Button activated UART message

Question

"I'm trying to use the button that when it is pressed, it triggers to send a string message via UART, but upon running the code, it always detect the button to be "always pressed" which is not ideal. I need help fixing my code. Thank you very much.

I'm using EasyPIC fusion v7. The device is P32MX795F512L @80MHz. I'm using the MikroC pro for PIC32 as my IDE."

"Button and UART library is used for this code."

char read;
unsigned int oldstate;


void main() {
  AD1PCFG = 0XFFFF;             //SETTING AN INPUTS AS DIGITAL I/O
  JTAGEN_bit = 0;               //disable JTAG


  UART2_Init(9600);
  Delay_ms(1000);
  TRISA = 1;                    //setting All portA as inputs

  UART_Set_Active(&UART2_Read, &UART2_Write, &UART2_Data_Ready, &UART2_Tx_Idle);    //Sets UART2 as active
  UART_Write_Text("UART is now ready.");
  UART_Write(13);
  UART_Write(10);

   while(1)
   {
    if (Button(&PORTA, 15, 1 ,1))            //detect if button is pressed
    {
        UART_Write_Text("Button is pressed.");
        UART_Write(13);
        UART_Write(10);
        oldstate = 1;

    }

    if (oldstate && Button(&PORTA, 15, 1, 0)) //detect from logic 1 to 0
    {
        UART_Write_Text("Button is pressed again.");
        UART_Write(13);
        UART_Write(10);
        oldstate = 0;

    }
   }
}

"I've only learned how UART works and how the Button library works.

I expect the output to be 'Button is pressed' when the button is pressed first, then "Button is pressed again" when I press the same button again.

The output for the code is always 'Button is pressed' and prints continuously"

Does it stop when the button is pressed? Could be the type of button. — Deanie, Jul 09 '19 at 04:06
Are you sure you are not simply missing the (single) iteration printing the "again" line among all the others? You reset `oldstate` in the block printing "again". — Hulk, Jul 09 '19 at 04:56
it does not stop. My solution for now is to specify which port is the input (RA15). I'll try your suggestions tomorrow. Thank you very much — Dorokun192, Jul 09 '19 at 09:12

score 0 · Answer 1 · answered Jul 09 '19 at 05:26

0

The main problem with your code is that this line:

UART_Write_Text("Button is pressed again.");

should in fact read:

UART_Write_Text("Button is RELEASED.");

Other than that you might be having a hardware issue if you're not tying your RA15 pin to GND through a pull-down resistor. You could also use the internal pull-ups on your PIC.

answered Jul 09 '19 at 05:26

Marcos G.

3,371
2
8
16

The solution that I've found for now is setting a specific port (RA15) as an input. But tomorrow I'll try setting the internal pull-up resistors. Thank you very much. – Dorokun192 Jul 09 '19 at 09:10
Now that you mention it, I think your `TRISA=1` should be `TRISA = 0b11111111` or `TRISA = 0xFF`, right? – Marcos G. Jul 09 '19 at 09:16
Yes, I have learned the difference between using "1" and using "0xFFFF". It seems that I have to use "1" only when specifying the Port that I'm going to use. Thank you very much for your support! – Dorokun192 Jul 09 '19 at 23:32

score 0 · Answer 2 · answered Jul 09 '19 at 11:02

0

The problem is here:

TRISA = 1;

This is simular to TRISA = 0x0001; and will only make the Port A0 an input. You had to write:

TRISA = 0xFFFF;`

answered Jul 09 '19 at 11:02

Mike

4,041
6
20
37

1

Yes this is what I've learned for now. I think the proper way of using "1" is to specify the port that I'll use (TRISA15_bit = 1). Thank you very much for this answer. I'm still learning and I think I have a long way to go – Dorokun192 Jul 09 '19 at 23:30

Button activated UART message

2 Answers2