用fork（）和pipe（）调用竞争条件

Question

我正在为一个父级播放连接N个子进程的作业编写一个程序。该程序使用管道在进程之间传递游戏移动。

但是，我遇到了修复存在于我的程序中的竞争条件的问题。在游戏结束后，子进程会挂起一个read（）调用。这只有在有多个子进程时才会发生。

我已经尝试了几个东西，比如命名的信号灯，但我对叉子，管道和IPC还是很新的东西。我已与相关的代码要点（我试图把它清理干净是最好的，我可以为可读性）位置：

Gist with relevant code

任何帮助，将不胜感激

编辑

以下是添加了声明的要点的相关来源。

int main (int argc, char const *argv[]) 
{ 
    int dimension = 8, children = 2, i; 
    int child_play_to_win = 0; 
    int fd[children][4]; 
    pid_t childpid[children]; 
    Board** boards = (Board**) malloc(sizeof(Board*) * children); 
    GameMove* lastMove, *tmpMove; 
    char buf[80]; 
    for(i = 0; i < children; i++) { 
    generate_board(&(boards[i]), dimension); 
    int tmp[2]; 
    pipe(tmp); 

    // child read 
    fd[i][0] = dup(tmp[0]); 
    // parent write 
    fd[i][1] = dup(tmp[1]); 

    pipe(tmp); 
    // parent read 
    fd[i][2] = dup(tmp[0]); 
    // child write 
    fd[i][3] = dup(tmp[1]); 

     childpid[i] = fork(); 

    if(childpid[i] == -1) { 
     perror("fork"); 
     exit(1); 
    } 
    if(childpid[i] == 0) { 
     srand(getpid()); 
     close(fd[i][1]); 
     close(fd[i][2]); 
     while(!boards[i]->finished) { 
     // Read in move from parent 
     printf("child[%d] about to read\n", getpid()); 
     read(fd[i][0], &buf, sizeof(GameMove)); 

     // repeat parent move on this board 

     if(gameNotFinished) { 
      // make child move 

      // write move back to parent 

      write(fd[i][3], lastMove, sizeof(GameMove)); 

      // If the board is finished (there was a win), 
      if (!gameNotFinihsed) { 
      // Child wins 
      close(fd[i][0]); 
      close(fd[i][3]); 
      printf("child[%d] ending\n", getpid()); 
      break; 
      } 
     } 
     else { 
      // Parent won 
      close(fd[i][0]); 
      close(fd[i][3]); 
      break; 
     } 
     } 
    dealloc(boards[i]); 
    exit(0); 
    } 
} 

    // When this hits children amount, all games are done 
    int games_complete = 0; 
    // Make first move to all children 
    for (i = 0; i < children; i++) { 
    close(fd[i][0]); 
    close(fd[i][3]); 
    lastMove = placePieceAtBestPosition(boards[i], 1); 
    printf("parent writing to child[%d]\n", childpid[i]); 
    write(fd[i][1], lastMove, sizeof(GameMove)); 
    } 
    while (games_complete != children) { 
    for (i = 0; i < children; i++) { 
     // Read move from child 
     read(fd[i][2], &buf, sizeof(GameMove)); 

     // repeat child move 

     // Check for a child win... 
     if (!checkForWin(boards[i], 2)) { 
     // No win yet, place piece at best position 

     lastMove = placePieceAtBestPosition(boards[i], 1); 

     // check for win again 
     boards[i]->finished = checkForWin(boards[i], 1); 
     // Write move back to child 
     write(fd[i][1], lastMove, sizeof(GameMove)); 

     // If we won, close everything up and increment 
     // the games_complete counter. 
     if(boards[i]->finished) { 
      close(fd[i][1]); 
      close(fd[i][2]); 
      games_complete++; 
     } 
     } else { 
    // write back child move if there was a win 
     write(fd[i][1], lastMove, sizeof(GameMove)); 
     close(fd[i][1]); 
     close(fd[i][2]); 
     printf("Parent lost!):\n"); 
     games_complete++; 
     } 
    } 
    } 

Answer 1

我想我知道你的问题是什么。当你分叉每个孩子时，你关闭它的管道的父侧。然而，每个孩子仍然打开所有以前孩子的管道的父方。因此，只有最后创建的子项将其管道的父级关闭。

建议你改变：

close(fd[i][1]); 
close(fd[i][2]); 

喜欢的东西：

for (j = 0; j <=i; j++) { 
    close(fd[j][1]); 
    close(fd[j][2]); 
}