Question

我有覆盆子pi和arduino。到目前为止，我已经能够使用串行通信将数据发送到arduino，但它只发送一个变量，我有多个变量（2），我想发送到arduino（x，y坐标）。有谁知道这是否可行。我希望从pi发送的第一个数字是x，第二个数字是y，下一个是下一个coord等的x。

我已经尝试编辑我用来发送一个变量的代码，但它不起作用。

任何帮助都很棒

Answer 1

考虑以下方法同时发送2个变量：

    int xpos, ypos;

    char x_tx_buffer[20], y_tx_buffer[20];
    char x_dummy_buffer[20];
    char y_dummy_buffer[20];
    char *p_x_tx_buffer, *p_y_tx_buffer;

    sprintf(x_dummy_buffer,"%d", xposs);
    sprintf(y_dummy_buffer,"%d", yposs);

    p_x_tx_buffer = &x_tx_buffer[0];
    *p_x_tx_buffer++ = x_dummy_buffer[0];
    *p_x_tx_buffer++ = x_dummy_buffer[1];
    *p_x_tx_buffer++ = x_dummy_buffer[2];
    *p_x_tx_buffer++ = x_dummy_buffer[3];

    p_y_tx_buffer = &y_tx_buffer[0];
    *p_y_tx_buffer++ = y_dummy_buffer[0];
    *p_y_tx_buffer++ = y_dummy_buffer[1];
    *p_y_tx_buffer++ = y_dummy_buffer[2];
    *p_y_tx_buffer++ = y_dummy_buffer[3];
    uart0_filestream = open("/dev/ttyAMA0", O_RDWR | O_NOCTTY | O_NDELAY);      //Open in non blocking read/write mode
    if (uart0_filestream == -1)
    {
        //ERROR - CAN'T OPEN SERIAL PORT
        printf("Error - Unable to open UART.  Ensure it is not in use by another application\n");
    }
    if (uart0_filestream != -1)
    {
        int countx = write(uart0_filestream, &x_tx_buffer[0], (p_x_tx_buffer - &x_tx_buffer[0]));       //Filestream, bytes to write, number of bytes to write
        int county = write(uart0_filestream, &y_tx_buffer[0], (p_y_tx_buffer - &y_tx_buffer[0]));       //Filestream, bytes to write, number of bytes to write
        if (countx < 0 || county < 0)
        {
            printf("UART TX error\n");
        }
    }
    close(uart0_filestream);

您一次最多可以发送8个字节。记住这一点，你可以修改about代码，在同一个uart0_filestream中发送你的x和y值。

祝你好运。

Arduino和Raspberry Pi串口通讯+多个变量

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