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Multiple calculations in 4 processes using fork()

  • Last Updated : 10 Nov, 2017

Write a program to create 4 processes: parent process and its child process which perform various tasks :

  • Parent process count the frequency of a number
  • 1st child sort the array
  • 2nd child find total even number(s) in a given array
  • 3rd child calculate the sum of even numbers in an array

Example –

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Input :  
2, 4, 6, 7, 9, 0, 1, 5, 8, 3

Output : 
Parent process :
the key to be searched is 7
the frequency of 7 is 1
1st child process :
the sorted array is
0 1 2 3 4 5 6 7 8 9 
2nd child process :
 Total even no are: 5 
3rd child process :
the sum is :45

Explanation – Here, we had used fork() function to create 4 processes three child and one parent process. So, here we use two fork() function which create 4 process n1=fork() and n2 = fork()

  • if n1 and n2 is greater than zero then it is parent process which counts the frequency of a number.
  • if n1 is equal to zero and n2 is greater than zero then it is 1st child process which sorts the given array.
  • if n1 is greater than zero and n2 is equal to zero than it is 2nd child process which finds the total even numbers in the array.
  • if n1 and n2 both equal to zero then it is 3rd child calculates the sum of all elements in an array.



Code –




// C++ code to demonstrate the calculation
// in parent and its 3 child processes using fork()
#include <iostream>
#include <unistd.h>
  
using namespace std;
  
int main()
{
  
    int a[10] = { 2, 4, 6, 7, 9, 0, 1, 5, 8, 3 };
    int n1, n2, i, j, key, c, temp;
    n1 = fork();
    n2 = fork();
  
    // if n1 is greater than zero
    // and n2 is greater than zero
    // then parent process executes
    if (n1 > 0 && n2 > 0) {
  
        int c = 0;
        cout << "Parent process :" << endl;
  
        // key to be searched is 7
        key = 7;
        cout << "the key to be searched is " << key << endl;
  
        for (i = 0; i < 10; i++) {
  
            if (a[i] == key)
                // frequency of key
                c++;
        }
  
        cout << "the frequency of " << key << " is " << c << endl;
    }
  
    // else if n1 is zero
    // and n2 is greater than zero
    // then 1st child process executes
    else if (n1 == 0 && n2 > 0) {
  
        cout << "1st child process :" << endl;
  
        for (i = 0; i < 10; i++) {
  
            for (j = 0; j < 9; j++) {
  
                if (a[j] > a[j + 1]) {
  
                    temp = a[j];
                    a[j] = a[j + 1];
                    a[j + 1] = temp;
                }
            }
        }
  
        cout << "the sorted array is" << endl;
  
        for (i = 0; i < 10; i++) {
  
            cout << a[i] << " ";
        }
  
        cout << endl;
    }
  
    // else if n1 is greater than zero
    // and n2 is  zero
    // then 2nd child process executes
    else if (n1 > 0 && n2 == 0) {
  
        int f = 0;
        cout << "2nd child process :" << endl;
  
        for (i = 0; i < 10; i++) {
  
            // counting total even numbers
            if (a[i] % 2 == 0) {
  
                f++;
            }
        }
  
        cout << " Total even no are: " << f << " ";
        cout << endl;
    }
  
    // else if n1 is zero
    // and n2 is zero
    // then 3rd child process executes
    else if (n1 == 0 && n2 == 0) {
  
        cout << "3rd child process :" << endl;
  
        int sum = 0;
        // summing all given keys
        for (i = 0; i < 10; i++) {
  
            sum = sum + a[i];
        }
  
        cout << "the sum is :" << sum << endl;
    }
  
    return 0;
}

Output –

Parent process :
the key to be searched is 7
the frequency of 7 is 1
1st child process :
the sorted array is
0 1 2 3 4 5 6 7 8 9 
2nd child process :
 Total even no are: 5 
3rd child process :
the sum is :45

This article is contributed by Shivani Baghel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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