An array of integers can be assigned to a memory address in the .data section of
ID: 3743776 • Letter: A
Question
An array of integers can be assigned to a memory address in the .data section of a MIPS assembly language program as show below. Here the length of the array is stored first, and then the elements of the array numbers next. You are given the following C program that will ask a user to enter two integers and they should be stored into variables named, index and num2. Using the first integer, index, it should examine each integer in the array up to the index to see if it is divisible by the second entered integer, num2. (You can divide each integer by the entered integer to see if its remainder is zero) and if it is, then the number should be multiplied by the second number, num2. If the first entered integer, index is less than 0, then no integer in the array should be changed. If the first entered integer, index is more than 10, then all integers in the array should be examined to see if they should be changed (by multiplying by the second number). After modifying the array, each integer in the array should be printed. Implement a MIPS assembly language program to perform the functionality of the following C program and print the updated array content, by listing each integer in it. For instance, if a user enters 5, then enters 3, then the output will be the following: -81 9 46 -7 11 72 -5 14 -18 12 35 i.e., the numbers that are located until the index 5: (-27, 3, 46, -7, 11, 24) are examined to see if each of them is divisible by the second entered number, 3. In this case, -27, 3, 24 that are divisible by 3, thus they are multiplied by the second entered number 3, then we get (-81, 9, 46, -7, 11, 72, -5, 14, -18, 12, 35) If your program causes an infinite loop, press Control and 'C' keys at the same time to stop it. Name your source code file assignment5.s.
.data numbers_len: .word 11
numbers: .word -27, 3, 46, -7, 11, 24, -5, 14, -18, 12, 35
The following shows how it looks like in a C program:
void main()
{
int numbers[11] = {-27, 3, 46, -7, 11, 24, -5, 14, -18, 12, 35};
int index, num2;
int j;
printf("Enter an integer: "); //read an integer from a user input and store it in index
scanf("%d", &index);
printf("Enter another integer: "); //read an integer from a user input and store it in num2
scanf("%d", &num2);
for (j = 0; j < 11 && j <= index; j = j+1)
{
if (numbers[j] % num2 == 0) { numbers[j] = numbers[j]*num2; } }
printf("Result Array Content: ");
for (j = 0; j < 11; j = j+1)
{ printf("%d ", numbers[j]); }
return; }
The following is a sample output (user input is in bold): Enter an integer: 5 Enter another integer: 3 Result Array Content: -81 9 46 -7 11 72 -5 14 -18 12 35 -------------------------------------------------- The following is another sample output: -------------------------------------------------- Enter an integer: 20 Enter another integer: -7 Result Array Content: -27 3 46 49 11 24 -5 -98 -18 12 -245 --------------------------------------------------
Explanation / Answer
#data section declares the variables
.data
numbers_len: .word 11
numbers: .word -27,3,46,-7,11,24,-5,14,-18,12,35
#declare strings
msg1: .asciiz "Enter an integer: "
newLine: .asciiz " "
#Text section
.text
main:
#display prompt
la $a0, msg1
li $v0, 4
syscall
#Read integer
li $v0, 5
syscall
move $t0, $v0
#load values
lw $a0, numbers_len
la $a1, numbers
li $t1, 0
li $t2, 0
#loop to arrange array
loop:
#Check condition
beq $t1, $a0, PrintLoop
#Load current array value
lw $t3, 0($a1)
#Check array value < user value
#then go to swap
blt $t3, $t0, swap
#if not, move to next array element
addi $t1, $t1, 1
addi $a1, $a1, 4
#jump to loop
j loop
#swap section
swap:
#Get the index
add $t5, $t2, $t2
add $t5, $t5, $t5
#Load array base address
la $t4, numbers
#Find the correct position
add $t6, $t4, $t5
#Take the element from the position
lw $t7, 0($t6)
#Insert the element present in $t3
sw $t3, 0($t6)
#Set the values
li $t5, 0
li $t6, 0
#Find the index
add $t5, $t1, $t1
add $t5, $t5, $t5
#Find the address
add $t6, $t4, $t5
#place element
sw $t7, 0($t6)
#Increase $t1
addi $t1, $t1,1
#Increase $t2
addi $t2, $t2, 1
#Move to next array element
addi $a1, $a1, 4
#jump to loop
j loop
#prints the array
PrintLoop:
#Load the base array address
la $a1, numbers
#Load number of elements
lw $t0, numbers_len
#Set $t1 to 0
li $t1, 0
#loop
loop2:
#Check condition
beq $t1, $t0, ExitPro
#Load current array element
lw $a0, 0($a1)
#Print array element
li $v0,1
syscall
#Print newline
la $a0, newLine
li $v0, 4
syscall
#move to next array element
addi $a1, $a1, 4
addi $t1, $t1, 1
#jump to loop2
j loop2
#End program
ExitPro:
li $v0, 10
syscall
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