sll $s2, $s1, 2
add $s2, $fp, $s2
# Store array value at calculated address
sw $v0, 0($s2)
# i++
addi $s1, $s1, 1
j ReadInput
ReadInputDone:
# Scale(array, length)
move $a0, $fp
move $a1, $s1
jal Scale
# max = FindMax(array, length)
jal FindMax
# Back up return val, print boilerplate
move $t0, $v0
li $v0, 4
la $a0, Maxstr
syscall
move $a0, $t0
li $v0, 1
syscall
la $a0, newline
li $v0, 4
syscall
# min = FindMin(array,length)
move $a0, $fp
jal FindMin
# Back up return val, print boilerplate
move $t0, $v0
li $v0, 4
la $a0, Minstr
syscall
move $a0, $t0
li $v0, 1
syscall
la $a0, newline
li $v0, 4
syscall
# Exit program
li $v0, 10
syscall
# Scale(array, length)
# Scales each array element by index+1. This does not alter any s or a registers,
# Alters the array in place.
Scale:
# Backup return address and fp on stack.
# Not strictly necessary here, but usually good to do this by habit
# when you're learning
sw $fp, 0($sp)
sw $ra, 4($sp)
addi $fp, $sp, 8
move $sp, $fp
# Load arguments into scratch registers
move $t0, $a0
move $t1, $a1
# Loop initialization as above
# i = 0
move $t2, $zero
# while i < 5
ScaleInput:
slt $t3, $t2, $t1
beq $t3, $zero, ScaleInputDone
# Calculate element offset, store address in $t3
sll $t3, $t2, 2
add $t3, $t0, $t3
# Load array element at $t3, multiply it by current index+1
# Grab it from the multiplication register (assume no overflow)
# then store the result back in the array
lw $t5, 0($t3)
addi $t4, $t2, 1
mult $t5, $t4
mflo $t5
sw $t5, 0($t3)
# i++
addi $t2, $t2, 1
j ScaleInput
ScaleInputDone:
# Unwind stack, restore frame pointer and
# return address. Again, not necessary here, but good
# practice
lw $ra, -4($fp)
move $sp, $fp
lw $fp, -8($fp)
jr $ra
# int Max(array, length); result returned in $v0,
# No s or a registers are altered
FindMax:
# Backup return address and fp on stack.
sw $fp, 0($sp)
sw $ra, 4($sp)
addi $fp, $sp, 8
move $sp, $fp
# Load arguments into scratch registers
move $t0, $a0
move $t1, $a1
# set currMin = array[0]
lw $v0, 0($t0)
# Loop initialization
# i = 1
li $t2, 1
# while i < 5
MaxLoop:
slt $t3, $t2, $t1
beq $t3, $zero, MaxLoopDone
# Calculate element offset, store address in $t3
sll $t3, $t2, 2
add $t3, $t0, $t3
# Load array element at $t3, check if it's the new max
lw $t4, 0($t3)
sgt $t5, $t4, $v0
beq $t5, $zero, notGreater
# If so, set return value to it
move $v0, $t4
notGreater:
# i++
addi $t2, $t2, 1
j MaxLoop
MaxLoopDone:
# Unwind stack
lw $ra, -4($fp)
move $sp, $fp
lw $fp, -8($fp)
jr $ra
# int Min(array, length); result returned in $v0,
# No s or a registers are altered
FindMin:
# Backup return address and fp on stack.
sw $fp, 0($sp)
sw $ra, 4($sp)
addi $fp, $sp, 8
move $sp, $fp
# Load arguments into scratch registers
move $t0, $a0
move $t1, $a1
# set currMin = array[0]
lw $v0, 0($t0)
# Loop initialization
# i = 1
li $t2, 1
# while i < 5
MinLoop:
slt $t3, $t2, $t1
beq $t3, $zero, MinLoopDone
# Calculate element offset, store address in $t3
sll $t3, $t2, 2
add $t3, $t0, $t3
# Load array element at $t3, check if it's the new min
lw $t4, 0($t3)
slt $t5, $t4, $v0
beq $t5, $zero, notLesser
# If so, set return value to it
move $v0, $t4
notLesser:
# i++
addi $t2, $t2, 1
j MinLoop
MinLoopDone:
# Unwind stack
lw $ra, -4($fp)
move $sp, $fp
lw $fp, -8($fp)
jr $ra
As I mention in the code, the monkeying around with storing the $ra and $fp is a bit paranoid. If I was REALLY paranoid I'd store all the s registers as well. However, it's generally good practice and can save you a lot of headaches when you decide to add "function calls" in the middle of a function.
This is basically the approach to assembly where you write the program in C in your head and then translate that rather literally into assembly. Hence why I treat scaling, finding the max, etc as "functions".
A couple assemblyish notes: I repeatedly use sll $register, $register, 2 instead of multiplying by 4 (word size on MIPS32). This is because doing this is fewer instructions due to not having to cal li followed by mult followed by mflo. You can do it that way, and I used to, but once you get used to using sll and other bit fiddling operations a lot it's just cleaner to use sll as well as easier to read.
I do use a few pseudoinstructions, it's not a big deal, most modern MIPS assemblers and simulators that I'm aware of support them (including SPIM). I dance dangerously with the line after branch instructions (which are theoretically always executed regardless of the branch result), but in this case it wouldn't usually matter and most simulators and assemblers usually inject a no-op for you when it does anyway.
I know assembly can be a bit of a pain to read, so feel free to ask if you have any questions.

Strange bug in MIPS code

I am making a program for a user to input a range of numbers, and calculate the min, max and median. Right now, I am only trying to collect the numbers and echo them back to make sure I am getting them. Here is the problem:
I input numbers like this:
1
2
3
4
5
And when the array is printed out I get:
15345
It does not matter what numbers are used, the 2nd element in the array is always replace by the last element.
Here is my mips code, I know it is kind of long, but it is the shortest workable example I can make.
Note: you must enter 9999 for the program to exit the loop.
.data
welcomeString: .asciiz "Please input one number at a time, and then press enter.\n"
intArray: .word 4000
size: .word 0
.text
main:
li $v0, 4
la $a0, welcomeString
syscall
la $a0, intArray
jal gather_numbers
la $a0, intArray
jal print_array
####################################################################################
gather_numbers:
addi $sp, $sp, -12
sw $a0, 0($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $t1, 12($sp)
move $s0, $a0 #the address of the array
lw $s1, size # load the size
li $t1, 0 # so it enters the loop
start_gather_numbers: beq $t1, 9999, exit_gather_numbers
li $v0, 5 # read the integer
syscall
sw $v0, 0($s0)
move $t1, $v0 # put the value into t1 to be tested
addi $s0, $s0, 4 #increment the address
addi $s1, $s1, 1 # increment the size
j start_gather_numbers
exit_gather_numbers: addi $s1, $s1, -1 # fix the size
sw $s1, size # store the size
lw $a0, 0($sp) # pop the stack
lw $s0, 4($sp)
lw $s1, 8($sp)
lw $t1, 12($sp)
addi $sp, $sp, 12
####################################################################################
####################################################################################
print_array:
addi $sp, $sp, -16
sw $a0, 0($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $t0, 12($sp)
sw $t1, 16($sp)
move $s0, $a0 # the address of the array
lw $s1, size # load the size of the array
li $t0, 0 # i = 0
start_print_array: bge $t0, $s1, exit_print_array
lw $t1, 0($s0) # load the int to print
li $v0, 1 # print the integer
move $a0, $t1
syscall
addi $s0, $s0, 4
addi $t0, $t0, 1
j start_print_array
exit_print_array: lw $a0, 0($sp)
lw $s0, 4($sp)
lw $s1, 8($sp)
lw $t0, 12($sp)
lw $t1, 16($sp)
addi $sp, $sp, 16
There are a few things going wrong here.
Firstly all of your functions are missing the terminating jr $ra.
Also, your stack manipulation is wrong. You are consistently allocating 4 less bytes than you use. If you want to put 5 words on the stack you should expand the stack by 20 not 16.
Most importantly though here is your intArray directive. You have used .word 4000 I'm guessing to allocate an array of ints, but rather you have allocated space for 1 word with the value 4000.
To instead allocate an array of 1000 ints you could use .space 4000 or equally .word 0:1000.
When I made these changes your program began to function as desired.

length of array in mips

How to find length of an array in mips,spim?
I wrote this for practice. I tested it, and it works well. You probably already figured this out, but if not, there it is.
.data
array1: .word 1,2,3,4,5,6,7,8,9
.text
main:
la $a0,array1
jal lenArray
move $a0,$v0
syscall $print_int
exit:
li $a0,10
syscall
lenArray: #Fn returns the number of elements in an array
addi $sp,$sp,-8
sw $ra,0($sp)
sw $a0,4($sp)
li $t1,0
laWhile:
lw $t2,0($a0)
beq $t2,$0,endLaWh
addi $t1,$t1,1
addi $a0,$a0,4
j laWhile
endLaWh:
move $v0,$t1
lw $ra,0($sp)
lw $a0,4($sp)
addi $sp,$sp,8
jr $ra

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