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();
delay(3000);
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!
            return;
    }
}

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
            return;
    }
}

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

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

and

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:

if (t1 - t2 + LONG_ENOUGH_DURATION < LONG_ENOUGH_DURATION)
    // 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;
    ATOMIC_BLOCK (ATOMIC_RESTORESTATE) {
        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)
    setMillis(-3000);

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.

Answer from Edgar Bonet on Stack Exchange
Top answer
1 of 4
187

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();
delay(3000);
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!
            return;
    }
}

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
            return;
    }
}

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

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

and

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:

if (t1 - t2 + LONG_ENOUGH_DURATION < LONG_ENOUGH_DURATION)
    // 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;
    ATOMIC_BLOCK (ATOMIC_RESTORESTATE) {
        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)
    setMillis(-3000);

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.

2 of 4
39

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.

🌐
Arduino Forum
forum.arduino.cc › projects › general guidance
Millis() roll over - General Guidance - Arduino Forum
June 27, 2025 - I need to delay an action by between 20 to 120 seconds on a esp32. I understand the millis() function can do this but I am unsure how the 49 day roll over works. It is my understanding that there is no telling what the initial value is. It is only used as a reference to compare against.
Discussions

Timing with millis() and rollover
I read that it's best to subtract the previous timestamp from the current one and compare it to an interval, but what I am wondering is how that works with rollover. As long as you subtract, it works perfectly. If I subtract the previous event from the current time, would that be a nevegive number? 3000-4,294,967,000= -4,294,964,000 which would not evaluate correctly for duration An unsigned integer cannot contain negative numbers. With uint32_t (unsigned long in arduino, return type of millis()), 3000 - 4294967000 = 3296 due to how unsigned subtraction overflows in basically all CPU cores - you can try it yourself. Or would it be like "reverse rollover" and be 3000-4,294,967,000=3296 which would be easily comparable to a duration time. Bingo, got a winner :) I frequently use if ((timeout - millis()) & ~(~0UL >> 1)) { timeout += time; doThings(); } in my code, which is a simple way to check if the result of the subtraction would have been negative if signed numbers were used. You could read this as if ((timeout - millis()) & 0x80000000) or if (((long) (timeout - millis())) < 0) if you prefer. More on reddit.com
🌐 r/arduino
7
3
February 24, 2021
millis() rollover
I’ve got a project where I need to keep the device running for months at a time without becoming unstable. I’m using millis() to keep a timer so actions happen on a schedule (default every 5 seconds). unsigned long time… More on community.sparkfun.com
🌐 community.sparkfun.com
8
0
October 3, 2012
Millis() roll over
It was discussed and demonstrated hundred times here in this forum. Use the unsigned subtraction like shown in the "Blink Without Delay" example and it will work. If you would like to know more about how and why ... search for "millis rollover" - in my opinion everthing was already said and ... More on forum.arduino.cc
🌐 forum.arduino.cc
7
0
June 27, 2025
How to test millis() rollover on ESP32
Hi. I have a sketch that deals with time intervals and I and would like to test if I have coded everything correctly. So I'd like to test the behavior when the number of milliseconds rolls over (when it exceeds `4294967295 and restarts from 0, which happens after about 49 days since startup). More on forum.arduino.cc
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0
0
November 21, 2021
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Norwegian Creations
norwegiancreations.com › 2018 › 10 › arduino-tutorial-avoiding-the-overflow-issue-when-using-millis-and-micros
Arduino Tutorial: Avoiding the Overflow Issue When Using millis() and micros() – Norwegian Creations
October 11, 2018 - Here we will get a buggy behavior after approximately 50 days when millis() will go from returning a very high number (close to (2^32)-1) to a very low number. This is known as overflow or rollover.
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Tech Explorations
techexplorations.com › home › tech explorations guides › /arduino/ › programming › 12. ​how to deal with the millis rollover
12. ​How to deal with the millis rollover - Tech Explorations
March 5, 2024 - In the first subtraction, both current millis and last millis are before the rollover. In the second subtraction, the millis register has rolled over, and the current millis is 3000. Feel free to try out different value to see how they are handled. Arduino Step by Step Getting Started is our ...
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Reddit
reddit.com › r/arduino › timing with millis() and rollover
r/arduino on Reddit: Timing with millis() and rollover
February 24, 2021 -

My project is mounted on a wall in a few miles away, so I can't test for myself, and I can figure out how to ask the question from Google correctly. I am working timing on my project that requires a 15 minute window. I am worried that rollover will mess with it. I read that it's best to subtract the previous timestamp from the current one and compare it to an interval, but what I am wondering is how that works with rollover. Since rollover happens after about 49 days, say there is an event that happens right before and so its timestamp is 4,294,967,000, then rollover happens and the current time is something like 3000.

If I subtract the previous event from the current time, would that be a nevegive number? 3000-4,294,967,000= -4,294,964,000 which would not evaluate correctly for duration

Or would it be like "reverse rollover" and be 3000-4,294,967,000=3296 which would be easily comparable to a duration time.

Top answer
1 of 5
3
I read that it's best to subtract the previous timestamp from the current one and compare it to an interval, but what I am wondering is how that works with rollover. As long as you subtract, it works perfectly. If I subtract the previous event from the current time, would that be a nevegive number? 3000-4,294,967,000= -4,294,964,000 which would not evaluate correctly for duration An unsigned integer cannot contain negative numbers. With uint32_t (unsigned long in arduino, return type of millis()), 3000 - 4294967000 = 3296 due to how unsigned subtraction overflows in basically all CPU cores - you can try it yourself. Or would it be like "reverse rollover" and be 3000-4,294,967,000=3296 which would be easily comparable to a duration time. Bingo, got a winner :) I frequently use if ((timeout - millis()) & ~(~0UL >> 1)) { timeout += time; doThings(); } in my code, which is a simple way to check if the result of the subtraction would have been negative if signed numbers were used. You could read this as if ((timeout - millis()) & 0x80000000) or if (((long) (timeout - millis())) < 0) if you prefer.
2 of 5
1
I am surprised I never see the simple solution I use, which works every time. I use signed longs for millis timestamps, and then subtract the timestamps and look only at whether the difference is negative. long target=millis()+1000; (do something) if (((long)millis() - a) >= 0) { ... whatever ... } -- basically, subtracting signed numbers will have a potential "underflow" problem but it is identical to the millis overflow problem except with opposite symptoms. The two symptoms cancel out and cease to exist, and this works every time without ever having to separately check for overflow.
🌐
Davidmac
davidmac.pro › posts › 2020-12-22-arduino-millis-roll-over
Arduino millis roll-over, and why it can ruin your project
December 22, 2020 - When the maximum number is reached (0xFFFFFFFF) and more time passes, it will roll-over back to 0 (0x00000000) and start again. It won't cause the Arduino to crash, lock-up, or anything like that, it'll just happen.
🌐
Faludi
faludi.com › 2007 › 12 › 18 › arduino-millis-rollover-handling
Arduino Millis() Rollover Handling – Rob Faludi
This can solve problems with servo routines, steppers, timed pauses and a variety of other calculations. In addition, because my millisRollover() function counts the number of times rollover has happened, it is now possible to record total Arduino runtime with a counter that’s good for over ...
Find elsewhere
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Wordpress
bigdanzblog.wordpress.com › 2016 › 05 › 15 › writing-code-to-handle-arduinos-millis-roll-over
Writing Code to Handle Arduino’s millis() Roll Over | Big Dan the Blogging Man
May 16, 2016 - It will take 49.7 days (4294967286 / 1000 / 60 / 60 / 24) for this to occur. If you want your Arduino project to run continuously for more than 49 days you will want to take this roll over into account.
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Bald Engineer
baldengineer.com › home › arduino: how do you reset millis() ?
Arduino: How do you reset millis() ? - Bald Engineer
March 31, 2022 - In HEX the maximum value is 0xFFFFFFFF. Add one more and it “rolls over” to zero. Hence the name “millis() rollover.” · Let’s be very clear: when millis() rolls over, your Arduino will not lock up. In fact the Arduino’s ATmega processors very rarely lock up.
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SparkFun Community
community.sparkfun.com › development boards › arduino
millis() rollover - Arduino - SparkFun Community
October 3, 2012 - I’ve got a project where I need to keep the device running for months at a time without becoming unstable. I’m using millis() to keep a timer so actions happen on a schedule (default every 5 seconds). unsigned long timer = 0; int pollRate = 5; // in seconds void setup() { analogReference(EXTERNAL); timer = millis(); } void loop() { if(millis() - timer >= (pollRate * 1000)) { // Do my stuff timer = millis(); } } I read that millis() will run on for about 50 days before it rol...
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Arduino Forum
forum.arduino.cc › projects › programming
How to test millis() rollover on ESP32 - Programming - Arduino Forum
November 21, 2021 - Hi. I have a sketch that deals with time intervals and I and would like to test if I have coded everything correctly. So I'd like to test the behavior when the number of milliseconds rolls over (when it exceeds `4294967295 and restarts from 0, which happens after about 49 days since startup).
🌐
Arduino Forum
forum.arduino.cc › projects › programming
Using % of millis() to Mitigate Rollover - Programming - Arduino Forum
April 18, 2017 - Hello, In my void loop() I want to branch off and do something once every 15 seconds. It isn't critical that it be done exactly every 15 seconds, but I want to not do it once I've done it for another 15 seconds. I thought the easiest way to do this would be to use millis() and compare to 15000, i.e. millis() when I ran the branch, to millis() each time I cross the check in my void loop() and when it's 15000 more, run the branch again.
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Black Electronics
black-electronics.com › posts › worried-about-millis-timer-overflow
Black Electronics | Worried about millis() timer overflow?
March 26, 2016 - millis() function return the time in millisecond since arduino was powered up. Sometimes also refered as arduino time in forums. this function returns unsigned long integer which is a 32 bit number which can hold up 0 to 4,294,967,295 (2^32 ...
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Arduino Forum
forum.arduino.cc › projects › programming
Is a more correct millis() rollover possible? - Programming - Arduino Forum
March 8, 2018 - I've done the math and the Using millis() for timing sticky post is only somewhat correct in stating that it will work through a rollover. It depends on your definition of what will work. Essentially the program will not crash. However if the currentMillis - ULONG_MAX (4294967295) = period.
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Arduino Forum
forum.arduino.cc › projects › programming
Millis overflow - Programming - Arduino Forum
September 11, 2012 - Please i would like to know does millis overflow (go back to zero), after approximately 50 days as i found here or it resets after 9 hours and 32 minutes. as i found here http://www.faludi.com/2007/12/18/arduino-milli…
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Arduino Forum
forum.arduino.cc › projects › programming
When variables rollover and safe millis() IF condition - Programming - Arduino Forum
July 23, 2016 - Hello recently I read this article www.baldengineer.com/arduino-how-do-you-reset-millis.html about how to be safe about generic millis timing if (millis() - previousMillis >= interval) as author predict it will rollover after 49 days, so solution ...
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Arduino Forum
forum.arduino.cc › forum 2005-2010 (read only) › software › troubleshooting
What is the millis() rollover time? - Troubleshooting - Arduino Forum
September 9, 2009 - Looking through documentation it appears that the timer count resets to zero after approximately 9 hours and 32 minutes. looking further: http://www.faludi.com/2007/12/18/arduino-millis-rollover-handling/ It appears that we now may have a * Improved millis(): it now overflows after 49 days instead of 9 hours.
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DeepBlue
deepbluembedded.com › home › blog › arduino millis() function (timer vs delay) tutorial
Arduino millis() Function (Timer vs delay) Tutorial
August 17, 2023 - The maximum value for the Arduino millis() function is 232-1 which is 4,294,967,295. This turns out to be 49.71 days before the millis() variable reaches overflow and rollovers back to zero and starts counting up again.
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Arduino Forum
forum.arduino.cc › projects › programming
Is a more correct millis() rollover possible? - #8 by odometer - Programming - Arduino Forum
March 9, 2018 - alto777: I found this to fix my problem. programming - How can I handle the millis() rollover? - Arduino Stack Exchange It includes a technique for testing a solution without waiting 56 days. I have another technique. It is to use micros() rather than millis(). That way, rollover happens within an hour and a quarter.