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.
Related
.data
userInput: .asciiz "give an integer to check if prime: "
prime: .asciiz "Prime"
notPrime: .asciiz "No prime"
.text
main:
li $v0, 4
la $a0, userInput
syscall
li $v0, 5
syscall
move $t0, $v0
addi $t1, $t1, 1
addi $t2, $t2, 2
ble $t0, $t1, no
beq $t0, $t2, yes
div $t0, $t2
mflo $s0
mfhi $s1
bgt $s1, $zero, yes
yes:
li $v0, 4
la $a0, prime
syscall
j exit
no:
li $v0, 4
la $a0, notPrime
syscall
j exit
exit:
li $v0, 10
syscall
For some reason, the program always jumps to the "yes" even when there's a remainder (mfhi) and I can't seem to find the problem on why it does it. When I give input 4 it should jump to the "no" label since the remainder is 0 so it isn't greater than the constant 0.
What are you trying to accomplish with this part of your code:
bgt $s1, $zero, yes
yes:
Play the computer and decide where to go if the gt is true vs. if the gt is false.
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.
I'm trying to practice my coding skill in MIPS (this is my first time ever learning an assembly language). I wrote this code below to sum up two user's inputs, and it is correct. However, the code is quite long..so, is there any way to optimize this code so it will be shorter? i need some suggestions. Thanks
.data
n1: .asciiz "enter your first number: "
n2: .asciiz "enter your second number: "
result: .asciiz "result is "
.text
#getting first input.
la $a0, n1
li $v0, 4
syscall
li $v0, 5
syscall
move $t0, $v0
#getting second input.
la $a0, n2
li $v0, 4
syscall
li $v0, 5
syscall
move $t1, $v0
#calculate and print out the result.
la $a0, result
li $v0, 4
syscall
add $t3, $t0, $t1
move $a0, $t3
li $v0, 1
syscall
#end program.
li $v0, 10
syscall
I also have wrote a program to calculate a factorial number. There are better ways to do this?
.data
str: .asciiz "Enter a number: "
result: .asciiz "The result is: "
.text
la $a0, str
li $v0, 4
syscall
li $v0, 5
syscall
move $s0, $v0 # move N into s0
li $s2, 1 # c = 1
li $s1, 1 # fact = 1
LOOP:
blt $s0, $s2, PRINT # if (n < c ) print the result
mul $s1, $s1, $s2 # fact = fact * c
add $s2, $s2, 1 # c = c + 1
j LOOP
PRINT:
la $a0, result
li $v0, 4
syscall
add $a0, $s1, $zero
li $v0, 1
syscall
li $v0, 10
syscall
The programs look ok, except for the use of the temporary registers $t0,...,$t9. These registers are not guaranteed to be preserved when another function is called, or when a syscall is issued. The $s0,...,$s7 registers are preserved across calls.
You need to replace: move $t0, $v0 with move $s0, $v0; move $t1, $v0 with move $s1, $v0 and add $t3, $t0, $t1 with add $s3, $s0, $s1.
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...
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