Unix进程PIPE

Question

我有以下任务要做：

乒乓。两个进程将发挥乒乓球比赛。     第一个进程将生成一个介于5000和15000之间的随机数，该数字将发送到另一个进程。     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。     每个流程都会打印收到的值。

所以我写了下面的代码：

#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进程发生了同样的事情

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，并附有评论中的解释：

#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