First i know ive asked tons of questions but ive come a long way since i last asked.
Im using mars simulator for some side project learning.
Ive always been interested in assembly and mips is my choice.
Anyways my program seems to crash after i try and move my values and add them up and i cant figure out why.
Thanks for looking and tips will be great.
Have a good day
.data
msg0: .asciiz "enter a number: \n"
msg1: .asciiz "enter another number: \n"
result: .asciiz "result is: \n"
.text
li $v0, 4
la $a0, msg0 #load first message
syscall
li $v0 5
move $t0, $v0 #user input
syscall #store number
li $v0, 4
la $a0, msg1 #second msg
syscall
li $v0, 5
move $t1, $v0 #second input and store number
syscall
li $v0, 4
la $a0, result #rsult message
syscall
add $t3,$t1,$t0 #and my problem is here for some reason?
syscall
You need to add a label that are function names. In the code I have provided below, the function name is main. Also, the last system call serves no purpose now. What you need to do is indicate that your code has finished before making the last system call by li $v0,10.
.data
msg0: .asciiz "enter a number: \n"
msg1: .asciiz "enter another number: \n"
result: .asciiz "result is: \n"
.text
.globl main
main:
li $v0, 4
la $a0, msg0 #load first message
syscall
li $v0 5
syscall
move $t0, $v0 #first input and store number
li $v0, 4
la $a0, msg1 #second msg
syscall
li $v0, 5
syscall
move $t1, $v0 #second input and store number
li $v0, 4
la $a0, result #result message
syscall
add $t3,$t1,$t0
move $a0, $t3
li $v0, 1
syscall #print answer
li $v0,10
syscall
Related
I'm learning MI trying to write a factorial program in MIPS Assembly code for which
3! = 6 (my output shows it 36)(input next to output)
4! = 24(my output shows it as 424)
Here is my code. What should I do to get rid of printing the input?
.data
num: .asciiz "\nPlease enter a number: "
num2: .asciiz"\nPlease give your second number : "
respon: .asciiz "\nThe factorial of the entered number is: "
nl: .asciiz"\n"
.text
fact:
beqz $a0,return1
li $v0, 1
li $t0, 1
fact_loop:
bgt $t0, $a0, end_fact_loop
mul $v0, $v0, $t0
addi $t0, $t0, 1
j fact_loop
end_fact_loop:
jr $ra
return1:
li $v0, 1
jr $ra
main:
li $v0, 4
la $a0, num
syscall
li $v0, 5
syscall
move $t0, $v0
li $v0, 4
la $a0, respon
syscall
li $v0, 1
move $a0, $t0
syscall
jal fact
move $t0, $v0
li $v0, 1
move $a0, $t0
syscall
li $v0, 4
la $a0, nl
syscall
################################
li $v0, 4
la $a0, num2
syscall
li $v0, 5
syscall
move $t0, $v0
li $v0, 4
la $a0, respon
syscall
li $v0, 1
move $a0, $t0
syscall
jal fact
move $t0,$v0
li $v0, 1
move $a0, $t0
syscall
li $v0, 10
syscall
To get rid of printing the input you have to remove the syscall you are issuing to print them.
That is, before jal-ing fact (both times) you are issuing these instructions:
li $v0, 1
move $a0, $t0
syscall
Just, remove the first and third instruction and keep the move as you use it on your fact routine:
move $a0, $t0
A few suggestions:
Add a comment to each functional chunk of code.
Use a single-step to debug so you can see what's happening where.
Simplify the code to a minimal failing example. If you have two chunks of code both exhibiting the same problem, remove one of them and focus on the first without distractions and noise from the second.
Prune unnecessary code. For example, the return1: block and beqz $a0,return1 are superfluous because the main factorial loop will exhibit the same behavior automatically -- you don't need to handle $a0 == 0 specially.
Having followed these tips, I wound up with the following program:
.data
num: .asciiz "\nPlease enter a number: "
num2: .asciiz "\nPlease give your second number : "
respon: .asciiz "\nThe factorial of the entered number is: "
nl: .asciiz "\n"
.text
fact:
li $v0, 1
li $t0, 1
fact_loop:
bgt $t0, $a0, end_fact_loop
mul $v0, $v0, $t0
addi $t0, $t0, 1
j fact_loop
end_fact_loop:
jr $ra
main:
# print prompt "Please enter a number: "
li $v0, 4
la $a0, num
syscall
# get user input for first fact call
li $v0, 5
syscall
move $t0, $v0
# call fact(n)
move $a0, $t0
jal fact
move $t0, $v0
# print "factorial is..."
li $v0, 4
la $a0, respon
syscall
# print the result of fact(n)
li $v0, 1
move $a0, $t0
syscall
# print newline
li $v0, 4
la $a0, nl
syscall
# exit program
li $v0, 10
syscall
The problem of double-printing is caused by this code:
li $v0, 1
move $a0, $t0
syscall
The move $a0, $t0 is necessary to set up the fact call, but we don't want li $v0, 1 and syscall, which prints the argument without a newline before the result of fact, causing your unwanted output.
A good technique to prevent this in the future is ensuring your argument set-up for the function call to fact happens right before the jal.
I'll leave it as an exercise to adapt this to your second chunk of code, which has the same extra syscall to service 1 (print integer).
I created an array in MIPS and this language is very new to me. Im suppose to show the number at position one instead i get position 2 which is 3
#create a simple idea of an array
.data
num1: .word 1, 2, 3 #allows input for slot one in array
label1: .asciiz "Input the first number:"
label2: .asciiz "Input the second number:"
label3: .asciiz "Input the third number:"
label4: .asciiz "The number in slot 1 of the array is:"
.text
main:
#intiate the first label
li $v0, 4
la $a0, label1
syscall
#input first number into the array
li $v0, 5
syscall
sw $v0, num1
#initiate the second label
li $v0,4
la $a0, label2
syscall
#input second number into the array
li $v0, 5
syscall
sw $v0, num1
#initiate the third label
li $v0, 4
la $a0, label3
syscall
#input third number into the array
li $v0, 5
syscall
sw $v0, num1
#intiate the fourth label
li $v0, 4
la $a0, label4
syscall
#output position 1
li $v0, 1
lw $a0, num1
syscall
#end program
li $v0,10
syscall
I'm new to MIPS and I've written a program to add two user-selected numbers but it's not working. It says Instruction references undefined symbol at 0x00400014. Can someone please go through my code and explain what's wrong? Thank you.
.data
msg1: .asciiz "Enter addend 1: "
msg2: .asciiz "Enter addend 2: "
msg3: .asciiz "Sum = "
.text
li $v0, 4
la $a0, msg1
syscall
li $v0, 5
syscall
move $t0, $v0
li $v0, 4
la $a0, msg2
syscall
li $v0, 5
syscall
move $t1, $v0
add $t2, $t0, $t1
li $v0, 4
la $a0, msg3
syscall
li $v0, 1
move $a0, $t2
syscall
li $v0, 10
syscall
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)
What does the following QtSPIM/MIPS code do. Describe by referring to the functions of various blocks of the code (Block1, Block2, …).
Answer the questions in front of certain instructions.
Block1:
.text
main:
li $s4, 0
la $a0, input
li $v0, 4
syscall
li $v0, 5
syscall
move $t0, $v0 #Why do we need this?
Block2:
la $a0, output
li $v0, 4
syscall
move $a0, $t0
li $v0, 1
syscall
la $a0, newline # How will the output change if newline is replaced by newspace defined as “ “?
li $v0, 4
syscall
Block3:
Do:
move $a0, $s4
li $v0, 1
syscall
la $a0, newline
li $v0, 4
syscall
slt $s1, $s4, $t0
addi $s4, $s4, 1
bgt $s1, $zero, Do # Why can’t we use “beq $s1, 1, Do “?
Block4:
Exit:
li $v0, 10
syscall
.data
input: .asciiz "Input a number: "
output: .asciiz "Let me count till "
newline: .asciiz "\n "
So I need help with identifying what the code does... I have the concept I just want o make sure Im properly figuring this out... So I have this so far:
move $t0, $v0 #Why do we need this? --- to store number in t0
la $a0, newline # How will the output change if newline is replaced by newspace defined as “ “?
--- load address of string to be printed into $a0 and adds the space if replaced by newspace
the last one I cant quite figure out...