导航功能将其所有输出打印两次

Question

当我在 main 中单独调用函数时，它们工作正常，只需打印一次所需的消息。调用这两个函数，导致导航打印其消息两次。

我查找了不同的解决方案。我尝试使用 fflush(stdout) 无济于事，我尝试在主体中使用条件以确保它们单独运行。

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <wait.h>
#include "RAND_API.h"
#define MAXLEN 500

void lifeSupport();
void navigation();

int main()
{


    lifeSupport();
    navigation();
    return 0;
}

void lifeSupport()
{
        //Create pipe


        ret = write( myPipe[1], lifeSup1, strlen( lifeSup1 ) + 1 );
        //sleep( getRandExponential() * 10 );
        ret = write( myPipe2[1], lifeSup2, strlen( lifeSup2 ) + 1 );
        //sleep( getSRand() * 6 );
        sleep( getRandExponential() * 5 );
        ret = write( myPipe3[1], lifeSup3, strlen( lifeSup3 ) + 1 );
        sleep( getRandExponential() * 4 );
        ret = write( myPipe4[1], lifeSup4, strlen( lifeSup4 ) + 1 );
        close( myPipe[1] );
        close( myPipe2[1] );
        close( myPipe3[1] );
        close( myPipe4[1] );
    }
    else
    {
        //wait(NULL);
        close( myPipe[1] );
        close( myPipe2[1] );
        close( myPipe3[1] );
        close( myPipe4[1] );
        ret = read( myPipe[0], buffer, MAXLEN );
        printf( "%s", buffer );
        ret = read( myPipe2[0], buffer, MAXLEN );
        printf( "%s", buffer );
        ret = read( myPipe3[0], buffer, MAXLEN );
        printf( "%s", buffer );
        ret = read( myPipe4[0], buffer, MAXLEN );
        printf( "%s", buffer );
    }

}

void navigation()
{
       //create pipe

        ret = write( apipe[1], nav1, strlen( nav1 ) + 1 );
        ret = write( apipe2[1], nav2, strlen( nav2 ) + 1 );
        //sleep( getSRand() * 6 );
        ret = write( apipe3[1], nav3, strlen( nav3 ) + 1 );

        close( apipe[1] );
        close( apipe2[1] );
        close( apipe3[1] );

    }
    else
    {
        close( apipe[1] );
        close( apipe2[1] );
        close( apipe3[1] );

        ret = read( apipe[0], buff, MAXLEN );
        printf( "%s\n", buff );

        ret = read( apipe2[0], buff, MAXLEN );
        printf( "%s\n", buff );

        ret = read( apipe3[0], buff, MAXLEN );
        printf( "%s\n", buff );

        close( apipe[0] );
        close( apipe2[0] );
        close( apipe3[0] );

    }

}

输出：

Life support system initiiated
Adjusting breathing gas levels
Adjusting enviroment
Life support system terminating
Initiating navigation system

Initiating navigation system
Making adjustments


Making adjustments
Adjustments done. Navigation system terminating.


Adjustments done. Navigation system terminating.

预期输出：

Life support system initiiated
Adjusting breathing gas levels
Adjusting enviroment
Life support system terminating
Initiating navigation system
Making adjustments

Adjustments done. Navigation system terminating.

Answer 1

当您调用 lifeSupport() 时，它会分叉一个 child。 child 将消息写入所有 myPipeX 管道，parent 从中读取并打印消息，然后 parent 和 child return 到 main().

然后两个进程都调用navigation()。他们各自创造了另一个 child。两个 children 写入 apipeX 管道，两个 parents 从它们读取，并且它们各自打印消息。因此，您从 navigation().

获得了所有消息的两份副本

在每个函数中，child 应该在完成后调用 exit()，而不是 returning。只有 parent 应该 return 到 main().

导航功能将其所有输出打印两次

Navigation function is printing all its output twice

c

linux

fork

pipe