Re: Is comparison of doubles stable?

Hi,

This is somewhat tricky question:
- we have two doubles d1 and d2
- and d1 < d2
Is it guaranteed (and where exactly in C++/IEEE 754 standards?)
that this relationship will stay if I move d1 and/or d2 around in
memory?

I'll try to illustrate this in example:

double foo() { ... } // this is a "generator"

deque<double> v_sorted(1, 0.0); // populated with one value of 0.0
for(size_t i = 0; i < 1000; ++i)
{
   double d = foo();

   if (v_sorted.back() < d) v_sorted.push_back(d);
}

As you see I am trying to extract ordered sequence (without
duplicates, values > 0) from stream of values produced by foo().

Now -- is it guaranteed that I'll end up with what I seek (ordered,
no duplicates)? I feel answer is "yes", but then more important
question arises -- where are those guarantees?

Thank you.
Michael.

P.S. Why answer can be "no" -- because on x87 FPU register is 80bits
(+ hidden bit) and in-memory double is 64 bits. I.e. if compiler
does not round double produced by foo() before comparison we might
end up with d1 < d2. But later we unload value into memory in
push_back() call which will force rounding and d2 might become
equal to d1.