siteee.blogg.se

1. #Arduino while loop timeout serial#
2. #Arduino while loop timeout update#
3. #Arduino while loop timeout code#

Instead, the address can be saved in the eeprom and printed next time the program starts.

#Arduino while loop timeout serial#

Of course, if the program has just crashed it is going to be a bit risky trying to send the address out the serial port.

#Arduino while loop timeout code#

That’s the address where the Arduino was executing code when the watchdog fired…and a good place to look for the bug. Why? When an interrupt is fired the micro pushes the current program counter onto the stack. It is the two interrupt modes that are useful for detecting lockups. Then, if wdt_reset() isn’t called within another timeout period, the program will reset and start again If wdt_reset() isn’t called within the timeout period, the watchdog timer will first raise an interrupt.

#Arduino while loop timeout update#

Update the bootloader using Arduino IDE version 1.0.4 (or later) and all should be well.

If you turn on the watchdog timer on this board, it can get stuck in the bootloader. Otherwise wrong value can be read to main code.A word of caution here: there is a bug in the early Mega2560 bootloader. Volatile long count_2 = 0 //interrupt counter 2Īccess to variables changed in ISR from main code have to be does in atomic style. volatile long count_1 = 0 //interrupt counter 1 This is hint for compiler that variable can change any time and compiler do not use optimization for access to variable. Variables used in interrupts and in main code have to be declared as volatile. It is means values is printing approximately two times per second

how many increment is done in 1s? Estimate one loop duration to 10µs, so it is 100000 increments.Type of count_3 is int, so every 2^16(65536) increments overflow to 0 and then is again 1. next time count_3 is incremented only in if statement.count_3 is incremented in if statement and then is incremented in while statement, some data is printed.if sensorVal goes to LOW you must know how long it takes.loop() is called again, again and again in never ending loop.Int sensorVal = digitalRead(4) //read pin 4Ĭount_3++ //increase count so program enters while loop onceĬount_3++ //increase count3 so it never prints again

put your main code here, to run repeatedly: Serial.begin(9600) // open the serial port at 9600 bps:ĪttachInterrupt(digitalPinToInterrupt(2), increment_1, RISING) //Enable InterruptĪttachInterrupt(digitalPinToInterrupt(3), increment_2, RISING) //Enable Interrupt put your setup code here, to run once: Int count_3 = 0 //counter to trigger serial print How can it enter the while loop when it is larger than 1?! I must be missing something obvious. I've printed the values of count_3 inside the while loop to show that it is larger than 1 and it still prints when count_3 is larger than 1. I wrote a while loop to make sure that the values are printed when count_3 =1, but even when count_3 is larger than 1 the values still print. When I pull pin 4 low I want to print the number of counts from pins 2 and 3 once and not multiple times. I wrote code for an Arduino Mega to count the number of times a rising edge occurs on pins 2 and 3 (interrupts).