Trouble with simple Mips function - mips

Attempting to write a Mips function with parameters $a0 - (addr) str, $a1 - (int) n that returns the addr of the string that's the last Length - n of the input str.
However I'm getting unwanted results and fear an infinite loop or logical errror. This is what I've got so far...
.globl suffix
suffix:
li $t0, 0 #sets $t0 to 0
jal length #length is another function that sets $v0 to the length of the
input string, pretty sure it works
li $t1, 0
add $t1, $t1, $v0
sub $t1, $t1, $a1
lb $v0, ($a0)
other_loop:
beq $t1, $t0, other_exit
sb $v0, ($a0)
addi $t0, $t0, 1
addi $a0, $a0, 1
j other_loop
other_exit:
jr $ra
I'm worried I'm incorrectly calling my Length function(which sets $v0 to the length of the input str $va0
-Any feedback would be much appreciated.

Related

A nested function in MIPS

My code has two parts; the first part is making a function that takes in two numbers and return their products. I believe I did this part right.
The second part is where I'm not sure what's the problem is. In this part I need to make a function that find the factorial number, and within this function, I have to use the multiplication function which I made in the first part. Please have a look at my code and tell me what am I doing wrong.
.data
Fa_message: .asciiz "\nFAIL TEST\n"
Pa_message: .asciiz "\nPASS TEST\n"
number1: .word 4
number2: .word 5
KnownAnswers: .word 20
START: .word 16
.text
main:
# taking in the numbers for calculation.
lw $a0, number1 # $a0 =4
lw $a1, number2 # $a1 =5
lw $t0, KnownAnswers # $t0 =20
jal func_multiply # calling the mulyiply function
move $t4,$v0 # store the product for any further comparisons
bne $t0, $t4, FailT # did it fail the test?
beq $t0, $t4, PassT # did it pass the test?
func_multiply: # the mulyiply function
mul $v0, $a0, $a1 # $v0 = number1 * number2
jr $ra
FailT: # print "\nFAIL TEST\n"
li $v0,4
la $a0, Fa_message
syscall
PassT: # print "\nPASS TEST\n"
li $v0,4
la $a0, Pa_message
syscall
###---------------------(PART-2)-------------------
lw $a0, number1 # load the number for the factorial procedure
beq $a0, $zero, factorialDone # (if the number = 0), !0 = 1
mul $a1, $a1, $zero # initializing $a1
mul $a2, $a1, $zero # initializing $a2
addi $a1, $a0, -1 # $a1 = (the entered number - 1)
addi $a2, $a0, 0 # $a2 = the entered number
jal findfactorial
###
#Stop
li $v0, 10
syscall
findfactorial:
jal func_multiply # calling the mulyiply function # mul $v0, $a0, $a1 # $v0 = number1 * number2
move $t4,$v0 # store the product in t4 for any further usage
addi $a0, $a0, -1 # $a1 = $a1-1
addi $a1, $a0, -1
bne $a1, $zero, findfactorial # enter a loop if $a1 does not equal 0
jr $ra
factorialDone:
addi $v0, $v0, 1
syscall
The jal instruction modifies the $ra register. So if function A calls a function B, then A has to save and restore the value that $ra had when entering A so that it can return to the correct place. This is typically done using the stack as temporary storage.
You can push a register on the stack (save it) like this:
addi $sp, $sp, -4
sw $ra, ($sp)
And pop a register off the stack (restore it) like this:
lw $ra, ($sp)
addi $sp, $sp, 4
Then there's your findfactorial loop. You're discarding the result of all the previous iterations, so your result will always be 1*2 == 2. The loop ought to look something like this:
findfactorial:
jal func_multiply
move $a0,$v0
addi $a1, $a1, -1
bne $a1, $zero, findfactorial
This way you first multiply 4 by 3, then 12 by 2, etc.
There are some other isues in your code. For example, if you jump to FailT you don't immediately exit the program after printing the message - you just keep executing the code after PassT.
Also I'm not sure what this is supposed to do:
factorialDone:
addi $v0, $v0, 1
syscall
If you wanted to execute syscall 1 (print_int), then this is incorrect because it doesn't set up $v0 properly (it should be li $v0,1). And if you wanted this to print the result of your factorial computation then that's not going to happen, because you have a jr $ra right before that, so the only time you end up at factorialDone is if number1 contained 0. Also, you'd have to set up $a0 with the value you want to print.

Strncpy in MIPS has a weird behavior

Here's my code for strncpy. In theory it should work, but when I run tests on it it gives out garbage.
Arguments:
$a0 = pointer to destination array
$a1 = source string
$a2 = number of characters to copy
Returns: the destination array
strncpy:
beqz $a2, out
lb $t0, 0($a1) #load byte
beqz $t0 out
subiu $a2, $a2, 1
sb $t0, 0($a0)
addiu $a0, $a0, 1
addiu $a1, $a1, 1
j strncpy
out:
lb $0 0($a0)
move $v0 $a0
jr $ra
coppy original address of destination array ($a0) and load it in "out:" in $a0->$v0. (in your version you would always get the char behind the last insert... in addition to the above mentioned problem in out:)
addi $a3 $a0 0
strncpy:
(...)
out:
move $a0 $a3
move $v0 $a0

MIPS Mult two numbers and result it negative number

I am new for MIPS, there is my code:
.text
main:
add $s0, $zero, 1 # set $s0 to 0 for the initial value for the following loop
Loop: jal srand # Call function srand
addi $s0, $s0, 1 # $s0 = i++
slti $s1, $s0, 6 # $s1 < 6
bne $s1, $zero, Loop # go to loop in i < 6
beq $s1, 6, end_loop # go out the loop when i == 6
end_loop:
li $v0, 10
syscall
srand: # This function will set the numbers for future calculation
lw $t0, a # Load a value to $t0 1103515245
lw $t1, c # Load c value to $t1 12345
lw $t2, m # Load m value to $t2 2147483648
multu $t0,$s0 # result for multiplication (Xn*a) and store the result to $t3
add $t4, $t3, $t1 # result for add multiplication (Xn*a+c) and store the result to $t4
move $a0, $t4 # Debug function
li $v0, 1 # Debug function
syscall # Debug function
la $a0, msg
li $v0, 4
syscall
jr $ra
There is a problem, when the code goes to this "multu $t0,$s0" commend and result will be wrong.
1103515245 * 2 returned a negative number -2087936806
anyone know how to fix it ??
thanks
As chrylis said, it looks like integer overflow. If that's unclear to you, you should read up on two's complement integer representations.
Basically, the value of the highest-order bit is defined to be negative. Suppose you have 32-bit integers. Then 0x800000 will have the value of -2**32, and the other bits have their normal value, so you end up with an expression that looks like -2**32 + [the sum of the other bits], in particular 0xFFFFFFFF has the value of -1, 0xFFFFFE is -2, and so on.

Count number of lowercase letters

So I have created this program to count the number of lowercase letters in a string. The problem I am having is that when I reach the end of the string and the nl character is reached, the line beq $t0, $t1, end is not being executed; it just continues indefinitely. I'm not sure what I'm doing incorrectly.
.data
msg1: .word 0:24
.text
.globl main
main:
addu $s0, $0, $ra
li $v0, 8
la $a0, msg1
la $a1, 100
syscall
loop:
lb $t0, 4($a0)
li $t1, 0x0a
beq $t0, $t1, end
continue:
li $t1, 'a'
blt $t0, $t1, count
li $t1, 'z'
bgt $t0, $t1, count
count:
addi $t4, $t4, 1
j loop
end:
li $v0, 1
addu $a0, $t2, $0
syscall
jr $ra
You compare 4($a0) with 0x0a on each iteration of the loop, but you never change $a0 in the loop, so you are not advancing through your string and never look at the \n at the end of the string.
There also are a few other bugs in your code.
Use this at the start of your loop:
loop:
lb $t0, 0($a0)
addiu $a0, $a0, 1
...

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