Arduino not adding decimals correctly on double variable

Question

I have encountered a strange behavior today. I can't understand why the sequence is altered after 26.1000. I get a number 0.0001 less than the desired value (26.2124 instead of 26.2125). I need precision and can't get it.

Here is the code:

double _pos = 0;

void setup() {
  Serial.begin(250000);
}

void loop() {
  _pos += (0.1125);
  Serial.println(_pos,4);
  while (_pos >26.5) delay(1);
}

and here is the result

If you want precision it's often better to work in fixed point, that way you can be sure of the significance of the least significant bits. — ratchet freak, Nov 13 '17 at 9:31

Michel Keijzers · Accepted Answer · 2017-11-13 09:49:09Z

1

I do not see any specific difference after 26.2124.

However, in your case you are using 6 digits (e.g. 26.5499). To use an unsigned long/uint32_t you can store easily 6 digits, but only using whole numbers.

However, by adding 1125 every time and dividing the result by 10,000.0 for printing you get exact results.

unsigned long _pos = 0;

void setup() {
  Serial.begin(250000);
}

void loop() {
  _pos += 1125;
  Serial.println((float) _pos / 10000.0,4);
  while (_pos > 265000) delay(1);
}

answered Nov 13 '17 at 9:49

Michel Keijzers

9,2396 gold badges22 silver badges45 bronze badges

Thanks, I used this approach, the only problem is that numbers can be negative too (something I did not clarify on the original question). – FedeKrum Nov 15 '17 at 8:13
Than remove unsigned, so you can have negative numbers. – Michel Keijzers Nov 15 '17 at 9:48

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Majenko · Accepted Answer · 2017-11-13 11:16:25Z

The 8 bit Arduinos don't have precision, or anything even close to precision when it comes to working with decimal values.

There is no such thing as double on an 8-bit Arduino - it is only an alias for float. By using double you aren't gaining any more precision than using float.

Even double isn't perfectly precise. It's more precise than float (on systems that support double), but still far from precise.

People that care about specific levels of precision code using variables of that precision. That's called using fixed point mathematics, and it's what DSPs work in. It's fast, efficient, and precise up to the point that you define it to be.

Fixed point mathematics works by operating in multiples of a pre-defined fraction. That fraction is your precision, or granularity and is the smallest value that can be represented. However, working with large numbers and fine granularity results in very large variables to store all that data - which is why for most things people use float: because it compresses the data down into a very small space - and like a highly compressed MP3 it loses a lot of the data.

Edgar Bonet · Accepted Answer · 2017-11-13 10:28:52Z

The number 0.1125 would written in binary as 0.0001110011001100..., with an infinite repetition of “1100”. As the float data type has only 24 bits of precision, it can obviously not represent such a number. Instead, when you write 0.1125 in code, you get the closest representable number, namely 0.11249999701976776123046875, which is 0.00011100110011001100110011 in binary. Note that AVR-based Arduinos do not really support double precision: you get a float32 whether you declare a float or a double.

Furthermore, when you keep adding this number to itself, quite often the sum is not a representable number. Thus, in addition to the initial rounding error in the representation of 0.1125, you get an additional rounding error with the addition. These errors accumulate and you end up with a value that is significantly off what you expect.

You can avoid accumulating errors by working with integers, and converting o float only when needed:

void loop()
{
    static long count;
    count++;
    Serial.println(count * 0.1125, 4);
    while (count > 235) delay(1);
}

Note that you still get two rounding errors: in the representation of 0.1125 and in the multiplication. But at least you are not accumulating a large number of them.

dannyf · Accepted Answer · 2017-11-13 11:10:47Z

0

I can't understand why the sequence is altered after 26.2124.

because floats / doubles are not precise representation of their respective values.

answered Nov 13 '17 at 11:10

dannyf

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