Comparing two double values --- Returning false or inconsistent

Are you a software developer or not? You should have indicated the type of a and b, because this is critically important. If the type is single, a == b, so you need to use the type double. The type single is not used often because of its precision insufficient for most problems, prefer using double.

Do you understand that it is not possible to create a type to represent any real number as it is understood in mathematics, because a real number may require infinite amount of memory?

—SA

It might also depend on your floating point options... as there are options to select between faster computation or more precise one (IEEE).

That being said, I have made a small test program

int i = 0;
float f = 0.601626;
double a = 1.0;
double b = a + f;

while (a < b)
{
    a *= 1.1;
    b = a + f;
    ++i;
}

That program fails after 386 iterations. At that point a is equal to 9 496 782 049 180 368.0 Thus it actually fails somewhere between 8.6e15 and 9.5e15 when adding 0.601626 to a double.

By the way 2^53 is equal to 9.001e15 (and it would probably fail at that point... Also a lot of imprecision would have been added if the additions are cumulatives).

Thus, I don't understand why the program does not works as expected as the OP numbers are around a million time smaller than that...

Also, how do you actuall print that number 12966127670.788626 as it has more than 15 significant digits so that number... Last 2 digits are probably incorrect. Are those values printed or inspected from code or the second value is what you think it should be.

By the way, are you always adding 0.601626 and in particular when it fails? And are you computation as simple as a simple addition. If they are somehow more complex (and you have inspected b, then there might some rounding errors you havent noticed).

Finally be aware that values printed are rounded so there can be a difference in the actual float value and what is displayed if you display 15 or more digits.

*** In you case, you might consider using 64 bits integers everywhere to avoid loosing some precision ***.

Binary arithmetic operators with mixed operand types convert the result to a common base type. See http://osr507doc.sco.com/en/tools/clang_conv_implicit.html#clang_conv_impl_arithmetic[^]
So your calculation of b will be temporarily converted to float, because duration is of type float:

C++

double b = trendpoint.m_Time.GetTotalSeconds() + duration;

is equivalent to

C++

double b = (double)((float)(trendpoint.m_Time.GetTotalSeconds()) + duration);

This means, if duration is very small compared to the first operand, it will just be cut off in this operation!

To solve this, explicitely cast your second operand to double, or, better yet, declare it type double right from the start.
P.S.: I just reread that article I linked to and apparently I misinterpreted it. Sorry, looks like this isn't itt after all.

Maybe you still should make duration type double, as I suggested, but I am no longer sure it would solve your problem.

I don't see where the time ever gets updated. It looks like you initialize the time once, and then compare the same values forever without ever getting the current time again.

It looks like you need to add a function.

while(a < b)
{
   //fetch next trendpoint;
    trendpoint.m_Time.GetCurrentTime();
    a = trendpoint.m_Time.GetTotalSeconds();
}