double or float comparison [duplicate]

Question

This question already has answers here:

How do you compare float and double while accounting for precision loss? (32 answers)

Closed 12 years ago.

I've seen posts like:

What is the most effective way for float and double comparison?

Compare two floats

And many other related posts.

I saw in d3js library, it uses the following comparison:

 return a < b ? -1 : a > b ? 1 : a >= b ? 0 : NaN;

Is it OK to use this in C/C++ to do the comparison of double and float?

• Drew Dormann: The questions you link attempt to explain why not. Was there something mentioned in them that seems unclear?
• Joseph Mansfield: Sufficient for what? Testing equality or testing equality with an epsilon? The answer is no, either way.
• user529758: @sftrabbit Then that branching you just used could be optimized away to a simple xor eax, eax :P
• Drew Dormann: "UPDATED:..." Then what you wrote is functionally no different from a == b.
• franji1: Algebraically, (1.0/3.0) * 3.0 = 1.0, but the same cannot be said of doubles or floating point arithmetic in a computer. Just imagine how a PC with limited memory can store an infinitely long rational number 0.333333333333333333333333.... It MUST terminate at some point. Let's say it ends at 0.33. When you multiply 0.33 * 3, you might get 0.99. You might even get 0.98. You could get 1.00. Two of those do NOT "equal" 1.0.

Answer 1

is the following code sufficient?

No.

If not, why?

Because it's the same as == (at least for non-edge cases, that is.)

• gongzhitaao: Internally, the < combined with > will be ==?
• user529758: @gongzhitaao define "internally" and "will be"
• user529758: @gongzhitaao Enter the concept of logic. It's pretty useful in real life too.
• Benjamin Lindley: @gongzhitaao: Given any 2 real numbers A and B, exactly one of the following is true: One, A is less than B. Two, A is greater than B. Three, A is equal to B. Your test eliminates both the first and second possibilities. That leaves only the third. That they are equal. There are of course differences between floating point values and actual real numbers, but excluding the edge cases, this logic still applies.
• MSalters: @H2CO3: I did. But you invoked logic to berate gongzhitaao. In fact, he had a very good question. If you apply logic to that question, you just do not end up with your answer. Logic is a mathematical discipline and doesn't ignore edge cases.

Answer 2

This is not sufficient because that will return true if either a or b is NaN

Answer 3

This is basically about the notion that using "nearly equal" can somehow make up for not knowing enough about floating-point calculations to get reliable results. "Nearly equal" is an advanced technique; it will often bite you because it is not transitive (that is, a "nearly equals" b and b "nearly equals" c does not mean that a "nearly equals" c). If you're a serious programmer, learn the basics of how floating-point works. Nobody's surprised that (1/2)*2) is not 1, but somehow people don't grasp that (1.0/10.0)*10.0 is also not 1 (unless your system has decimal floating point), and for exactly the same reason. That's a failure of education, but having been taught badly doesn't mean you can't learn it. (end of rant)

Answer 4

The idea of comparing with a tolerance is that two values that are almost exactly the same might want to be considered the same. If you have 2.000000000000 and 2.000000000001, should those be equal, or not? If they should be then your code will fail.

• gongzhitaao: not equal according to my code.

Answer 5

Your mistake is, you read the article which has

 diff = A - B;
 return (diff < EPSILON) && (-diff > EPSILON);

and, you put EPSILON = 0 and turn it into

if (!(a > b) && !(a < b))

You shouldn't suppose EPSILON as 0. Because you should consider a bit error is float/double comparison which we call it EPSILON.

Answer 6

You really need to read this document:

https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/02Numerics/Double/paper.pdf

It is Golberg's paper = whatever every programmer should know about floating point.