Reputation: 201
I'm using the tiva c 1294, the idea is to blink the led a given number of times, I already have done it here , but right now isn't working, all it does is blink one of the users leds (PF4) and when pressing instead of blinking 5 times and then turn off, it turns on the other led (PF0). When you stop pressing the switch 1 it gets back to blink continuously the first led.
#include <stdint.h>
#include <stdbool.h>
#include "inc/hw_memmap.h"
#include "driverlib/debug.h"
#include "driverlib/gpio.h"
#include "driverlib/sysctl.h"
#include <inc/tm4c1294ncpdt.h>
uint32_t i,j; //int 1
int main(void) {
SYSCTL_RCGCGPIO_R=0X1100; // set clock portn
i=SYSCTL_RCGCGPIO_R; // delay (more than 3 cycles)
j=0;
GPIO_PORTN_DIR_R=0X03; //enable the GPIO pin for the LED-PN0, set the direction as output, and
GPIO_PORTN_DEN_R=0X03; //enable the GPIO pin for digital function
GPIO_PORTJ_AHB_DIR_R=0;
GPIO_PORTJ_AHB_DEN_R=0X03;
GPIO_PORTJ_AHB_PUR_R=0X01;
while(1){
GPIO_PORTN_DATA_R &=~0X02; //turn led off
while (GPIO_PORTJ_AHB_DATA_R & 0X01){
GPIO_PORTN_DATA_R |=0X01; //turn led on
SysCtlDelay(2666666);
GPIO_PORTN_DATA_R &=~0X01; //turn led off again
SysCtlDelay(2666666);
}
for (j=0; i<5; i++)
{
GPIO_PORTN_DATA_R |=0X01; //turn led on
SysCtlDelay(2666666);
GPIO_PORTN_DATA_R &=~0X01; //turn led off again
SysCtlDelay(2666666);
}
GPIO_PORTN_DATA_R |=0X02; //clear the interrupt flag before return
}
}
Upvotes: 0
Views: 1186
Reputation: 12600
Instead of
for (j=0; i<5; i++)
write
for (j=0; j<5; j++)
Do you see your typo?
EDIT
Let's look at your endless loop. You wrote it like this (I corrected its white spaces a bit):
while (1) {
GPIO_PORTN_DATA_R &= ~0X02; //turn led off
while (GPIO_PORTJ_AHB_DATA_R & 0X01) {
GPIO_PORTN_DATA_R |= 0X01; //turn led on
SysCtlDelay(2666666);
GPIO_PORTN_DATA_R &= ~0X01; //turn led off again
SysCtlDelay(2666666);
}
for (j = 0; j < 5; j++) {
GPIO_PORTN_DATA_R |= 0X01; //turn led on
SysCtlDelay(2666666);
GPIO_PORTN_DATA_R &= ~0X01; //turn led off again
SysCtlDelay(2666666);
}
GPIO_PORTN_DATA_R |= 0X02; //clear the interrupt flag before return
}
First we correct the last comment and add LED numbers based on their bit position. Then let's replace GPIO_PORTN_DATA_R |= 0x01
with SwitchLedOn(0)
and GPIO_PORTN_DATA_R &= ~0x01
with SwitchLedOff(0)
. And let's do the same with GPIO_PORTN_DATA_R
/ 0x02
and SwitchLedO*(1)
. Next we can replace GPIO_PORTJ_AHB_DATA_R & 0X01
with IsSwitchUp(0)
.
while (1) {
SwitchLedOff(1); //turn led 1 off
while (IsSwitchUp(0)) {
SwitchLedOn(0); //turn led 0 on
SysCtlDelay(2666666);
SwitchLedOff(0); //turn led 0 off again
SysCtlDelay(2666666);
}
for (j = 0; j < 5; j++) {
SwitchLedOn(0); //turn led 0 on
SysCtlDelay(2666666);
SwitchLedOff(0); //turn led 0 off again
SysCtlDelay(2666666);
}
SwitchLedOn(1); //turn led 1 on
}
Now it's clear to see. As the source is written there are two equal statement sequences that blink LED 0 one time. The inner while
-loop repeats this as long as switch 0 is not pressed. If the switch is pressed after a blink period the loop is left and the for
-loop starts. It lets the same LED blink 5 times. Then LED 1 is switched on. But just for a very short time because the outer while
-loop is repeated which switches off LED 1 right at the beginning.
If the switch is still pressed the inner while
-loop will be skipped going straight to the for
-loop. If it is nor pressed anymore the inner while
-loop will wait for the next press.
Anyway, we can not see at LED 0 where we are in the code.
Upvotes: 2