2. What is the output of the following program? namespace ConsoleApplication1 struct TMA public int macks: class Progran static void Main(string) acaA) TMA X - new IMAL); Checktref x); "); public static void Check(ref THA y) YJAsks - 78;
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i forgot this part of the quostion:
Note: Provide a copy of the code and screen shot for the output in the solutions’. Explain the working of the program in your own words.
can u help me with it
- You are said to store data of Hospitals in a city. For that purpose, you are asked to develop a structure of maxHeap using these structures in c++. You have to submit Hospital. h file only class Hospital { string HospitalName;string Address;string HospitalID;public:Hospital(); Hospital(string ,string ,string); Hospital(string HospitalID); // conversion Constructor string getHospitalName()const; string getHospitalId()const; string getAddress()const; void setHospitalName(string); void setHospitalId(string);void setAddress(string); // Relational operators with respect to Hospital Idbool operator<(Hospital)const;bool operator<=(Hospital)const;bool operator>(Hospital)const; bool operator>=(Hospital)const; bool operator==(Hospital)const;bool operator!=(Hospital)const; // output stream operatorfriend ostream& operator<<(ostream&, const Hospital& ref); s}; ostream& operator<<(ostream&, const Hospital& ref); class MaxHeap { public:class Node…I want the below code modified that works for multi user in below code there are only one user but i need multi user up to 10? Code #include <iostream> #include <time.h> #include <conio.h> Using namespace std; class Bank {public:}; class ATMAccountHolders:Bank {string accountHolders; string accountHoldersAddress, branch; int accountNumber; double startBalance; double accountBalance; double amount; int count; Public: void deposit(); void withdraw(); void accountExit(); ATMAccountHolders() {accountNumber = 7787; accountHolders = " Ammad Naseer"; accountHoldersAddress = " House no.112"; startBalance = 6000.00; accountBalance = 6000.00; branch = " Islamabad"; amount = 20000; } }; void ATMAccountHolders::deposit() { system("cls"); cout<<" ATM ACCOUNT DEPOSIT SYSTEM "; cout<<"\n\nThe Names of the Account Holders are :"<<accountHolders<<"\n\n"; cout<<"\tThe Account Holders' address is…using Java: Assign courseStudent's name with Smith, age with 20, and ID with 9999. Use the printAll member method and a separate println statement to output courseStudents's data. Sample output from the given program: Name: Smith, Age: 20, ID: 9999 // ===== Code from file PersonData.java =====public class PersonData {private int ageYears;private String lastName; public void setName(String userName) {lastName = userName;} public void setAge(int numYears) {ageYears = numYears;} // Other parts omitted public void printAll() {System.out.print("Name: " + lastName);System.out.print(", Age: " + ageYears);}}// ===== end ===== // ===== Code from file StudentData.java =====public class StudentData extends PersonData {private int idNum; public void setID(int studentId) {idNum = studentId;} public int getID() {return idNum;}}// ===== end ===== // ===== Code from file StudentDerivationFromPerson.java =====public class StudentDerivationFromPerson {public static void main (String [] args)…
- fix the code below to do a reverse number, you can add max 2 lines of code and also change max 2 lines of code. code: public class Main { public static void main(String[] args) { solution(29020714); solution(-364); solution(514820); solution(16091900); } static void solution(int N) { int enablePrint = 0; while (N > 0) { if (enablePrint == 0 && N % 10 != 1) { enablePrint = N / 10; } if (enablePrint > 0) { System.out.print(enablePrint); } N /= 10; } System.out.println(); } }Below is a java program to find roots of quadratic equation. Please explain each line of code if what it does in the program when executed. public class Main { public static void main(String[] args) { double a = 2.3, b = 4, c = 5.6; double root1, root2; double determinant = b * b - 4 * a * c; if (determinant > 0) { root1 = (-b + Math.sqrt(determinant)) / (2 * a); root2 = (-b - Math.sqrt(determinant)) / (2 * a); out.format("root1 = %.2f and root2 = %.2f", root1, root2); } else if (determinant == 0) { root1 = root2 = -b / (2 * a); out.format("root1 = root2 = %.2f;", root1); } else { double real = -b / (2 * a); double imaginary = Math.sqrt(-determinant) / (2 * a); out.format("root1 = %.2f+%.2fi", real, imaginary); out.format("\nroot2 = %.2f-%.2fi", real, imaginary); } } }Question is in the picture. #include <iostream> using namespace std; class Product { private: double price; int weight; public: void info() { cout <<price<<", "<<weight; } }; class Fruit { public: string type; void setInfo(double p, int w) { price = p; weight = w; } }; int main() { Fruit obj; obj.type = "Apple"; obj.setInfo(4.99, 10); obj.info(); }
- Determine the output for the following code. Box in your final output result. public class Beta extends Baap { public int h = 44; public int getH( ) { System.out.println("Beta " + h); return h; } public static void main(String[ ] args) { Baap b = new Beta(); System.out.println(b.h + " " + b.getH( )); Beta bb = (Beta) b; System.out.println(bb.h + " " + bb.getH( )); } } public class Baap { public int h = 4; public int getH( ) { System.out.println("Baap " + h); return h; } }Hello. Correct my code. Write a program to approximate the derivatives of e x , cos(x) and sin(x) at x = 0 using the limits. #include<stdio.h>#include<math.h>#define E 2.718 int main(void) { FILE* fout = fopen("derivativeout.txt", "w"); int n, i; float h, der1, der2, der3; fprintf(fout, "h Dexp(x) Dcos(x) Dsin(x)\n"); fprintf(fout, "--------------------------------------\n"); printf("Enter n for h = 1/2 down to 1/2^n: "); scanf("%d", &n); for(i=1; i<=n; i++) { h = 1.0/pow(2.0, n); der1 = (exp(h)-1.0)/h; der2 = (cos(h)-1.0)/h; der3 = (sin(h))/h; fprintf(fout, "%f %f %f %f\n", h, der1, der1, der3); } fclose(fout); return 0;}What is the output of the given code? #include<iostream> using namespace std; class Base { public: Base() { cout<<"Constructing Base \n"; } virtual~Base() { cout<<"Destructing Base \n"; } }; class Derived: public Base { public: Derived() { cout<<"Constructing Derived \n"; } ~Derived() { cout<<"Destructing Derived \n"; } }; int main(void) { Derived *d = new Derived(); Base *b = d; delete b; return 0; }
- What is the output of given code? #include<iostream>using namespace std;class Base {public:Base(){ cout<<"Constructing Base \n"; }virtual~Base(){ cout<<"Destructing Base \n"; }};class Derived: public Base {public:Derived(){ cout<<"Constructing Derived \n"; }~Derived(){ cout<<"Destructing Derived \n"; }}; int main(void){Derived *d = new Derived();Base *b = d;delete b;return 0;}Can you help me with this problem please. Here is my code: #include <iostream> using namespace std; const double GALLONS_IN_A_CUBIC_FEET = 7.48; class swimmingPool { public: void set(double l = 0, double w = 0, double d = 0, double fi = 0, double fo = 0, double wInPool = 0); void setLength(double l); void setWidth(double w); void setDepth(double d); void setWaterFlowRateIn(double fi); void setWaterFlowRateOut(double fo); void addWater(double time, double fillRate); void drainWater(double time, double drainRate); double poolTotalWaterCapacity(); double getLength(); double getWidth(); double getDepth(); double getWaterFlowRateIn(); double getWaterFlowRateOut(); double getTotalWaterInPool(); int timeToFillThePool(); int timeToDrainThePool(); double waterNeededToFillThePool(); swimmingPool(double l = 0, double w = 0, double d = 0, double fi = 0, double fo = 0, double…Assume that you are applying “testing” and “inspection”, the two defect finding methods in the following buggy code example to find defects. Explain how “testing” and “inspection” differ in finding defects quickly and the efforts required for the example. public class Employee { private String id; private int name; private double salary; public Employee(String id,int name,double salary) { } public Employee() { } } Public class Customer { public static void main(String s[]) { Employee e1=new Employee(“10060OTU”, “S1”, 2000.50); Employee e2=new Employee(); System.out.println (“The salary of Employee S1 is”+ e1.salary); }