Jump to level 1Compute: z = √xEx: If the input is 4.0 2.0, then the output is:4.01 #include 2 #include <3 #include 4 using namespace std;56 int main() {7double x;8 double y;9 double z;1011 cin >> X;cin >> y;121300 ChSEEEEEEEE1415V* Your code goes here */16 cout << fixed <<<

Include the necessary libraries for input/output, mathematical functions, and output…

Jump to level 1 Compute: z = √x Ex: If the input is 4.0 2.0, then the output is: 4.0 1 #include 2 #include 3 #include 4 using namespace std; 5678SOH 6 int main() { double x; double y; double z; 9 10 11 संलललललल cin >>> X; 12 cin >> y; 13 14 15 345678 16 17 18 19 20 } V* Your code goes here */ cout << fixed <<

Jump to level 1 Compute: z = √x Ex: If the input is 4.0 2.0, then the output is: 4.0 1 #include 2 #include 3 #include 4 using namespace std; 5678SOH 6 int main() { double x; double y; double z; 9 10 11 संलललललल cin >>> X; 12 cin >> y; 13 14 15 345678 16 17 18 19 20 } V* Your code goes here */ cout << fixed <<

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Compute: z = √x Y Ex: If the input is 4.0 3.0, then the output is: 1.0 #include #include #include using namespace std; int main() { double x; double y; double z; } cin >> X; cin >> y; /* Your code goes here */ cout << fixed <<<
change this code to C++ #include <stdio.h> #include <stdlib.h> /** * structure to hold each point */ struct Point3D { float x; float y; float z; }; /** * Parses all the vertices from the path. * @param path The input file path. * @param points The points storage. * @param numPoints The number of points. * @return points Dynamically allocated array of points. * @return The center of mass. */ struct Point3D parseInput(const char *path, struct Point3D **points, int *numPoints) { // to hold each record char f_or_v; float x; float y; float z; // to store the file pointer FILE *inFile; // to iterate the points array int it; // structure to store the center of mass struct Point3D centerOfMass; // opening the file inFile = fopen(path, "r"); // counting the number of points *numPoints = 0; while(!feof(inFile)) { fscanf(inFile, "%c %f %f %f\n", &f_or_v, &x, &y, &z); *numPoints += f_or_v == 'v'; } // dynamically allocating space // for the points *points =…
#include <iostream> #include <cmath> #include <iomanip> using namespace std; int calc_qty(int qty); void receipt( int qty, int qty2); int calc_qty(int qty); void coffee_size(int qty, int size, int choice, int qty2); void slices(); void beanounces(); int main() { int qty = 0, order = 0, size = 0, qty2 = 0; receipt (qty,qty2); calc_qty (qty); do { //displays the menu cout<<" Please select an item on the menu you would like to buy."<<endl; cout<<"Enter 1 for Coffee\n"<<"Enter 2 for cheese cake\n"<<"Enter 3 for coffee beens\n"<<endl; cin>>choice; } switch (choice) { case 1: coffee_size(qty, size, order, qty2); break; case 2: slices(); break; case 3: beanounces(); break; case 4: cout<<"Enjoy your day"<<endl;…
int main(){ long long int total; long long int init; scanf("%lld %lld", &total, &init); getchar(); long long int max = init; long long int min = init; int i; for (i = 0; i < total; i++) { char op1 = '0'; char op2 = '0'; long long int num1 = 0; long long int num2 = 0; scanf("%c %lld %c %lld", &op1, &num1, &op2, &num2); getchar(); long long int maxr = max; long long int minr = min; if (op1 == '+') { long long int sum = max + num1; maxr = sum; minr = sum; long long int res = min + num1; if (res > maxr) { max = res; } if (res < minr) { minr = res; } } else { long long int sum = max * num1; maxr = sum; minr = sum; long long int res = min * num1;…
Consider the producer-consumer problem where the producer produces items to be consumed by the consumer. A solution of the problem is to be implemented using threads. The main process will run producer and consumer as two different threads. The producer will put the items it generates into a buffer of length 15. The consumer reads the elements of the buffer to get the items. A solution is given below. Each item produced and consumed are also printed to the screen. Run the program several times and find a sample run when the items produced are not correctly consumed. Explain why the program does not always work correctly.
main.cpp 1 #include 2 using namespace std; 3 4 int main() 5- { int a,b; cin>>a>>b; int i = a; int total 7 8 9. 0; %3D 10 11 i++; 12 13 cout<
Code so far is this. // @brief Program 1 - ASCII Building// REPLACE ME AND THE LINES AROUND ME!// Adapted from prior programs developed by Professor Reed. /* Running the program looks like:Choose from among the following options:1. Exit the program2. Display buildingYour choice -> 1*/ #include <iostream> // for cin and cout#include <iomanip> // for setw() and setfill()using namespace std; // so that we don't need to preface every cin and cout with std:: void printFirstTwoBuildingSection(int n, int startSpacing){int start = n / 2, end = 0;if (n <= 2){start = 0;}for (int i = 0; i < n; i++){for (int j = 0; j < startSpacing; j++){cout << " ";}cout << "|";if (i < (n / 2)){for (int k = 0; k < i; k++){cout << " ";}cout << "\\";for (int k = 0; k < start; k++){cout << " ";}cout << "/";start -= (n / 2);for (int k = 0; k < i; k++){cout << " ";}cout << "|" << endl;}else{for (int k = n - i - 1; k > 0;…
def square(x): return x * X def halve(x): return x // 2 def twice(f,x): """Apply f to the result of applying f to x >>> twice (square,3) 81 >>> twice (square, 4) 256 >>> twice (halve, 32) 8 >>> twice (halve, 80) 20 *** YOUR CODE HERE ***"
#include <stdio.h>#include <stdlib.h>#include <unistd.h>#include <sys/types.h>#include <sys/wait.h> int main(){ int i, s; int a[] = {23,4,8,33,25,7,12,6,49,50,11,20,21,24,35}; pid_t pid; pid = fork(); if (pid == 0) { printf("I'm a child, I'll be printing the even numbers:\n"); for(i=0;i<(sizeof(a)/sizeof(int));i++) if(a[i]%2 == 0) printf("%d ",a[i]); } else { printf("I'm the parent, I'll be printing the odd numbers:\n"); if (pid > 0) pid = waitpid(pid, &s, 0); for(i=0;i<(sizeof(a)/sizeof(int));i++) if(a[i]%2 != 0) printf("%d ",a[i]); } return 0;} the output seem to have some error whereby it cannot output the number for the first if
// Assume all libraries are included 3 void func (int a, int &b, int &c); // int main () { int i = 5, j = 4, k = 33; 4 7 %3D 8. func (i, j, k); func (j, i, k); cout << j « k << i <« endl; 10 11 12 13 return 0; 14 } // 15 16 void func (int a, int &b, int &c) { 17 18 = 2*c + b; 19 b a; 20 C = a + 3*b; } // 21 22 I|||
#include #include #include #define SIZE 5 int nums [SIZE] = [5,0,6,1,2); int main() { int i; pid_t pid; pid= fork(); if (pid == 0) { for (i = 0; i 0) { wait (NULL); } for (i = 0; i < SIZE; i++) printf("PARENT: %d\n", nums [i] + 1); /* LINE Y */ return 0; From the above-given program, please state what will be the output at lines X and Y. Explanation in detail required for output at both lines X and Y.
//main.cpp #include<iostream>#include<cmath>#include<string>#include<cstring>#include<algorithm>#include<list>using namespace std; #include "Dog.h" void popDogs(list<Dog>&); /** Change NOTHING above this line **/ //TODO, overload << operator//here's a startostream& operator<<(ostream& os, Dog& d){ //Insert code here so that when a Dog is output, it is done as: //dogName weighs weight pounds. //Like: //Spot weighs 28 pounds. //Don't forget the period at the end. Do NOT include an endl at the end of the line. os<< d.dogName<< " weighs " << d.weight << " pounds."; return os; } int main(){ //TODO, instantiate list of Dog variable called kennel. list<Dog> kennel; //TODO, call the popDogs function passing the kennel variable declared above popDogs(kennel); //This code will sort the list of Dog using the less-than operator kennel.sort();…