If you are going to build a custom fast PC for gaming and you need a lot of storage for big video files and videos, but you have
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Answer: Hello your question lacks some details attached below is the missing detail
answer :
a) True , B) False C) True D) True
Explanation:
a) True ; The critical section cannot be executed by more than one process at a time
b) False : The code does not satisfy the progress condition, because while loops are same hence no progress
c ) True : The code satisfies the bounded waiting condition, because of the waiting condition of the while loop
d) True : No matter how many times this program is run, it will always produce the same output, this is because of the while loop condition
Answer:
I am writing the program in C++ programming language.
#include<iostream> // to use input output functions
using namespace std; // to identify objects like cin cout
class ProblemSolution { //class name
private:
// private data members that only be accessed within ProblemSolution class
int real, imag;
// private variables for real and imaginary part of complex numbers
public:
// constructor ProblemSolution() to initialize values of real and imaginary numbers to 0. r is for real part and i for imaginary part of complex number
ProblemSolution(int r = 0, int i =0) {
real = r; imag = i; }
/* print() method that displays real and imaginary part in output of the sum of complex numbers */
void print(){
//prints real and imaginary part of complex number with a space between //them
cout<<real<<" "<<imag;}
// computes the sum of complex numbers using operator overloading
ProblemSolution operator + (ProblemSolution const &P){ //pass by ref
ProblemSolution sum; // object of ProblemSolution
sum.real = real + P.real; // adds the real part of the complex nos.
sum.imag = imag + P.imag; //adds imaginary parts of complex nos.
//returns the resulting object
return sum; } //returns the sum of complex numbers
}; //end of the class ProblemSolution
int main(){ //start of the main() function body
int real,imag; //declare variables for real and imaginary part of complex nos
//reads values of real and imaginary part of first input complex no.1
cin>>real>>imag;
//creates object problemSolution1 for first complex number
ProblemSolution problemSolution1(real, imag); //creates object
//reads values of real and imaginary part of first input complex no.2
cin>>real>>imag;
//creates object problemSolution2 for second complex number
ProblemSolution problemSolution2(real,imag);
//creates object problemSolution2 to store the addition of two complex nos.
ProblemSolution problemSolution3 = problemSolution1 + problemSolution2;
problemSolution3.print();} //calls print() method to display the result of the //sum with real and imaginary part of the sum displayed with a space
Explanation:
The program is well explained in the comments mentioned with each statement of the program. The program has a class named ProblemSolution which has two data members real and imag to hold the values for the real and imaginary parts of the complex number. A default constructor ProblemSolution() which initializes an the objects for complex numbers 1 and 2 automatically when they are created.
ProblemSolution operator + (ProblemSolution const &P) is the operator overloading function. This performs the overloading of a binary operator + operating on two operands. This is used here to add two complex numbers. In order to use a binary operator one of the operands should be passed as argument to the operator function. Here one argument const &P is passed. This is call by reference. Object sum is created to add the two complex numbers with real and imaginary parts and then the resulting object which is the sum of these two complex numbers is returned.
In the main() method, three objects of type ProblemSolution are created and user is prompted to enter real and imaginary parts for two complex numbers. These are stored in objects problemSolution1 and problemSolution2. Then statement ProblemSolution problemSolution3 = problemSolution1 + problemSolution2; creates another object problemSolution3 to hold the result of the addition. When this statement is executed it invokes the operator function ProblemSolution operator + (ProblemSolution const &P). This function returns the resultant complex number (object) to main() function and print() function is called which is used to display the output of the addition.
