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ANTONII [103]
3 years ago
10

Write a MIPS assembly language program that adds the following two integers and displays the sum and the difference. In the .dat

a section, define two variables num1 and num2 both words. Initialize num1 to 78591 10 and num2 to D5A 16 (use 0xD5A to initialize, Note that D5A is a hexadecimal number). Your main procedure/function should load the values of num1 and num2 into two temporary registers, and display them on the console window. Then add the values together, and use syscall for the print_int system call function to display the sum on the console window. Also compute the difference of two numbers and display it on the console window. (Reference: see Assignment 1) To print an integer on the console window, you must put the integer to be printed in the $a0 register, and the value 1 in the $v0 register. Then perform a syscall operation. This makes a call to the SPIM operating system which will display the integer in $a0 on the console window. Name your source code file assignment2.s.
Computers and Technology
1 answer:
lara [203]3 years ago
4 0
You can modify this code

.data
msg1: .asciiz "Enter the first number: "
msg2: .asciiz "\nEnter the second number: "
result: .asciiz "\nThe result of addition is: "

.text
li $v0,4
la $a0,msg1
syscall

li $v0,5
syscall
move $t1,$v0

li $v0,4
la $a0,msg2
syscall

li $v0,5
syscall
move $t2,$v0

Add $t3,$t1,$t2

li $v0,4
la $a0,msg3
syscall

li $v0,1
move $a0,$t3
syscall

li $v0,10
syscall
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5 0
2 years ago
True or False? Any edition or version of Windows can join a domain.
Veseljchak [2.6K]

Answer: TRUE! 100%

If I was helpful i would to be rated brainliest please thank you!

Explanation:

8 0
2 years ago
Given six memory partitions of 300 KB, 600 KB, 350 KB, 200 KB, 750 KB, and 125 KB (in order), how would the first-fit, best-fit,
Inga [223]

Answer:

In terms of efficient use of memory: Best-fit is the best (it still have a free memory space of 777KB and all process is completely assigned) followed by First-fit (which have free space of 777KB but available in smaller partition) and then worst-fit (which have free space of 1152KB but a process cannot be assigned). See the detail in the explanation section.

Explanation:

We have six free memory partition: 300KB (F1), 600KB (F2), 350KB (F3), 200KB (F4), 750KB (F5) and 125KB (F6) (in order).

Using First-fit

First-fit means you assign the first available memory that can fit a process to it.

  • 115KB will fit into the first partition. So, F1 will have a remaining free space of 185KB (300 - 115).
  • 500KB will fit into the second partition. So, F2 will have a remaining free space of  100KB (600 - 500)
  • 358KB will fit into the fifth partition. So, F5 will have a remaining free space of 392KB (750 - 358)
  • 200KB will fit into the third partition. So, F3 will have a remaining free space of 150KB (350 -200)
  • 375KB will fit into the remaining partition of F5. So, F5 will a remaining free space of 17KB (392 - 375)

Using Best-fit

Best-fit means you assign the best memory available that can fit a process to the process.

  • 115KB will best fit into the last partition (F6). So, F6 will now have a free remaining space of 10KB (125 - 115)
  • 500KB will best fit into second partition. So, F2 will now have a free remaining space of 100KB (600 - 500)
  • 358KB will best fit into the fifth partition. So, F5 will now have a free remaining space of 392KB (750 - 358)
  • 200KB will best fit into the fourth partition and it will occupy the entire space with no remaining space (200 - 200 = 0)
  • 375KB will best fit into the remaining space of the fifth partition. So, F5 will now have a free space of 17KB (392 - 375)

Using Worst-fit

Worst-fit means that you assign the largest available memory space to a process.

  • 115KB will be fitted into the fifth partition. So, F5 will now have a free remaining space of 635KB (750 - 115)
  • 500KB will be fitted also into the remaining space of the fifth partition. So, F5 will now have a free remaining space of 135KB (635 - 500)
  • 358KB will be fitted into the second partition. So, F2 will now have a free remaining space of 242KB (600 - 358)
  • 200KB will be fitted into the third partition. So, F3 will now have a free remaining space of 150KB (350 - 200)
  • 375KB will not be assigned to any available memory space because none of the available space can contain the 375KB process.
8 0
3 years ago
Write a static method middleValue that takes three int parameters, and returns a int . It should return the middle value of the
anygoal [31]

We use if-else structure to check the each possible scenario and return the median accordingly in the middleValue() method. The main is also provided so that you can test the method.

Comments are used to explain the each line.

You may see the output in the attachment.

public class Main

{

public static void main(String[] args) {

   

    //call the method for different scenarios

    System.out.println(middleValue(1, 2, 3));

    System.out.println(middleValue(1, 3, 2));

    System.out.println(middleValue(2, 1, 3));

    System.out.println(middleValue(2, 3, 1));

    System.out.println(middleValue(3, 1, 2));

    System.out.println(middleValue(3, 2, 1));

 

}

       //method that takes three int and returns an int

public static int middleValue(int n1, int n2, int n3) {

    //set the median as n1

    int median = n1;

   

    //check the situation where the n1 is the highest

    //if n2 is greater than n2 -> n1 > n2 > n3

    //if not -> n1 > n3 > n2

    if(n1 > n2 && n1 > n3){

        if(n2 > n3)

            median = n2;

        else

            median = n3;

    }

   

    //check the situation where the n2 is the highest

    //if n3 is greater than n1 -> n2 > n3 > n1

    //if not -> n2 > n1 > n3

    //note that we set the median as n1 by default, that is why there is no else part

    else if(n2 > n1 && n2 > n3){

        if(n3 > n1)

            median = n3;

    }

   

    //otherwise, n3 is the highest

    //if n2 is greater than n1 -> n3 > n2 > n1

    //if not -> n3 > n1 > n2

    //note that we set the median as n1 by default, that is why there is no else part

    else{

        if(n2 > n1)

            median = n2;

    }

   

    return median;

}

}

You may see another if-else question at:

brainly.com/question/13428325

6 0
3 years ago
What two pieces of information would you need in order to measure the masses of stars in an eclipsing binary system?
Nostrana [21]

Answer:

the time between eclipses and the average distance between the stars.

Explanation:

8 0
3 years ago
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