C# 程序在串口读取时冻结打开新窗体

Question

我正在开发一个从串行端口接收数据的接口，一旦它获得特定值，它就会显示另一种形式，但是当显示这种形式时，window 冻结你能帮我找到一个解决方案？

public string[] data=new string[14]; public SerialPort port = new SerialPort("COM17",2400, Parity.None,8,StopBits.One); private Queue recievedData = new Queue();

public MainForm()
{
 InitializeComponent();
  
    port.Open();
    port.DataReceived += new SerialDataReceivedEventHandler(serialPort1_DataReceived);
    int count = port.BytesToRead;
    byte[] ByteArray = new byte[count];

}




public void serialPort1_DataReceived(object sender, SerialDataReceivedEventArgs e)
{


    byte[] data = new byte[port.BytesToRead];
    port.Read(data, 0, data.Length);
    data.ToList().ForEach(b => recievedData.Enqueue(b));
    processData();
   

}

public void processData()
{
 byte[] byteArray = new byte[14];
 if (recievedData.Count > 13)
 {
 for (int i = 0; i < byteArray.Length; i++)
 {
  byteArray[i] = recievedData.Dequeue();
 }

    }
    for (int i = 0; i < 14; i++)
    {
        data[i] = byteArray[i].ToString();
    }
  
    int value0 = Convert.ToInt16(data[0]);
    BitArray b0 = new BitArray(new int[] { value0 });
    bool[] bits0 = new bool[b0.Count];
    b0.CopyTo(bits0, 0);

    int value1 = Convert.ToInt16(data[1]);
    BitArray b1 = new BitArray(new int[] { value1 });
    bool[] bits1 = new bool[b1.Count];
    b1.CopyTo(bits1, 0);

    int value2 = Convert.ToInt16(data[2]);
    BitArray b2 = new BitArray(new int[] { value2 });
    bool[] bits2 = new bool[b2.Count];
    b2.CopyTo(bits2, 0);

    int value3 = Convert.ToInt16(data[3]);
    BitArray b3 = new BitArray(new int[] { value3 });
    bool[] bits3 = new bool[b3.Count];
    b3.CopyTo(bits3, 0);

    int value4 = Convert.ToInt16(data[4]);
    BitArray b4 = new BitArray(new int[] { value4 });
    bool[] bits4 = new bool[b4.Count];
    b4.CopyTo(bits4, 0);

    int value5 = Convert.ToInt16(data[5]);
    BitArray b5 = new BitArray(new int[] {value5});
    bool[] bits5 = new bool[b5.Count];
    b5.CopyTo(bits5, 0);

    int value6 = Convert.ToInt16(data[6]);
    BitArray b6 = new BitArray(new int[] { value6 });
    bool[] bits6 = new bool[b6.Count];
    b6.CopyTo(bits6, 0);

    int value7 = Convert.ToInt16(data[7]);
    BitArray b7 = new BitArray(new int[] { value7 });
    bool[] bits7 = new bool[b7.Count];
    b7.CopyTo(bits7, 0);

    int value8 = Convert.ToInt16(data[8]);
    BitArray b8 = new BitArray(new int[] { value8 });
    bool[] bits8 = new bool[b8.Count];
    b8.CopyTo(bits8, 0);

    int value9 = Convert.ToInt16(data[9]);
    BitArray b9 = new BitArray(new int[] { value9});
    bool[] bits9 = new bool[b9.Count];
    b9.CopyTo(bits9, 0);

    int value10 = Convert.ToInt16(data[10]);
    BitArray b10 = new BitArray(new int[] { value10 });
    bool[] bits10 = new bool[b10.Count];
    b10.CopyTo(bits10, 0);

    int value11 = Convert.ToInt16(data[11]);
    BitArray b11 = new BitArray(new int[] { value11 });
    bool[] bits11 = new bool[b11.Count];
    b11.CopyTo(bits11, 0);

    int value12 = Convert.ToInt16(data[12]);
    BitArray b12 = new BitArray(new int[] { value12 });
    bool[] bits12 = new bool[b12.Count];
    b12.CopyTo(bits12, 0);

    int value13 = Convert.ToInt16(data[13]);
    BitArray b13 = new BitArray(new int[] { value13 });
    bool[] bits13 = new bool[b13.Count];
    b13.CopyTo(bits13, 0);

  
    if (bits13[2]&&reentrant==0)
    {
     reentrant = 1;
     
        Form1 formita = new Form1(this);
        formita.Show();

    }

    string direction = "";

    if (bits1[3]==true)
    {
        direction = "-";
    }

    string firstdigital = bittobcd(bits1[2], bits1[1], bits1[0], bits2[3], bits2[2], bits2[1], bits2[0]);
    string seconddigital = bittobcd(bits3[2], bits3[1], bits3[0], bits4[3], bits4[2], bits4[1], bits4[0]);
    string thirddigital = bittobcd(bits5[2], bits5[1], bits5[0], bits6[3], bits6[2], bits6[1], bits6[0]);
    string fourthdigital = bittobcd(bits7[2], bits7[1], bits7[0], bits8[3], bits8[2], bits8[1], bits8[0]);


    Writing(direction+firstdigital+seconddigital+thirddigital+fourthdigital);
  
}

string bittobcd(bool a, bool b, bool c, bool d, bool e, bool f, bool g)
{
 string number="";

 if (a == false && b == false && c == false && d == false && e == true && f == false && g == true)
 {
  number = "1";
 }
 else if(a == true && b == false && c == true && d == true && e == false && f == true && g == true)
 {
  number = "2";
 }
 else if(a == false && b == false && c == true && d == true && e == true && f == true && g == true)
 {
  number = "3";
 }
 else if(a == false && b == true && c == false && d == false && e == true && f == true && g == true)
 {
  number = "4";
 }
 else if(a == false && b == true && c == true && d == true && e == true && f == true && g == false)
 {
  number = "5";
 }
 else if(a == true && b == true && c == true && d ==true  && e == true && f == true && g == false)
 {
  number = "6";
 }
 else if(a == false && b == false && c == true && d == false && e == true && f == false && g == true)
 {
  number = "7";
 }
 else if(a == true && b == true && c == true && d == true && e == true && f == true && g == true)
 {
  number = "8";
 }
 else if(a == false && b == true && c == true && d == true && e == true && f == true && g == true)
 {
  number = "9";
 }
 else if(a == true && b == true && c == true && d == true && e == true && f == false && g == true)
 {
  number = "0";
 }
 else if(a == true && b == true && c == false && d == true && e == false && f == false && g == false)
 {
  number = "L";
 }
 else if(a == false && b == false && c == false && d == false && e == false && f == false && g == false)
 {
  number = " ";
 }

 return number;
}

public void Writing(string text)
{
    if (textBox2.InvokeRequired)
    {
        Action safeWrite = delegate { Writing($"{text}"); };
        textBox2.Invoke(safeWrite);
    }
    else
        textBox2.Text = text;
}

}

public MainForm mainy2； public Form1(主窗体) { 初始化组件(); 主要 2 = 主要； }

    private void Form1_Load(object sender, EventArgs e)
    {
        this.Invoke((MethodInvoker)delegate
        {

            mainy2.Hide();
        });

    }

Answer 1

正如 Ralf 提到的，serialPort1_DataReceived 很可能会在后台线程上引发。但是你打电话给

Form1 formita = new Form1(this);
formita.Show();

虽然这不是严格禁止的，但这很可能不是一个好主意，因为您不会有多个表单与不同的 UI 线程相关联。

我假设冻结的原因是阻塞调用。它也可能导致各种同步问题：

if (textBox2.InvokeRequired)
    {
        Action safeWrite = delegate { Writing($"{text}"); };
        textBox2.Invoke(safeWrite);
    }

意味着后台线程将阻塞，直到 UI 线程完成更新。而且可能还会有一堆其他的同步问题。

我建议只从 UI 线程访问任何 UI 类。我认为这至少应该有所帮助。 IE。将上面的代码更改为

textBox2.BeginInvoke(() => {
Form1 formita = new Form1(this);
formita.Show();
});

C# 程序在串口读取时冻结打开新窗体

C# Program gets frozen opening a new form while serial port is reading

c#

multithreading

forms

visual-studio

serial-port