请各位大侠看看为什么这个程序没有正常退出

函数源码：
#include <stdio.h>
#include <unistd.h>
int main()
{
    int x = 0, fd[2] = {0};
        char buf[30] = {0}, s[30] = {0};
        pipe(fd);
        while (-1 == (x = fork()));
        if (0 == x) {
                #if 0
            sprintf(buf, "This is message from child\r\n");
                write(fd[1], buf, 30);
                #endif
               
        } else {
            #if 0
            wait(0);
                read(fd[0], s, 30);
                printf("childPID is %d, message%s", x, s);
                #endif
            sprintf(buf, "This is message from Father\r\n");
                write(fd[1], buf, 30);

        }
        if (0 == x) {
            wait(0);
                read(fd[0], s, 30);
                printf("childPID is %d, message%s", x, s);
                //exit(0);
        }
        exit(0);
}

为什么在linux编译运行后，没有正常退出呢？
[root@localhost c_testy]# childPID is 0, messageThis is message from Father//程序卡在这个地方，没有出现下面的命令行提示符呢
[root@localhost c_testy]#

樓主你的程序太過高深..
俺看不懂啊..

僵尸

In UNIX System terminology, a process that has terminated, but whose parent has not yet waited for it, is called a zombie.

If we write a long-running program that forks many child processes, they become zombies unless we wait for them and fetch their termination status.

請教樓上.apue上是這麼說的.

但是,父進程終止后,子進程不是被init領養了嗎...

楼主发上来的时候可以把#if 0 #endif中间的东西都去掉。

wait(0); ???

怎么能Wait 0呢？

Parent:Wait(X)

回复 folklore
我现在改成如下，怎么还是卡在那不动呢，为什么啊

    int main()
{
    int x = 0, fd[2] = {0};
        char buf[30] = {0}, s[30] = {0};
        pipe(fd);
        while (-1 == (x = fork()));
        if (0 == x) {
               
        } else {
            sprintf(buf, "This is message from Father\r\n");
                write(fd[1], buf, 30);

        }
        if (0 == x) {
            //wait(0);
                read(fd[0], s, 30);
                printf("childPID is %d, message%s", x, s);
                exit(0);
        }
        exit(0);
}

本帖最后由 zhangsuozhu 于 2010-07-26 13:53 编辑

呵呵。刚才看错了！

手上没有Unix系统，没发现什么错误,试试