XOR operation

[LadyInRed 0]

Hi.

I want to made a simple program who turn on/off a led.

In the function below i must use the XOR operator in line  : GPIOPinWrite(GPIO_PORTF_BASE, GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3, 8);

 

How can i write the XOR operation ? 

 

void Timer0BIntHandler()

{

        TimerIntClear(TIMER0_BASE, TIMER_TIMB_TIMEOUT);

        counter++;

        if(counter == NUMBER_OF_INTS)

        {

              GPIOPinWrite(GPIO_PORTF_BASE, GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3, 8);

              counter = 0;

        }

}

[Lyon 3]

Hi,

As far as I know, a LED has only one pin to act on (beside the other which is at fixed potential, either Hi or Lo) - why do you "must" and on what pin is the LED mounted on? what about the other pins in your expression?

L

[pabigot 355]

I don't believe driverware provides an API that supports XOR, only setting or clearing all the specified pins. The following hack will invert all three LEDs on the ek-tm4c123gxl in one step:

    ((uint32_t *)GPIO_PORTF_BASE)[GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3] ^= ~0;

or the equivalent:

    HWREG(GPIO_PORTF_BASE + (GPIO_O_DATA + ((GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3) << 2))) ^= ~0;

which uses material from <inc/hw_types.h> and <inc/hw_gpio.h> and is closer to how GPIOPinWrite() is implemented.

[Lyon 3]

Hi,

Here is my solution, for changing pins 1 and 2. Adapt for your case:

 

static uint8_t ledonoff;

GPIOPinWrite(GPIO_PORTF_BASE, \

                        (GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3), \

                         ledonoff^=(GPIO_PIN_1 | GPIO_PIN_2));

L

[LadyInRed 0]

I found a solution : 

 

void Timer0BIntHandler()

{

        TimerIntClear(TIMER0_BASE, TIMER_TIMB_TIMEOUT);

 

        COUNTER++;

   

        GPIOPinWrite(GPIO_PORTF_BASE, GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3, LED);

 

        if(COUNTER == NUMBER_OF_INTERRUPTS)

        {

              LED ^= 0x08;

              COUNTER = 0;

        }

}

[LadyInRed 0]

Where LED = 0x08;

 

 

void Timer0BIntHandler()

{

        TimerIntClear(TIMER0_BASE, TIMER_TIMB_TIMEOUT);

 

        COUNTER++;

   

        GPIOPinWrite(GPIO_PORTF_BASE, GPIO_PIN_1 | GPIO_PIN_2 | GPIO_PIN_3, LED);

 

        if(COUNTER == NUMBER_OF_INTERRUPTS)

        {

              LED ^= 0x08;

              COUNTER = 0;

        }

}

[spirilis 1,265]

Yup that works.

[pabigot 355]

Agreed, that works, assuming LED is a statically allocated variable and nothing else changes either it or the corresponding pin's output value. I dislike caching a last-known state when the actual current state can easily be read, but that's a combination of personal choice and the application requirements.

[Rickta59 589]

The GPIO peripherals on some other brands of ARM controllers (NXP LPC8xx chips come to mind)  have a register that lets you toggle pins atomically by setting a register.  See the NOT0 register on page 91 for details:

 

http://www.nxp.com/documents/user_manual/UM10601.pdf

 

I tried to find something like this for the TI chip but nothing jumped out at me.  

 

-rick