Re: Is comparison of doubles stable?

"Bo Persson" <>
Sun, 18 Jul 2010 14:28:36 CST
Michael Kilburn wrote:


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

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.

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.

The answer will be Yes if you use proper settings for you compiler, so
that it either doesn't use the 80 bit registers or forces the
appropriate rounding before comparisons. Some compilers have a default
setting that favors faster code, even if it means being slightly

Bo Persson

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