How to advance to next node in MIPS 32? - mips

I have this function:
makenode:
li $v0, 9 #allocate memory
li $a0, 8 #8 bytes
syscall #$v0 <---- address
move $s1, $v0 #$s1 = address of first node
sw $t4, 0($s1) #$t4 is a string
sw $s2, 4($s1) #copy address of 2nd node
#into first node
Can it be done like this? That 4($s1) points to $2 even though it hasn't been created yet?
and since this is a function (which will be called again), how do I make it that it will work on the next run? Do I increment $s1 so it now represents $s2?

Related

Mips function return a number

I've been trying to learn how to program with Assembly MIPS32 on Mars.
I got a question though, i want to make a function (i don't know if that is how its called) that returns an integer.
Basically i would like to make a method that asks the user to insert a number and add that number with another one that was previously requested
This is what i came up with but i don't know how to return the value to $s2
New:
la $a0, prompt2 # Print string "Input: "
li $v0,4
syscall
li $v0,5 #Read int x
syscall
jr $ra
Add:
j New
add $s1, $s1, $s2 #add Int $s1 and $s2 and save them to $s1
j loop
if anyone has any suggestions or fixes please respond.
Thanks in advance
It is common on MIPS processors to use $v0 for function return values. Of course, you're free to come up with any calling convention you want for your own functions, but unless you have a very good reason not to I'd say stick with $v0 for return values.
Also, to call a function you'd typically use jal, because that's what sets up the $ra register so that you later can return with jr $ra.
If you don't already have it, I suggest that you download MIPS32™ Architecture For Programmers Volume II: The MIPS32™ Instruction Set and read up on how jal and jr work (or any other instruction that you feel that you don't fully understand).
So the code you described in your question could look something like:
New:
la $a0, prompt2 # Print string "Input: "
li $v0,4
syscall
li $v0,5 #Read int x
syscall
# The result of syscall 5 is in $v0
jr $ra
Add:
jal New
add $s1, $s1, $v0 # add the value we just read to $s1
j loop

MIPS, how to invert a series of numbers in a while loop (a rapid method)

I coding an exercise but I have find a little problem... My exercise ask me that convert a character (keyboard input) in:
ASCII code
Binary code
Hex code
without the MIPS syscall. 2 points on 3 are solved for now but in the point 2 I have find my problem... When I insert a letter for example, the binary code is invertited (see below)
Give me a character: a
ASCII: 97
Binary 10000110 (The real is 01100001)
Exist an easy method to invert this number? I leave the code below.
print_binary:
#move the function parameter in t0
move $t0, $a0
#Show a message
li $v0, 4
la $a0, m3
syscall
while:
#check if my counter is 8(conversion on 8 bit)
beq $t2, 8, end_loop
#Divide the number
div $t1, $t0, 2
#move hi in t4 to show in video
mfhi $t4
#Counter
add $t2, $t2, 1
#move for another division
move $t0, $t1
#Show hi for every division
li $v0, 1
move $a0, $t4
syscall
j while
end_loop:
jr $ra
I don't use the array very well, so I would avoid this method if possible...
Thank you

Passing Arguments: MIPS

I'd like to pass a character as an argument to a function in MIPS. Do I do this by storing the character into register $a0, use jal to move to the function, then extract what's in $a0 into a separate register using lw?
If someone could give me an example of passing an argument or two in MIPS, I'd really appreciate it. I've found a lot of info on the MIPS calling conventions, but not any simple and succinct examples.
No need to use lw which is for extracting words from memory. You can simply use $a0 in the sub-routine.
Take a look at this example of a "print-char" function:
.text
main:
#save $ra on stack
addi $sp $sp -4
sw $fp 0($sp)
move $fp $sp
addi $sp $sp -4
sw $ra -4($fp)
#call sub-routine
addi $a0 $zero 'A'
jal printchar
#restore and shrink stack
lw $ra -4($fp)
lw $fp 0($fp)
addi $sp $sp 8
jr $ra
#prints a char and then a new line
printchar:
#call print-char syscall
addi $v0 $zero 11
syscall
addi $a0 $zero 10
syscall
jr $ra
As demonstrated, you the value of the $a0 register is just used in the sub-routine as it returns the value that it was given before the jal.
Also demonstrated is proper expansion and shrinking of the stack as is necessary for calling a sub-routing. As you will notice, the sub-routine does not perform this operation as it does not call a sub-routine and there-fore does not need to save the $ra. Stack manipulations would also be required in the sub-routine if it were to use an $s register as the MIPS calling convention specifies these as callee saved.
a very easy way to go about it would be to load the argument in a temp register and then just use:
move $a0,$t0
this way the argument stored in the temp register will be given as an argument

MIPS, using a while loop to calculatethe sum of odd integers 1-9

The following is my code in MIPS to calculate the sum of odd integers using a while loop.
.data
num: .space 4
.text
.globl main
main:
li $t1, 1
li $t2, 9 # make $t2 9 to break the loop
li $t3, 1
loop:
beq $t3, 11, Exit # check to see if $t3 = 11 if so exit
addi $t3, $t3, 2 # change $t3 to next odd number by adding 2
add $t1, $t1, $t3 # add $t3 to $t1 (1+3,...3+5...etc...)
j loop #jump back to the start of the loop
Exit:
li $v0, 1 # system call code to print an int
lw $a0, num # address of int to print
syscall # print the int
jr $ra #exit
This is my first real experience with MIPS and I'm not sure what is going wrong in this code. I put the print inside the while loop to see if it was ever calculating, but the result is always 1.
So, my result at the end is just 111111.
Edit: Removed the prints inside of the loop with the same result.
And OS is Windows 7 64x
Update: Having num as a variable was over complicating things. Code has been revised as follow and works. Thank you for the help!
enter code here
.data
.text
.globl main
main:
addi $t1, $0, 1
addi $t2, $0, 3
loop: bge $t2, 11, Exit # check to see if $t3 >= 11 if so exit
add $t1, $t1, $t2 # add $t2 to $t1 (1+3,...3+5...etc...)
addi $t2, $t2, 2 # change $t2 to next odd number by adding 2
j loop #jump back to the start of the loop
Exit:
li $v0, 1 # system call code to print an int
move $a0,$t1 # address of int to print
syscall # print the int
jr $ra #exit
la $t1, num
You're clearly getting into trouble here, since you're overwriting your accumulator with the address of num every time you're making the syscall. You're losing the current state of your calculation each time.
You'll need to either save your registers, or simply use different ones. Since I don't know what OS it is that you're using, I don't know if you more generally need to save registers over a syscall, but that could also be a source of errors.
I did similar problems for an architecture class and this seemed to be a recurrent problem among all students. When facing problems similar to this our professor's recommendation was to use a different register to temporarily store the register's address to avoid overwriting other desired values from our most commonly used regs.

Need help with MIPS program

I'm working on a mips program that will run on pcspim and i need a little help. The description of the program is: Write a program that reads a string (from a keyboard), stores it in the memory, and computes and prints the frequency of each character; and then it reverses the string and prints the reversed string.
so far i have is...
.data # Data declaration section
userString: .space 256
Prompt: .asciiz "\nEnter a word: "
newLine: .asciiz "\n"
.text
main: # Start of code section
li $v0, 4
la $a0, Prompt
syscall
li $v0, 8
la $a0, userString
li $a1, 256
syscall
jr $ra
la $a0, userString
move $t0, $a0
lb $t1, 0($t0)
li $v0, 4
move $a0, $t1
syscall # prints first letter of word
Right now i just wanted to see if i've actually stored the input into the userString array. So at the end i tried to print out the first letter. but it doesnt seem to be printing anything.
Any suggestion?
thank.
I've broken your code down into three parts: prompting, input, display. I assume the first two parts were given to you and the third is what you are focusing on right now. I'll explain what the first to parts are doing then explain what the third is doing right now and what you probably want it to do at this point.
.data # Data declaration section
userString: .space 256
Prompt: .asciiz "\nEnter a word: "
newLine: .asciiz "\n"
.text
# Part I
main: # Start of code section
li $v0, 4
la $a0, Prompt
syscall
# Part II
li $v0, 8
la $a0, userString
li $a1, 256
syscall
jr $ra
# Part III
la $a0, userString
move $t0, $a0
lb $t1, 0($t0)
li $v0, 4
move $a0, $t1
syscall # prints first letter of word
Part I
This is pretty straightforward, when we start executing the program counter will be set to the address of the main label. It loads the value 4 into $v0 (that seems to be the print string system call number), and then loads the address of the Prompt string into the first argument register $a0. The last bit just performs the system call that puts the string on the screen.
Part II
Now that the "Enter a word: " string has been printed on screen, we want to actually read what the user is typing. It looks like here we're using system call #8 (probably read string), so we load that value into $v0 in preparation for the syscall. Then, since we want to read the user string into userString, we load the address of that label into $a0 (the first argument for the read string function) then, since we are savvy programmers, we give the upper bound of how many bytes userString can hold (256) in $a1. Then we perform the system call, you type in a string at the keyboard and hit enter and we return to the next line of code.
That line is jr $ra, which means "jump to the location stored in register $ra (return address)". You probably don't want this, because it marks the end of the main function and likely you program exits back to the command line at this point, probably best to remove it.
Part III
Again, you're loading the address of userString into $a0 (and also moving it into $t0 in the next line). Now it gets weird, you load the first byte 0($t0) of userString into $t1. This is a ASCII character value (like 72 or something). Then you start up the system call stuff again with the print string system call (#4) and the argument of $t1. Which you think will print the first letter of the word, which I disagree with. Here's why. If the user types the string, "Hello, world!" this is what it looks like in memory:
userString: H e l l o , w o r l d !
offset: 0 1 2 3 4 5 6 7 8 9 10 11 12
So, loading 0($t0) moves the letter H into register $t1, then when you perform the system call it tries to print the string starting at H to the screen. However, there is not a string starting at letter H, it starts at the address of userString. So if you just move the address of userString into register $a0, then do system call #4 it should print userString to the screen.
#mjshultz
i've changed it up a little. Didnt think i needed 2 loops. Also i've increment it by four because i thought each character is 4 bytes so to go to the next letter i need to increment the offset by four.
.data # Data declaration section
userString: .space 256
Prompt: .asciiz "\nEnter a word: "
newSpace: .asciiz " "
newLine: .asciiz "\n"
.text
main: # Start of code section
li $v0, 4
la $a0, Prompt
syscall
la $a0, userString
li $a1, 256
li $v0, 8
syscall
la $a0, userString
move $s0, $a0
loop:
lb $t1, 0($s0)
li $v0, 1
move $a0, $t1
syscall
li $v0, 4
la $a0, newSpace
syscall
addi $s0, $s0, 4
blt $s0, 256, loop