#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 < SIZE; i++) { nums [i] = 1 + 1; printf("CHILD %d\n",nums [i]); /* LINE X */ else if (pid > 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.

#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 < SIZE; i++) { nums [i] = 1 + 1; printf("CHILD %d\n",nums [i]); /* LINE X / else if (pid > 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.

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

See similar textbooks

Similar questions

Question 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews O
int func(int a, int b) { return (a
int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;
C programming: I want to turn this code without the use of arrays... void myFunction(int myNumbers[5]) { for (int i = 0; i < 5; i++) { printf("%d\n", myNumbers[i]); }}int main() { int myNumbers[5] = {10, 20, 30, 40, 50}; myFunction(myNumbers); return 0;}
in c language !! typedef struct {long restaurant_id;char restaurant_name[10];char description[120];double rate;char cuisine[30]; opening_year;long capacity;char city[30];char address[60];char owner[30];} RECORD_t, *RECORD; RECdelete (unsigned long restaurant_id, RECORD *restaurant_array, unsigned long *p_size){int rec;int index;// index <- find the index of the RECORD with given restaurant_id in restaurant_array ????? (I DIDNT)if (index is valid)rec=restaurant_array[index];elserec=NULL;return rec;}
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;…
C programming pseudocode pointers help. { totalInAGivenHour+=*(rawReadings+ monthNo*30*14*4 + 14*4*nThDateOfMonth+ 4*nThHourOfDateFrom7AM + nThReadingTimeOfHour); // the pointer is to assign the values in the bracket and storing them in totalInAGivenHour }
#include using namespace std; int main() (double degree [6] [2] = {30,40,10,70, 20, 30, 60, 70, 30, 10, 10, 85); int buffer=degree[0][0]; for (int i=1;ibuffer) buffer=degree[i][0]; cout using namespace std; int main() { int j=1; while (j<=10) if ((j%3)==0) cout<
#include using namespace std; main () { int array [8] for (int x=; x> for ( cout<< return ; X< : ) : }
1. Implement a class IntArr using dynamic memory. a. data members: capacity: maximum number of elements in the array size: current number of elements in the array array: a pointer to a dynamic array of integers b. constructors: default constructor: capacity and size are 0, array pointer is nullptr user constructor: create a dynamic array of the specified size c. overloaded operators: subscript operator: return an element or exits if illegal index d. "the big three": copy constructor: construct an IntArr object using deep copy assignment overload operator: deep copy from one object to another destructor: destroys an object without creating a memory leak e. grow function: "grow" the array to twice its capacity f. push_back function: add a new integer to the end of the array g. pop_back function: remove the last element in the array h. getSize function: return the current size of the array (not capacity) i. Use the provided main function (next page) and output example Notes a. grow…
int* p: int a[3]{1, 2, 3}; p = a; What is the value of *(p+2)?
void fact(int num) { int k,f=1; for (k=1;k<=num; k++) f=f*k; cout<< f; int main() { int arri[3]={3,2,1} ; for(int i=0;i<3;i++) fact(arri[i]); } 261 O 621 O 7 462