在对这个bug进行了大量搜索和阅读之后，我发现没有办法知道进程是否在写入输出之前等待输入，也没有办法进入 Process 类来停止java程序冻结和停止工作，例如，当c++程序需要两个输入，而用户只输入一个输入时。
如何处理这种情况：我曾经 Thread 如下面的新代码所示，在这种情况下保持java程序的活动性并阻止它被冻结。

public byte runFile(String input, byte compilere, String fileName) {
    try {
        ProcessBuilder processBuilder = new ProcessBuilder();

        if (compilere == 0) // c++  code
            processBuilder.command("bash", "-c", "./a.out");
        else
            processBuilder.command("bash", "-c", "java " + fileName);

        Process process = processBuilder.start();

        BufferedOutputStream writer = new BufferedOutputStream(process.getOutputStream());
        BufferedReader stdInput = new BufferedReader(new InputStreamReader(process.getInputStream()));
        BufferedReader stdError = new BufferedReader(new InputStreamReader(process.getErrorStream()));

        if (!input.isEmpty()) {
            writer.write((input + "\n").getBytes());
            writer.flush();
        } else {
            textArea2.setText("Input field is empty!");
            return 1;
        }

        // Read the output from the command:
        thread1 = new Thread() {
            public void run() {
                try {
                    String s = null;
                    s = stdInput.readLine();
                    if (s != null) {
                        textArea2.setText(s);
                        while ((s = stdInput.readLine()) != null)
                            textArea2.setText(textArea2.getText() + "\n" + s);
                    }
                    thread2.start();
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        };

        // Read any errors from the attempted command:
        thread2 = new Thread() {
            public void run() {
                try {
                    String s = null;
                    s = stdError.readLine();
                    if (s != null) {
                        textArea2.setText(s);
                        while ((s = stdError.readLine()) != null)
                            textArea2.setText(textArea2.getText() + "\n" + s);
                    }

                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        };

        thread1.start();

        thread3 = new Thread() {
            @Override
            public void run() {
                try {
                    sleep(8000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                if (textArea2.getText().isEmpty()) {
                    textArea2.setText("The input is not complete or the your program is slow !");
                }
            }
        };
        thread3.start();

    } catch (Exception e) {
        textArea2.setText(textArea2.getText() + "\n" + e.toString());
        e.printStackTrace();
        return 1;
    }
    return 0;
}

编辑：在洛恩侯爵的评论之后，我处理了没有输入而c++应用程序需要输入的情况。