在后台分叉无法正常工作 [英] fork in background don't work correct

问题描述

我运行此编.在前景和背景中:

I run this prog. in foreground and background:

int main()    
{
    int pid;
    printf("App Start pid=%d!\n",getpid());

    while(1) {
        pid=fork();
        if(pid==0) {
            printf("Child\n\n");
            exit(0);
        }
        else if(pid>0) {
            printf("Parent\n");
        }    
        sleep(1);
    }
}

在前台:

$ ./fork

结果是:

App Start pid=1360!    
Parent    
Child    
Parent    
Child    
...

在后台:

$./fork > out.txt &
$cat out.txt    
App Start pid=1368!    
Child    
App Start pid=1368!    
Parent    
Child    
App Start pid=1368!    
Parent    
Parent    
Child    
...

为什么应用程序在后台重启"? 我不明白发生了什么. 如何使fork在后台应用程序中正常工作? 谢谢

Why does the app 'restart' in background? I don't understand what happening. How can i make fork to work correctly in background app? Thanks

推荐答案

这与输出缓冲区有关:新创建的进程正在覆盖其父级print编写的内容.请注意，该消息不会更改，即:始终为App Start pid=1368!.

This has to do with the output buffers: newly created process are writing over and over what their parent already printed. Notice that the message does not change, i.e.: it is always App Start pid=1368!.

在第一次调用printf()之后放置fflush(stdout):

Place fflush(stdout) after the first call to printf():

printf("App Start pid=%d!\n",getpid());
fflush(stdout);

这样，在创建子进程之前，将清空 输出缓冲区.

This way the output buffer will be flushed before creating the children processes.

请注意，通过启动fork程序而不重定向stdout(即:$ ./fork)，默认情况下stdout是行缓冲的.因此，每次stdout收到换行符时，都已经执行了刷新stdout.

Note that by starting the fork program without redirecting stdout (i.e.: $ ./fork), stdout is line-buffered by default. For that reason, a flush of stdout is already performed every time it receives a new-line character.

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