使用 fork()
在 4 个进程中进行多次计算
原文:https://www . geeksforgeeks . org/multiple-computing-4-processes-using-fork/
编写一个程序创建 4 个进程:执行各种任务的父进程及其子进程:
- How often the parent process calculates a number.
- The first child process sort array
- The second subprocess finds the even total number in the given array.
- The third subprocess calculates the sum of even numbers in an array.
示例–
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
解释–这里，我们已经使用了 fork()函数创建了 4 个进程三个子进程和一个父进程。因此，这里我们使用两个 fork()函数来创建 4 个进程 n1=fork()和 n2 = fork()
- If n1 and n2 are greater than zero, it is the frequency at which the parent process calculates a number.
- If n1 is equal to zero and n2 is greater than zero, then it is the first subprocess that sorts the given array.
- If n1 is greater than zero and n2 is equal to zero, the second subprocess will find the even total number in the array.
- If both n1 and n2 are equal to zero, then it is the third subelement to calculate the sum of all elements in the array.
代码–
// 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;
}
输出–
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
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