I need to read a sensor every five minutes, but since my sketch also has other tasks to do, I cannot just delay() between the readings. There is the Blink without delay tutorial suggesting I code along these lines:

void loop()
    unsigned long currentMillis = millis();

    // Read the sensor when needed.
    if (currentMillis - previousMillis >= interval) {
        previousMillis = currentMillis;

    // Do other stuff...

The problem is that millis() is going to roll over back to zero after roughly 49.7 days. Since my sketch is intended to run for longer than that, I need to make sure the rollover does not make my sketch fail. I can easily detect the rollover condition (currentMillis < previousMillis), but I am not sure what to do then.

Thus my question: what would be the proper/simplest way to handle the millis() rollover?

  • 10
    Editorial note: This is not exactly a question of mine, rather a tutorial in a question/answer format. I have witnessed a lot of confusion on the Internet (including here) about this topic, and this site seems like the obvious place to search for an answer. This is why I am providing this tutorial here. Commented Jun 12, 2015 at 11:18
  • 3
    I would do previousMillis += interval instead of previousMillis = currentMillis if I wanted a certain frequency of results.
    – Jasen
    Commented Feb 7, 2016 at 9:54
  • 5
    @Jasen: That's right! previousMillis += interval if you want constant frequency and are sure that your processing takes less than interval, but previousMillis = currentMillis for guaranteeing a minimum delay of interval. Commented Feb 7, 2016 at 10:02
  • We really need a FAQ for things like this.
    – user31481
    Commented Aug 11, 2017 at 10:47
  • One of the "tricks" I use is to lighten the load on the arduino by using the smallest int that contains the interval. For instance, for max 1 minute intervals, I write uint16_t previousMillis; const uint16_t interval = 45000; ... uint16_t currentMillis = (uint16_t) millis(); if ((currentMillis - previousMillis) >= interval) ...
    – frarugi87
    Commented Nov 30, 2018 at 14:42

4 Answers 4


Short answer: do not try to “handle” the millis rollover, write rollover-safe code instead. Your example code from the tutorial is fine. If you try to detect the rollover in order to implement corrective measures, chances are you are doing something wrong. Most Arduino programs only have to manage events that span relatively short durations, like debouncing a button for 50 ms, or turning a heater on for 12 hours... Then, and even if the program is meant to run for years at a time, the millis rollover should not be a concern.

The correct way to manage (or rather, avoid having to manage) the rollover problem is to think of the unsigned long number returned by millis() in terms of modular arithmetics. For the mathematically inclined, some familiarity with this concept is very useful when programming. You can see the math in action in Nick Gammon's article millis() overflow ... a bad thing?. For the problem at hand, what's important to know is that in modular arithmetics the numbers “wrap around” when reaching a certain value – the modulus – so that 1 − modulus is not a negative number but 1 (think of a 12 hour clock where the modulus is 12: here 1 − 12 = 1).

For those who do not want to go through the computational details, I offer here an alternative (hopefully simpler) way of thinking about it. It is based on the simple distinction between instants and durations. As long as your tests only involve comparing durations, you should be fine.

Note on micros(): Everything said here about millis() applies equally to micros(), except for the fact that micros() rolls over every 71.6 minutes, and the setMillis() function provided below does not affect micros().

Instants, timestamps and durations

When dealing with time, we have to make the distinction between at least two different concepts: instants and durations. An instant is a point on the time axis. A duration is the length of a time interval, i.e. the distance in time between the instants that define the start and the end of the interval. The distinction between these concepts is not always very sharp in everyday language. For example, if I say “I will be back in five minutes”, then “five minutes” is the estimated duration of my absence, whereas “in five minutes” is the instant of my predicted coming back. Keeping the distinction in mind is important, because it is the simplest way to entirely avoid the rollover problem.

The return value of millis() could be interpreted as a duration: the time elapsed from the start of the program until now. This interpretation, however, breaks down as soon as millis overflows. It is generally far more useful to think of millis() as returning a timestamp, i.e. a “label” identifying a particular instant. It could be argued that this interpretation suffers from these labels being ambiguous, as they are reused every 49.7 days. This is, however, seldom a problem: in most embedded applications, anything that happened 49.7 days ago is ancient history we do not care about. Thus, recycling the old labels should not be an issue.

Do not compare timestamps

Trying to find out which among two timestamps is greater than the other does not make sense. Example:

unsigned long t1 = millis();
unsigned long t2 = millis();
if (t2 > t1) { ... }

Naively, one would expect the condition of the if () to be always true. But it will actually be false if millis overflows during delay(3000). Thinking of t1 and t2 as recyclable labels is the simplest way to avoid the error: the label t1 has clearly been assigned to an instant prior to t2, but in 49.7 days it will be reassigned to a future instant. Thus, t1 happens both before and after t2. This should make clear that the expression t2 > t1 makes no sense.

But, if these are mere labels, the obvious question is: how can we do any useful time calculations with them? The answer is: by restricting ourselves to the only two calculations that make sense for timestamps:

  1. later_timestamp - earlier_timestamp yields a duration, namely the amount of time elapsed between the earlier instant and the later instant. This is the most useful arithmetic operation involving timestamps.
  2. timestamp ± duration yields a timestamp which is some time after (if using +) or before (if −) the initial timestamp. Not as useful as it sounds, since the resulting timestamp can be used in only two kinds of calculations...

Thanks to modular arithmetics, both of these are guaranteed to work fine across the millis rollover, at least as long as the delays involved are shorter than 49.7 days.

Comparing durations is fine

A duration is just the amount of milliseconds elapsed during some time interval. As long as we do not need to handle durations longer than 49.7 days, any operation that physically makes sense should also make sense computationally. We can, for example, multiply a duration by a frequency to get a number of periods. Or we can compare two durations to know which one is longer. For example, here are two alternative implementations of delay(). First, the buggy one:

void myDelay(unsigned long ms) {          // ms: duration
    unsigned long start = millis();       // start: timestamp
    unsigned long finished = start + ms;  // finished: timestamp
    for (;;) {
        unsigned long now = millis();     // now: timestamp
        if (now >= finished)              // comparing timestamps: BUG!

And here is the correct one:

void myDelay(unsigned long ms) {              // ms: duration
    unsigned long start = millis();           // start: timestamp
    for (;;) {
        unsigned long now = millis();         // now: timestamp
        unsigned long elapsed = now - start;  // elapsed: duration
        if (elapsed >= ms)                    // comparing durations: OK

Most C programmers would write the above loops in a terser form, like

while (millis() < start + ms) ;  // BUGGY version


while (millis() - start < ms) ;  // CORRECT version

Although they look deceptively similar, the timestamp/duration distinction should make clear which one is buggy and which one is correct.

What if I really need to compare timestamps?

Better try to avoid the situation. If it is unavoidable, there is still hope if it is known that the respective instants are close enough: closer than 24.85 days. Yes, our maximum manageable delay of 49.7 days just got cut in half.

The obvious solution is to convert our timestamp comparison problem into a duration comparison problem. Say we need to know whether instant t1 is before or after t2. We choose some reference instant in their common past, and compare the durations from this reference until both t1 and t2. The reference instant is obtained by subtracting a long enough duration from either t1 or t2:

unsigned long reference_instant = t2 - LONG_ENOUGH_DURATION;
unsigned long from_reference_until_t1 = t1 - reference_instant;
unsigned long from_reference_until_t2 = t2 - reference_instant;
if (from_reference_until_t1 < from_reference_until_t2)
    // t1 is before t2

This can be simplified as:

    // t1 is before t2

It is tempting to simplify further into if (t1 - t2 < 0). Obviously, this does not work, because t1 - t2, being computed as an unsigned number, cannot be negative. This, however, although not portable, does work:

if ((signed long)(t1 - t2) < 0)  // works with gcc
    // t1 is before t2

The keyword signed above is redundant (a plain long is always signed), but it helps make the intent clear. Converting to a signed long is equivalent to setting LONG_ENOUGH_DURATION equal to 24.85 days. The trick is not portable because, according to the C standard, the result is implementation defined. But since the gcc compiler promises to do the right thing, it works reliably on Arduino. If we wish to avoid implementation defined behavior, the above signed comparison is mathematically equivalent to this:

#include <limits.h>

if (t1 - t2 > LONG_MAX)  // too big to be believed
    // t1 is before t2

with the only problem that the comparison looks backwards. It is also equivalent, as long as longs are 32-bits, to this single-bit test:

if ((t1 - t2) & 0x80000000)  // test the "sign" bit
    // t1 is before t2

The last three tests are actually compiled by gcc into the exact same machine code.

How do I test my sketch against the millis rollover

If you follow the precepts above, you should be all good. If you nevertheless want to test, add this function to your sketch:

#include <util/atomic.h>

void setMillis(unsigned long ms)
    extern unsigned long timer0_millis;
        timer0_millis = ms;

and you can now time-travel your program by calling setMillis(destination). If you want it to go through the millis overflow over and over again, like Phil Connors reliving Groundhog Day, you can put this inside loop():

// 6-second time loop starting at rollover - 3 seconds
if (millis() - (-3000) >= 6000)

The negative timestamp above (-3000) is implicitly converted by the compiler to an unsigned long corresponding to 3000 milliseconds before the rollover (it is converted to 4294964296).

What if I really need to track very long durations?

If you need to turn a relay on and turn it off three months later, then you really need to track the millis overflows. There are many ways to do so. The most straightforward solution may be to simply extend millis() to 64 bits:

uint64_t millis64() {
    static uint32_t low32, high32;
    uint32_t new_low32 = millis();
    if (new_low32 < low32) high32++;
    low32 = new_low32;
    return (uint64_t) high32 << 32 | low32;

This is essentially counting the rollover events, and using this count as the 32 most significant bits of a 64 bit millisecond count. For this counting to work properly, the function needs to be called at least once every 49.7 days. However, if it is only called once per 49.7 days, for some cases it is possible that the check (new_low32 < low32) fails and the code misses a count of high32. Using millis() to decide when to make the only call to this code in a single "wrap" of millis (a specific 49.7 day window) could be very hazardous, depending on how the time frames line up. For safety, if using millis() to determine when to make the only calls to millis64(), there should be at least two calls in every 49.7 day window.

Keep in mind, though, that 64 bit arithmetic is expensive on the Arduino. It may be worth to reduce the time resolution in order to stay at 32 bits.

  • 4
    So, you are saying that the code written in the question will actually work correctly?
    – Jasen
    Commented Feb 7, 2016 at 9:48
  • 8
    @Jasen: Exactly! I've seem more than once people trying to “fix” the problem that did not exist in the first place. Commented Feb 7, 2016 at 10:00
  • 2
    I'm glad I found this. I've had this question before. Commented Jul 2, 2016 at 1:28
  • 3
    One of the best and most useful answers on StackExchange! Thanks a lot! :)
    – Falko
    Commented May 18, 2018 at 12:43
  • 2
    This is such an amazing answer to the question. I come back to this answer basically once a year because I am paranoid of messing rollovers. Commented Sep 4, 2019 at 16:41

TL;DR Short version:

An unsigned long is 0 to 4,294,967,295 (2^32 - 1).

So lets say previousMillis is 4,294,967,290 (5 ms before rollover), and currentMillis is 10 (10ms after rollover). Then currentMillis - previousMillis is actual 16 (not -4,294,967,280) since the result will be calculated as an unsigned long (which can't be negative, so itself will roll around). You can check this simply by:

Serial.println( ( unsigned long ) ( 10 - 4294967290 ) ); // 16

So the above code will work perfectly fine. The trick is to always calculate the time difference, and not compare the two time values.

  • How about a 15ms before rollover and a 10ms after rollover (ie 49.7 days after). 15 > 10, but the 15ms stamp is nearly a month and a half old. 15-10 > 0 and 10-15 > 0 unsigned logic, so that is no use here!
    – xyz
    Commented Jun 12, 2015 at 21:15
  • @prakharsingh95 10ms-15ms will become ~49.7 days - 5ms, which is the correct difference. The math works until millis() rolls over twice, but that is very unlikely to occur to the code in question. Commented Jun 12, 2015 at 22:16
  • Let me rephrase. Suppose you have two timestamps 200ms and 10ms. How do you tell which is(are) rolled over?
    – xyz
    Commented Jun 12, 2015 at 22:28
  • @prakharsingh95 The one stored in previousMillis has to have been measured before currentMillis, so if currentMillis is smaller than previousMillis a rollover occurred. The math happens to work out that unless two rollovers have occurred, you don't even need to think about it. Commented Jun 12, 2015 at 22:35
  • 1
    Ah, ok. if you do t2-t1, and if you can guarantee t1 is measured before t2 then it's equivalent to signed (t2-t1)% 4,294,967,295, hence the auto wraparound. Nice!. But what if there are two rollovers, or interval is > 4,294,967,295?
    – xyz
    Commented Jun 12, 2015 at 23:11

I loved this question, and the great answers it generated. First a quick comment on a previous answer (I know, I know, but I don't have the rep to comment yet. :-).

Edgar Bonet's answer was amazing. I've been coding for 35 years, and I learned something new today. Thank you. That said, I believe the code for "What if I really need to track very long durations?" breaks unless you call millis64() at least once per rollover period. Really nitpicky, and unlikely to be an issue in a real-world implementation, but there you go.

Now, if you really did want timestamps covering any sane time range (64-bits of milliseconds is about half a billion years by my reckoning), it looks simple to extend the existing millis() implementation to 64 bits.

These changes to attinycore/wiring.c (I'm working with the ATTiny85) seem to work (I'm assuming the code for other AVRs is very similar). See the lines with the //BFB comments, and the new millis64() function. Clearly it's going to be both bigger (98 bytes of code, 4 bytes of data) and slower, and as Edgar pointed out, you almost certainly can accomplish your goals with just a better understanding of unsigned integer math, but it was an interesting exercise.

volatile unsigned long long timer0_millis = 0;      // BFB: need 64-bit resolution

#if defined(__AVR_ATtiny24__) || defined(__AVR_ATtiny44__) || defined(__AVR_ATtiny84__)
    // copy these to local variables so they can be stored in registers
    // (volatile variables must be read from memory on every access)
    unsigned long long m = timer0_millis;       // BFB: need 64-bit resolution
    unsigned char f = timer0_fract;

    m += MILLIS_INC;
    f += FRACT_INC;
    if (f >= FRACT_MAX) {
        f -= FRACT_MAX;
        m += 1;

    timer0_fract = f;
    timer0_millis = m;

// BFB: 64-bit version
unsigned long long millis64()
    unsigned long long m;
    uint8_t oldSREG = SREG;

    // disable interrupts while we read timer0_millis or we might get an
    // inconsistent value (e.g. in the middle of a write to timer0_millis)
    m = timer0_millis;
    SREG = oldSREG;

    return m;
  • 4
    You are right, my millis64() only works if it is called more frequently than the rollover period. I edited my answer to point out this limitation. Your version does not have this issue, but it carries another drawback: it does 64-bit arithmetics in interrupt context, which occasionally increases the latency in responding to other interrupts. Commented Jun 8, 2018 at 17:14

Wrap the millis() in a class!


  1. Use id's instead of millis() directly.
  2. Compare reversals using id's. This is clean and rollover independent.
  3. For specific applications, to calculate exact difference between two id's, keep track of reversals and stamps. Calculate the difference.

Keeping track of reversals:

  1. Update a local stamp periodically faster than millis(). This will help you find out if millis() has overflown.
  2. The period of the timer determines accuracy
class Timer {

    static long last_stamp;
    static long *stamps;
    static int *reversals;
    static int count;
    static int reversal_count;

    static void setup_timer() {
        // Setup Timer2 overflow to fire every 8ms (125Hz)
        //   period [sec] = (1 / f_clock [sec]) * prescale * (255-count)
        //                  (1/16000000)  * 1024 * (255-130) = .008 sec

        TCCR2B = 0x00;        // Disable Timer2 while we set it up

        TCNT2  = 130;         // Reset Timer Count  (255-130) = execute ev 125-th T/C clock
        TIFR2  = 0x00;        // Timer2 INT Flag Reg: Clear Timer Overflow Flag
        TIMSK2 = 0x01;        // Timer2 INT Reg: Timer2 Overflow Interrupt Enable
        TCCR2A = 0x00;        // Timer2 Control Reg A: Wave Gen Mode normal
        TCCR2B = 0x07;        // Timer2 Control Reg B: Timer Prescaler set to 1024

        count = 0;
        stamps = new long[50];
        reversals = new int [10];
        reversal_count =0;

    static long get_stamp () {
        stamps[count++] = millis();
        return count-1;

    static bool compare_stamps_by_id(int s1, int s2) {
        return s1 > s2;

    static long long get_stamp_difference(int s1, int s2) {
        int no_of_reversals = 0;
        for(int j=0; j < reversal_count; j++)
        if(reversals[j] < s2 && reversals[j] > s1)
        return stamps[s2]-stamps[s1] + 49.7 * 86400 * 1000;       


long Timer::last_stamp;
long *Timer::stamps;
int *Timer::reversals;
int Timer::count;
int Timer::reversal_count;

ISR(TIMER2_OVF_vect) {

    long stamp = millis();
    if(stamp < Timer::last_stamp) // reversal
        Timer::reversals[Timer::reversal_count++] = Timer::count;
        ; // no reversal
    Timer::last_stamp = stamp;    
    TCNT2 = 130;     // reset timer ct to 130 out of 255
    TIFR2 = 0x00;    // timer2 int flag reg: clear timer overflow flag

// Usage

void setup () {

    long s1 = Timer::get_stamp();
    long s2 = Timer::get_stamp();

    Timer::compare_stamps_by_id(s1, s2); // true

    Timer::get_stamp_difference(s1, s2); // return true difference, taking into account reversals

Timer credits.

  • 10
    I edited the code to remove the maaaaany errors that prevented it to compile. This stuff will cost you about 232 bytes of RAM and two PWM channels. It will also start to corrupt the memory after you get_stamp() 51 times. Comparing delays instead of timestamps will certainly be more efficient. Commented Jun 13, 2015 at 13:00
  • Isn't that full static class effectively like namespace? Commented Aug 10, 2023 at 10:50

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