What will be the final value of EAX in this example? mov eax, 0 mov ecx, 10 L1: mov eax, 3 mov ecx, 5 L2: add eax, 5 loop L2 loop L1
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What will be the final value of EAX in this example?
mov eax, 0
mov ecx, 10
L1: mov eax, 3
mov ecx, 5
L2: add eax, 5
loop L2
loop L1
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- What will EAX contain following execution of the following code? mov eax, 0 ; sum := 0 mov ecx, 4 ; count := 4 for4: add eax, ecx ; add count to sum loop for4Q1. Srikanth is a school student. He studies in class 8th. One day his Math’s teacher came to the class and told them that he is going to start a new chapter “Measurement”. In order to give them a demo of that chapter, he told the students to find the volume of the class room using Volume( ). Input Constraint 15<=Length<=25 10<=Breadth<=20 20<=Height<=30 its the question to be solved in Dev C++Modify pipe4.cpp so that it accepts a message from the keyboard and sends it to pipe5. //pipe4.cpp (data producer) #include <unistd.h> #include <stdlib.h> #include <stdio.h> #include <string.h> int main() { int data_processed; int file_pipes[2]; const char some_data[] = "123"; char buffer[BUFSIZ + 1]; pid_t fork_result; memset(buffer, '\0', sizeof(buffer)); if (pipe(file_pipes) == 0) { //creates pipe fork_result = fork(); if (fork_result == (pid_t)-1) { //fork fails fprintf(stderr, "Fork failure"); exit(EXIT_FAILURE); } if (fork_result == 0) { //child sprintf(buffer, "%d", file_pipes[0]); (void)execl("pipe5", "pipe5", buffer, (char *)0); exit(EXIT_FAILURE); } else { //parent data_processed = write(file_pipes[1], some_data, strlen(some_data));…
- C++ Please explain the code below. It doesn't have to be long, as long as you explain what the important parts of the code do. You can also explain it line by line for best ratings. Thank you so much! The code is already correct, only explanation needed void addTail(int num) { node* newest = create_node(num); if (tail != NULL) { tail->next = newest; } tail = newest; if (head == NULL) { head = newest; } index++; } int add(int num) { addTail(num); return index; } int get(int pos) { node* currnode = head; int count = 0; while (currnode != NULL){ count++; if (count == pos){ return currnode -> element; }else{ currnode = currnode -> next; } } return -1; } int size() { return index; }…Mark each of the following statements as valid or invalid. If a statement is invalid, explain why. a. p = list->link;b. first = list;c. temp->link = nullptr;d. current->link = temp->info;e. p = *last;f. first = 90;g. p->link->info = current->info;h. current->info = temp->link;i. *list = *temp;j. temp->link = last->link->link;k. cout << trail->link->link->info;CHALLENGE ACTIVITY 5.10.2: Simon says Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares the two strings starting from index 0. For each match, add one paint to userScore. Upon a mismatch, exit the loop using a break statement Assume simon Pattern and userPattern are always the same length. Ex The following patterns yield a userScore of 4: simonPattern: RRGBRYYBGY userPattern: RRGBBRYBGY Learn how our autograde works 13 14 simonPattern- scnr.next(); userPattern - scnr.nextO: 15 16 17 for (i=0; i<10; i++) { 18 if O[ simornPattern: 19 20 21 if ( 22 23 24 userPattern: 25 26 returns System.out.println("userScore: userScore); 매매
- Write a program that displays the table show in the* sample executable.** the relationship is given by the quadratic equation* y = 2 + 3x - 2x2** Type of loop-> your favorite (you MUST use a loop)* Hints:* 1) Generate the first column of all the rows using a* Console.WriteLine() statement and a single expression* (use increment = 0.25)* 2) Generate successive columns one at a time by adding* another expression* 3) Format using a column specifier (do not use tabs)*Implement the counter increment before returning to the loop's beginning. After completing these stages, your code should resemble the following, implementing the counter control loop's fundamental structure: .text li $s0, 0 lw $s1, n start_loop: sle $t1, $s0, $s1 beqz $t1, end_loop # code block addi $s0, $s0, 1 b start_loop end_loop:.data n: .word 5Q1) Rewrite the given cursor using for loop: DECLARE emp rec employees°rowtype; CURSOR emps IS SELECT * FROM employees; BEGIN OPEN emps; LOOP FETCH emps INTO emp rec; EXT WHEN emps°onotfound; dbms_ output.put _line(emp _rec.first name|l' 'I|emp rec.salary); END LOOP: CLOSE emps; END;
- Hi need you to add the feature that is in the picture to the code down below, please def add():id = input("Enter ID:")f = open("data.txt", "r")id_list = []lines = f.readlines()if lines != "":for line in lines:token = line.split(" ")id_list.append(token[0])for i in id_list:if i == id:print("The student record is already in the database!Zero records added")returnf.close()file = open("data.txt", "a")name = input("Enter name: ")course = input("Enter course: ")absences = int(input("Enter absences: "))exam1 = int(input("Enter exam 1 grade: "))exam2 = int(input("Enter exam 2 grade: "))marks = int(input("Enter Total Marks: "))s = id+" "+name+" "+course+" "+str(absences)+" "+str(exam1)+" "+str(exam2)+" "+str(marks)+"\n"file.write(s)file.close() while True:print("1.Add a student")print("Any other numeric to terminate")choice = int(input())if choice == 1:add()else:breakThe Dining Philosopher solution using monitors shown in the code snippet below does not suffer from deadlock. However, starvation is still possible. Explain in details why. Each philosopher i invokes the operations pickup() and putdown() in the following sequence: DiningPhilosophers.pickup(i); EAT DiningPhilosophers.putdown(i);hi I need you to add this feature to the code down below, please def add():id = input("Enter ID:")f = open("data.txt", "r")id_list = []lines = f.readlines()if lines != "":for line in lines:token = line.split(" ")id_list.append(token[0])for i in id_list:if i == id:print("The student record is already in the database!Zero records added")returnf.close()file = open("data.txt", "a")name = input("Enter name: ")course = input("Enter course: ")absences = int(input("Enter absences: "))exam1 = int(input("Enter exam 1 grade: "))exam2 = int(input("Enter exam 2 grade: "))marks = int(input("Enter Total Marks: "))s = id+" "+name+" "+course+" "+str(absences)+" "+str(exam1)+" "+str(exam2)+" "+str(marks)+"\n"file.write(s)file.close()while True:print("1.Add a student")print("Any other numeric to terminate")choice = int(input())if choice == 1:add()else:break