Prerequisite – fork() in C, sorting in fork()
Problem statement – Write a program to search the key element in parent process and print the key to be searched in child process.
Examples –
Input :
Key = 10;
array[5] = {3, 8, 4, 10, 80};
Output:
Parent process
key is present in array
Child process
numbers to be search is 10
Explanation – Here, we had used fork( ) function to create two processes one child and one parent process.
- fork() returns value greater than 0 for parent process so we can perform the searching operation.
- for child process fork() returns 0 so we can perform the print the value of key to be searched.
- Here we are using a simple searching algorithm to search the key element in the given array.
- We are using the returned values of fork() to know which process is a child or which is a parent process.
Note – At some instance of time, it is not necessary that child process will execute first or parent process will be first allotted CPU, any process may get CPU assigned, at some quantum time. Moreover process id may differ during different executions.
Code –
// C++ program to demonstrate searching // in parent and printing result // in child processes using fork() #include <iostream> #include <unistd.h> using namespace std; // Driver code int main() { int key = 10; int id = fork(); // Checking value of process id returned by fork if (id > 0) { cout << "Parent process
"; int a[] = { 3, 8, 4, 10, 80 }; int n = 5; int flag; int i; for (i = 0; i < n; i++) { if (a[i] != key) { flag = 0; } else { flag = 1; } } if (flag == 1) { cout << "key is not present in array"; } else { cout << "key is present in array"; cout << "
"; } } // If n is 0 i.e. we are in child process else { cout << "Child process
"; cout << "numbers to be search is "; cout << key; } return 0; } |
Output –
Parent process key is present in array Child process numbers to be search is 10
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
This article is attributed to GeeksforGeeks.org
0
0
leave a comment
0 Comments