Unix进程管道

Question

我有以下任务要做：

乒乓球。两个进程将进行乒乓球比赛。 第一个进程将生成一个介于 5000 和 15000 之间的随机数，并将其发送给另一个进程。 此过程将减去一个随机值（介于 50 和 1000 之间）并将该数字发回， 进程之间的聊天将使用管道通道实现。 当值低于零时，游戏结束。 每个进程都会打印接收到的值。

所以我写了以下代码：

#include <stdio.h>
#include <time.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#include <stdlib.h>

int main()
{
    int p[2];
    int a, n;

    pipe(p);
    int pid = fork();
    if (pid == 0)
    {
        int r = 0;
        srand(time(NULL));
        while ( r < 50 || r > 1000)
        {
            r = rand();
        }

        if ((a=read(p[0], &n, sizeof(int)))==-1)
            perror("Error read pipe:");

        printf("Process 2 recived %d\n", a);
        a = a - r;

        close(p[0]); close(p[1]);
    }
    else if (pid > 0)
    {
        srand(time(NULL));
        while ( n < 5000 || n > 15000) {
            n = rand();
        }

        while (n > 0)
        {
            printf("Process 1 recived %d\n", n);
            if (write(p[1], &n, sizeof(int))==-1) 
                perror("Error write pipe:");

            wait(0);
        }
    }
    return 0;
}

当它被执行时，它进入一个无限循环，打印"Process 1 received 4"，我不知道为什么。

我创建了另一个管道，现在它正确打印了第一个接收到的值，但是从第二个过程开始，它发生了同样的事情

Process 1 recived 9083
Process 2 recived 4
Process 1 recived 4
and infinite loop

Answer 1

我给你一个更正的程序，在 cmets 中有解释：

#include <stdio.h>
#include <time.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#include <stdlib.h>

int main()
{
    int p[2];
    int p2[2];
    int a, n;

    pipe(p);
    pipe(p2);
    int pid = fork();
    if (pid == 0)
    {
        close(p[1]); // not writing in p, so closing p[1] immediately
        close(p2[0]); // not reading in p2, so closing p2[0] immediately

        srand(time(NULL));

        do {
            if ((a = read(p[0], &n, sizeof(int)))==-1)
                perror("Error read pipe:");

            if (a == 0) // nothing read means all processes have closed p[1]
                n = 0;
            else
                printf("Process 2 recived %d\n", n);

            int r = 50 + rand() % 950; // Only need one random number, and need one different each time
            n -= r;
            if (n > 0)
            {
                if(write(p2[1], &n, sizeof(int))==-1)
                    perror("Error write pipe:");
            }
        } while (n > 0);

        close(p[0]); close(p2[1]);
        exit(0); // or return(0) - as pointed out by Stian
    }
    else if (pid > 0)
    {
        close(p[0]); // not reading in p, so closing p[0] immediately
        close(p2[1]); // not writing in p2, so closing p2[1] immediately

        srand(time(NULL) + pid); // Adding pid so that processes each use a different seed
        n = rand() % 10000 + 5000; // Only need one random number

        while (n > 0)
        {
            if (write(p[1], &n, sizeof(int))==-1) 
                perror("Error write pipe:");

            // wait(0); Wrong use of wait()
            if ((a = read(p2[0], &n, sizeof(int)))==-1)
                perror("Error read pipe:");

            if (a == 0) // nothing read means all processes have closed p2[1]
                n = 0;
            else
                printf("Process 1 recived %d\n", n);

            int r = 50 + rand() % 950;
            n -= r;
        }

        close(p[1]); close(p2[0]);
    }

    wait(NULL); // Better use of wait(). Still not perfect, you need to check return value.
    return 0;
}

Answer 2

你需要两个管道。每个方向一个