Final answer:
MIPS Assembly Program:
.data
prompt: .asciiz
result: .asciiz "\\Result: "
tabSpace: .asciiz "\t"
.text
.globl main
main:
li $v0, 4
la $a0, prompt
syscall
li $v0, 5
syscall
blt $v0, 10, main
bgt $v0, 40, main
move $t0, $v0
sll $t1, $t0, 3
sll $t2, $t0, 4
add $t1, $t1, $t2
sll $t0, $t0, 1
add $t1, $t1, $t0
li $v0, 4
la $a0, result
syscall
li $v0, 1
move $a0, $t1
syscall
li $v0, 10
syscall
Step-by-step explanation:
To write a MIPS Assembly program that prompts the user to input an integer between 10 and 40 and then multiplies this number by 24 without using the multiply instruction, we can achieve this by using a series of addition operations.
Here is a MIPS Assembly code:
.data
prompt: .asciiz "Please enter an integer between 10 and 40: "
result: .asciiz "\\Result: "
tabSpace: .asciiz "\t"
.text
.globl main
main:
li $v0, 4 # syscall for print string
la $a0, prompt # load address of prompt
syscall
li $v0, 5 # syscall for read integer
syscall
blt $v0, 10, main # check if input is less than 10
bgt $v0, 40, main # check if input is greater than 40
move $t0, $v0 # move input number to $t0
sll $t1, $t0, 3 # multiply input by 8 using shift left logical
sll $t2, $t0, 4 # multiply input by 16 using shift left logical
add $t1, $t1, $t2 # add the results of both shifts
sll $t0, $t0, 1 # multiply input by 2 using shift left logical
add $t1, $t1, $t0 # final result is input multiplied by 24
li $v0, 4
la $a0, result
syscall
li $v0, 1 # syscall for print integer
move $a0, $t1 # move result to $a0
syscall
li $v0, 10 # syscall for exit
syscall
This code performs the required functionality. User input is obtained and checked to be within range. If it is not, the user is prompted again.
The multiplication by 24 is done by bit shifting the input to create multiples that are then added together, which provides the same result as multiplying by 24.