Hash function for floats - c++

I'm currently implementing a hash table in C++ and I'm trying to make a hash function for floats...
I was going to treat floats as integers by padding the decimal numbers, but then I realized that I would probably reach the overflow with big numbers...
Is there a good way to hash floats?
You don't have to give me the function directly, but I'd like to see/understand different concepts...
Notes:
I don't need it to be really fast, just evenly distributed if possible.
I've read that floats should not be hashed because of the speed of computation, can someone confirm/explain this and give me other reasons why floats should not be hashed? I don't really understand why (besides the speed)

It depends on the application but most of time floats should not be hashed because hashing is used for fast lookup for exact matches and most floats are the result of calculations that produce a float which is only an approximation to the correct answer. The usually way to check for floating equality is to check if it is within some delta (in absolute value) of the correct answer. This type of check does not lend itself to hashed lookup tables.
EDIT:
Normally, because of rounding errors and inherent limitations of floating point arithmetic, if you expect that floating point numbers a and b should be equal to each other because the math says so, you need to pick some relatively small delta > 0, and then you declare a and b to be equal if abs(a-b) < delta, where abs is the absolute value function. For more detail, see this article.
Here is a small example that demonstrates the problem:
float x = 1.0f;
x = x / 41;
x = x * 41;
if (x != 1.0f)
{
std::cout << "ooops...\n";
}
Depending on your platform, compiler and optimization levels, this may print ooops... to your screen, meaning that the mathematical equation x / y * y = x does not necessarily hold on your computer.
There are cases where floating point arithmetic produces exact results, e.g. reasonably sized integers and rationals with power-of-2 denominators.

If your hash function did the following you'd get some degree of fuzziness on the hash lookup
unsigned int Hash( float f )
{
unsigned int ui;
memcpy( &ui, &f, sizeof( float ) );
return ui & 0xfffff000;
}
This way you'll mask off the 12 least significant bits allowing for a degree of uncertainty ... It really depends on yout application however.

You can use the std hash, it's not bad:
std::size_t myHash = std::cout << std::hash<float>{}(myFloat);

unsigned hash(float x)
{
union
{
float f;
unsigned u;
};
f = x;
return u;
}
Technically undefined behavior, but most compilers support this. Alternative solution:
unsigned hash(float x)
{
return (unsigned&)x;
}
Both solutions depend on the endianness of your machine, so for example on x86 and SPARC, they will produce different results. If that doesn't bother you, just use one of these solutions.

You can of course represent a float as an int type of the same size to hash it, however this naive approach has some pitfalls you need to be careful of...
Simply converting to a binary representation is error prone since values which are equal wont necessarily have the same binary representation.
An obvious case: -0.0 wont match 0.0 for example. *
Further, simply converting to an int of the same size wont give very even distribution, which is often important (implementing a hash/set that uses buckets for example).
Suggested steps for implementation:
filter out non-finite cases (nan, inf) and (0.0, -0.0 whether you need to do this explicitly or not depends on the method used).
convert to an int of the same size(that is - use a union for example to represent the float as an int, not simply cast to an int).
re-distribute the bits, (intentionally vague here!), this is basically a speed vs quality tradeoff. But if you have many values in a small range you probably don't want them to in a similar range too.
*: You may wan't to check for (nan and -nan) too. How to handle those exactly depends on your use case (you may want to ignore sign for all nan's as CPython does).
Python's _Py_HashDouble is a good reference for how you might hash a float, in production code (ignore the -1 check at the end, since that's a special value for Python).

If you're interested, I just made a hash function that uses floating point and can hash floats. It also passes SMHasher ( which is the main bias-test for non-crypto hash functions ). It's a lot slower than normal non-cryptographic hash functions due to the float calculations.
I'm not sure if tifuhash will become useful for all applications, but it's interesting to see a simple floating point function pass both PractRand and SMHasher.
The main state update function is very simple, and looks like:
function q( state, val, numerator, denominator ) {
// Continued Fraction mixed with Egyptian fraction "Continued Egyptian Fraction"
// with denominator = val + pos / state[1]
state[0] += numerator / denominator;
state[0] = 1.0 / state[0];
// Standard Continued Fraction with a_i = val, b_i = (a_i-1) + i + 1
state[1] += val;
state[1] = numerator / state[1];
}
Anyway, you can get it on npm
Or you can check out the github
Using is simple:
const tifu = require('tifuhash');
const message = 'The medium is the message.';
const number = 333333333;
const float = Math.PI;
console.log( tifu.hash( message ),
tifu.hash( number ),
tifu.hash( float ),
tifu.hash( ) );
There's a demo of some hashes on runkit here https://runkit.com/593a239c56ebfd0012d15fc9/593e4d7014d66100120ecdb9
Side note: I think that in future using floating point,possibly big arrays of floating point calculations, could be a useful way to make more computationally-demanding hash functions in future. A weird side effect I discovered of using floating point is that the hashes are target dependent, and I surmise maybe they could be use to fingerprint the platforms they were calculated on.

Because of the IEEE byte ordering the Java Float.hashCode() and Double.hashCode() do not give good results. This problem is wellknown and can be adressed by this scrambler:
class HashScrambler {
/**
* https://sites.google.com/site/murmurhash/
*/
static int murmur(int x) {
x ^= x >> 13;
x *= 0x5bd1e995;
return x ^ (x >> 15);
}
}
You then get a good hash function, which also allows you to use Float and Double in hash tables. But you need to write your own hash table that allows a custom hash function.
Since in a hash table you need also test for equality, you need an exact equality to make it work. Maybe the later is what President James K. Polk intends to adress?

Related

Floating point Arithmetics

Today in my C++ programming lessons, my proff told me that one should never compare two floating point values directly.
So I tried this piece of code and found out the reason for his statement.
double l_Value=94.9;
print("%.20lf",l_Value);
And I found the results as 94.89999999 ( some relative error )
I understand that floating numbers are not stored in the way one presents it to the code. Squeezing those ones and zeros in binary form involves some relative rounding errors.
Iam looking for solutions to two problems.
1. Efficient way to compare two floating values.
2. How to add a floating value to another one. Example. Add 0.1111 to 94.4345 to get the exact value as 94.5456
Thanks in advance.
Efficient way to compare two floating values.
A simple double a,b; if (a == b) is an efficient way to compare two floating values. Yet as OP noticed, this may not meet the overall coding goal. Better ways depend on the context of the compare, something not supplied by OP. See far below.
How to add a floating value to another one. Example. Add 0.1111 to 94.4345 to get the exact value as 94.5456
Floating values as source code have effective unlimited range and precision such as 1.23456789012345678901234567890e1234567. Conversion of this text to a double is limited typically to one of 264 different values. The closest is selected, but that may not be an exact match.
Neither 0.1111, 94.4345, 94.5456 can be representably exactly as a typical double.
OP has choices:
1.) Use another type other than double, float. Various libraries offer decimal floating point types.
2) Limit code to rare platforms that support double to a base 10 form such that FLT_RADIX == 10.
3) Write your own code to handle user input like "0.1111" into a structure/string and perform the needed operations.
4) Treat user input as strings and the convert to some integer type, again with supported routines to read/compute/and write.
5) Accept that floating point operations are not mathematically exact and handle round-off error.
double a = 0.1111;
printf("a: %.*e\n", DBL_DECIMAL_DIG -1 , a);
double b = 94.4345;
printf("b: %.*e\n", DBL_DECIMAL_DIG -1 , b);
double sum = a + b;
printf("sum: %.*e\n", DBL_DECIMAL_DIG -1 , sum);
printf("%.4f\n", sum);
Output
a: 1.1110000000000000e-01
b: 9.4434500000000000e+01
sum: 9.4545599999999993e+01
94.5456 // Desired textual output based on a rounded `sum` to the nearest 0.0001
More on #1
If an exact compare is not sought but some sort of "are the two values close enough?", a definition of "close enough" is needed - of which there are many.
The following "close enough" compares the distance by examining the ULP of the two numbers. It is a linear difference when the values are in the same power-of-two and becomes logarithmic other wise. Of course, change of sign is an issue.
float example:
Consider all finite float ordered from most negative to most positive. The following, somewhat-portable code, returns an integer for each float with that same order.
uint32_t sequence_f(float x) {
union {
float f;
uint32_t u32;
} u;
assert(sizeof(float) == sizeof(uint32_t));
u.f = x;
if (u.u32 & 0x80000000) {
u.u32 ^= 0x80000000;
return 0x80000000 - u.u32;
}
return u.u3
}
Now, to determine if two float are "close enough", simple compare two integers.
static bool close_enough(float x, float y, uint32_t ULP_delta) {
uint32_t ullx = sequence_f(x);
uint32_t ully = sequence_f(y);
if (ullx > ully) return (ullx - ully) <= ULP_delta;
return (ully - ullx) <= ULP_delta;
}
The way I've usually done this is is to have a custom equality comparison function. The basic idea, is you have a certain tolerance, say 0.0001 or something. Then you subtract your two numbers and take their absolute value, and if it is less than your tolerance you treat it as equal. There are other strategies that may be more appropriate for certain situations, of course.
Define for yourself a tolerance level e (for example, e=.0001) and check if abs(a-b) <= e
You aren't going to get an "exact" value with floating point. Ever. If you know in advance that you are using four decimals, and you want "exact", then you need to internally treat your numbers as integers and only display them as decimals. 944345 + 1111 = 945456

quickly find the integer part of the base 2 logarithm

What is an efficient method to calculate the integer part of the base 2 logarithm of a floating point number? Something like
N = ceil( log2( f ))
or
N = floor( log2( f ))
for floating point f. I guess this is possible to realize very efficiently somehow as one probably only needs access to the floating point exponent.
EDIT2: I am not primarily interested in exactness. I could tolerate an error of +-1. I listed the two variants just as an example because one might be computationally cheaper than the other (but I dont know).
I need this for accuracy control of an algorithm where the parameter f is some tolerance and the log is needed to control the number of terms. Accurate calculation of the log is not important.
EDIT: this is not a duplicate of other the many other questions asking for the log2 of an integer argument (e.g. How to do an integer log2() in C++?) . This here is about floating point argument and a completely different story. Specifically I need it for f < 1, which is not possible at all with the integer methods
The standard library function frexp does exactly that: it decomposes a double into an integer exponent and a normalized mantissa.
If you are content with the floor of the logarithm, rather than rounding the logarithm to the nearest integer, you are probably better off with the newer standard library function ilogb.
Note that these two functions treat zeros and infinities differently, so they are not quite interchangeable.
Inspired by rici, who pointed me to frexp, I think I found the answer. In C99 and recent C++ we have the function ilogb which does exactly that what I need
int ilogb( float arg );
int ilogb( double arg );
and is equivalent to
(int)logb( arg )
It returns one less than frexp. So the frexp result corresponds to
floor(log2(arg)+1
and ilogb(arg) to
floor(log2(arg))
This is a horrible hack which extracts the exponent from the little-endian float although I make no guarantee about portability, etc.
#include <stdio.h>
int main(void) {
float f;
unsigned i;
unsigned *ip = (unsigned*)&f;
printf("Enter a float: ");
scanf("%f", &f);
i = *ip;
i = (i >> 23) & 0xFF;
i -= 127;
printf("%f %d\n", f, (int)i);
return 0;
}
Program output:
Enter a float: 0.125
0.125000 -3

Truncate floor into three decimal point C++

I want to truncate floor number to be 3 digit decimal number. Example:
input : x = 0.363954;
output: 0.364
i used
double myCeil(float v, int p)
{
return int(v * pow(float(10),p))/pow(float(10),p );
}
but the output was 0.3630001 .
I tried to use trunc from <cmath> but it doesn't exist.
Floating-point math typically uses a binary representation; as a result, there are decimal values that cannot be exactly represented as floating-point values. Trying to fiddle with internal precisions runs into exactly this problem. But mostly when someone is trying to do this they're really trying to display a value using a particular precision, and that's simple:
double x = 0.363954;
std::cout.precision(3);
std::cout << x << '\n';
The function your looking for is the std::ceil, not std::trunc
double myCeil(double v, int p)
{
return std::ceil(v * std::pow(10, p)) / std::pow(10, p);
}
substitue in std::floor or std::round for a myFloor or myRound as desired. (Note that std::round appears in C++11, which you will have to enable if it isn't already done).
It is just impossible to get 0.364 exactly. There is no way you can store the number 0.364 (364/1000) exactly as a float, in the same way you would need an infinite number of decimals to write 1/3 as 0.3333333333...
You did it correctly, except for that you probably want to use std::round(), to round to the closest number, instead of int(), which truncates.
Comparing floating point numbers is tricky business. Typically the best you can do is check that the numbers are sufficiently close to each other.
Are you doing your rounding for comparison purposes? In such case, it seems you are happy with 3 decimals (this depends on each problem in question...), in such case why not just
bool are_equal_to_three_decimals(double a, double b)
{
return std::abs(a-b) < 0.001;
}
Note that the results obtained via comparing the rounded numbers and the function I suggested are not equivalent!
This is an old post, but what you are asking for is decimal precision with binary mathematics. The conversion between the two is giving you an apparent distinction.
The main point, I think, which you are making is to do with identity, so that you can use equality/inequality comparisons between two numbers.
Because of the fact that there is a discrepancy between what we humans use (decimal) and what computers use (binary), we have three choices.
We use a decimal library. This is computationally costly, because we are using maths which are different to how computers work. There are several, and one day they may be adopted into std. See eg "ISO/IEC JTC1 SC22 WG21 N2849"
We learn to do our maths in binary. This is mentally costly, because it's not how we do our maths normally.
We change our algorithm to include an identity test.
We change our algorithm to use a difference test.
With option 3, it is where we make a decision as to just how close one number needs to be to another number to be considered 'the same number'.
One simple way of doing this is (as given by #SirGuy above) where we use ceiling or floor as a test - this is good, because it allows us to choose the significant number of digits we are interested in. It is domain specific, and the solution that he gives might be a bit more optimal if using a power of 2 rather than of 10.
You definitely would only want to do the calculation when using equality/inequality tests.
So now, our equality test would be (for 10 binary places (nearly 3dp))
// Normal identity test for floats.
// Quick but fails eg 1.0000023 == 1.0000024
return (a == b);
Becomes (with 2^10 = 1024).
// Modified identity test for floats.
// Works with 1.0000023 == 1.0000024
return (std::floor(a * 1024) == std::floor(b * 1024));
But this isn't great
I would go for option 4.
Say you consider any difference less than 0.001 to be insignificant, such that 1.00012 = 1.00011.
This does an additional subtraction and a sign removal, which is far cheaper (and more reliable) than bit shifts.
// Modified equality test for floats.
// Returns true if the ∂ is less than 1/10000.
// Works with 1.0000023 == 1.0000024
return abs(a - b) < 0.0001;
This boils down to your comment about calculating circularity, I am suggesting that you calculate the delta (difference) between two circles, rather than testing for equivalence. But that isn't exactly what you asked in the question...

What is a standard way to compare float with zero?

May question is: What is a standard way to compare float with zero?
As far as I know direct comparison:
if ( x == 0 ) {
// x is zero?
} else {
// x is not zero??
can fail with floating points variables.
I used to use
float x = ...
...
if ( std::abs(x) <= 1e-7f ) {
// x is zero, do the job1
} else {
// x is not zero, do the job2
...
Same approach I find here. But I see two problems:
Random magic number 1e-7f ( or 0.00005 at the link above ).
The code harder to read
This is such a common comparison, I wonder whether there is a standard short way to do this. Like
x.is_zero();
To compare a floating-point value with 0, just compare it:
if (f == 0)
// whatever
There is nothing wrong with this comparison. If it doesn't do what you expect it's because the value of f is not what you thought it was. Its essentially the same problem as this:
int i = 1/3;
i *= 3;
if (i == 1)
// whatever
There's nothing wrong with that comparison, but the value of i is not 1. Almost all programmers understand the loss of precision with integer values; many don't understand it with floating-point values.
Using "nearly equal" instead of == is an advanced technique; it often leads to unexpected problems. For example, it is not transitive; that is, a nearly equals b and b nearly equals c does not mean that a nearly equals c.
There is no standard way, because whether or not you want to treat a small number as if it were zero depends on how you computed the number and what it's for. This in turn depends on the expected size of any errors introduced by your computations, and perhaps on errors of physical measurement that determined your original inputs.
For example, suppose that your value represents the length of a journey in miles in some mapping software. Then you are happy to treat 1e-7 as equal to zero because in that context it is a very small number: it has come about because of a rounding error or other reason for slight inexactness.
On the other hand, suppose that your value represents the size of a molecule in metres in some electron microscopy software. Then you certainly don't want to treat 1e-7 as equal to zero because in that context it's a very large number.
You should first consider what would be a suitable accuracy to present your value: what's the error bar, or how many significant figures can you reasonably display. This will give you some idea with what tolerance it would be appropriate to test against zero, although it still might not settle the case. For the mapping software, you can probably treat a journey as zero if it's less than some fixed value, although the value itself might depend on the resolution of your maps. For the microscopy software, if the difference between two sizes is such that zero lies with the 95% error range on those measurements, that still might not be sufficient to describe them as being the same size.
I don't know whether my answer useful, I've found this in irrlicht's irrmath.h and still using it in engine's mathlib till nowdays:
const float ROUNDING_ERROR_f32 = 0.000001f;
//! returns if a equals b, taking possible rounding errors into account
inline bool equals(const float a, const float b, const float tolerance = ROUNDING_ERROR_f32)
{
return (a + tolerance >= b) && (a - tolerance <= b);
}
The author has explained this approach by "after many rotations, which are trigonometric operations the coordinate spoils and the direct comparsion may cause fault".

Is floating-point == ever OK?

Just today I came across third-party software we're using and in their sample code there was something along these lines:
// Defined in somewhere.h
static const double BAR = 3.14;
// Code elsewhere.cpp
void foo(double d)
{
if (d == BAR)
...
}
I'm aware of the problem with floating-points and their representation, but it made me wonder if there are cases where float == float would be fine? I'm not asking for when it could work, but when it makes sense and works.
Also, what about a call like foo(BAR)? Will this always compare equal as they both use the same static const BAR?
Yes, you are guaranteed that whole numbers, including 0.0, compare with ==
Of course you have to be a little careful with how you got the whole number in the first place, assignment is safe but the result of any calculation is suspect
ps there are a set of real numbers that do have a perfect reproduction as a float (think of 1/2, 1/4 1/8 etc) but you probably don't know in advance that you have one of these.
Just to clarify. It is guaranteed by IEEE 754 that float representions of integers (whole numbers) within range, are exact.
float a=1.0;
float b=1.0;
a==b // true
But you have to be careful how you get the whole numbers
float a=1.0/3.0;
a*3.0 == 1.0 // not true !!
There are two ways to answer this question:
Are there cases where float == float gives the correct result?
Are there cases where float == float is acceptable coding?
The answer to (1) is: Yes, sometimes. But it's going to be fragile, which leads to the answer to (2): No. Don't do that. You're begging for bizarre bugs in the future.
As for a call of the form foo(BAR): In that particular case the comparison will return true, but when you are writing foo you don't know (and shouldn't depend on) how it is called. For example, calling foo(BAR) will be fine but foo(BAR * 2.0 / 2.0) (or even maybe foo(BAR * 1.0) depending on how much the compiler optimises things away) will break. You shouldn't be relying on the caller not performing any arithmetic!
Long story short, even though a == b will work in some cases you really shouldn't rely on it. Even if you can guarantee the calling semantics today maybe you won't be able to guarantee them next week so save yourself some pain and don't use ==.
To my mind, float == float is never* OK because it's pretty much unmaintainable.
*For small values of never.
The other answers explain quite well why using == for floating point numbers is dangerous. I just found one example that illustrates these dangers quite well, I believe.
On the x86 platform, you can get weird floating point results for some calculations, which are not due to rounding problems inherent to the calculations you perform. This simple C program will sometimes print "error":
#include <stdio.h>
void test(double x, double y)
{
const double y2 = x + 1.0;
if (y != y2)
printf("error\n");
}
void main()
{
const double x = .012;
const double y = x + 1.0;
test(x, y);
}
The program essentially just calculates
x = 0.012 + 1.0;
y = 0.012 + 1.0;
(only spread across two functions and with intermediate variables), but the comparison can still yield false!
The reason is that on the x86 platform, programs usually use the x87 FPU for floating point calculations. The x87 internally calculates with a higher precision than regular double, so double values need to be rounded when they are stored in memory. That means that a roundtrip x87 -> RAM -> x87 loses precision, and thus calculation results differ depending on whether intermediate results passed via RAM or whether they all stayed in FPU registers. This is of course a compiler decision, so the bug only manifests for certain compilers and optimization settings :-(.
For details see the GCC bug: http://gcc.gnu.org/bugzilla/show_bug.cgi?id=323
Rather scary...
Additional note:
Bugs of this kind will generally be quite tricky to debug, because the different values become the same once they hit RAM.
So if for example you extend the above program to actually print out the bit patterns of y and y2 right after comparing them, you will get the exact same value. To print the value, it has to be loaded into RAM to be passed to some print function like printf, and that will make the difference disappear...
I'll provide more-or-less real example of legitimate, meaningful and useful testing for float equality.
#include <stdio.h>
#include <math.h>
/* let's try to numerically solve a simple equation F(x)=0 */
double F(double x) {
return 2 * cos(x) - pow(1.2, x);
}
/* a well-known, simple & slow but extremely smart method to do this */
double bisection(double range_start, double range_end) {
double a = range_start;
double d = range_end - range_start;
int counter = 0;
while (a != a + d) // <-- WHOA!!
{
d /= 2.0;
if (F(a) * F(a + d) > 0) /* test for same sign */
a = a + d;
++counter;
}
printf("%d iterations done\n", counter);
return a;
}
int main() {
/* we must be sure that the root can be found in [0.0, 2.0] */
printf("F(0.0)=%.17f, F(2.0)=%.17f\n", F(0.0), F(2.0));
double x = bisection(0.0, 2.0);
printf("the root is near %.17f, F(%.17f)=%.17f\n", x, x, F(x));
}
I'd rather not explain the bisection method used itself, but emphasize on the stopping condition. It has exactly the discussed form: (a == a+d) where both sides are floats: a is our current approximation of the equation's root, and d is our current precision. Given the precondition of the algorithm — that there must be a root between range_start and range_end — we guarantee on every iteration that the root stays between a and a+d while d is halved every step, shrinking the bounds.
And then, after a number of iterations, d becomes so small that during addition with a it gets rounded to zero! That is, a+d turns out to be closer to a then to any other float; and so the FPU rounds it to the closest representable value: to a itself. Calculation on a hypothetical machine can illustrate; let it have 4-digit decimal mantissa and some large exponent range. Then what result should the machine give to 2.131e+02 + 7.000e-3? The exact answer is 213.107, but our machine can't represent such number; it has to round it. And 213.107 is much closer to 213.1 than to 213.2 — so the rounded result becomes 2.131e+02 — the little summand vanished, rounded up to zero. Exactly the same is guaranteed to happen at some iteration of our algorithm — and at that point we can't continue anymore. We have found the root to maximum possible precision.
Addendum
No you can't just use "some small number" in the stopping condition. For any choice of the number, some inputs will deem your choice too large, causing loss of precision, and there will be inputs which will deem your choiсe too small, causing excess iterations or even entering infinite loop. Imagine that our F can change — and suddenly the solutions can be both huge 1.0042e+50 and tiny 1.0098e-70. Detailed discussion follows.
Calculus has no notion of a "small number": for any real number, you can find infinitely many even smaller ones. The problem is, among those "even smaller" ones might be a root of our equation. Even worse, some equations will have distinct roots (e.g. 2.51e-8 and 1.38e-8) — both of which will get approximated by the same answer if our stopping condition looks like d < 1e-6. Whichever "small number" you choose, many roots which would've been found correctly to the maximum precision with a == a+d — will get spoiled by the "epsilon" being too large.
It's true however that floats' exponent has finite limited range, so one actually can find the smallest nonzero positive FP number; in IEEE 754 single precision, it's the 1e-45 denorm. But it's useless! while (d >= 1e-45) {…} will loop forever with single-precision (positive nonzero) d.
At the same time, any choice of the "small number" in d < eps stopping condition will be too small for many equations. Where the root has high enough exponent, the result of subtraction of two neighboring mantissas will easily exceed our "epsilon". For example, 7.00023e+8 - 7.00022e+8 = 0.00001e+8 = 1.00000e+3 = 1000 — meaning that the smallest possible difference between numbers with exponent +8 and 6-digit mantissa is... 1000! It will never fit into, say, 1e-4. For numbers with relatively high exponent we simply have not enough precision to ever see a difference of 1e-4. This means eps = 1e-4 will be too small!
My implementation above took this last problem into account; you can see that d is halved each step — instead of getting recalculated as difference of (possibly huge in exponent) a and b. For reals, it doesn't matter; for floats it does! The algorithm will get into infinite loops with (b-a) < eps on equations with huge enough roots. The previous paragraph shows why. d < eps won't get stuck, but even then — needless iterations will be performed during shrinking d way down below the precision of a — still showing the choice of eps as too small. But a == a+d will stop exactly at precision.
Thus as shown: any choice of eps in while (d < eps) {…} will be both too large and too small, if we allow F to vary.
... This kind of reasoning may seem overly theoretical and needlessly deep, but it's to illustrate again the trickiness of floats. One should be aware of their finite precision when writing arithmetic operators around.
Perfect for integral values even in floating point formats
But the short answer is: "No, don't use ==."
Ironically, the floating point format works "perfectly", i.e., with exact precision, when operating on integral values within the range of the format. This means that you if you stick with double values, you get perfectly good integers with a little more than 50 bits, giving you about +- 4,500,000,000,000,000, or 4.5 quadrillion.
In fact, this is how JavaScript works internally, and it's why JavaScript can do things like + and - on really big numbers, but can only << and >> on 32-bit ones.
Strictly speaking, you can exactly compare sums and products of numbers with precise representations. Those would be all the integers, plus fractions composed of 1 / 2n terms. So, a loop incrementing by n + 0.25, n + 0.50, or n + 0.75 would be fine, but not any of the other 96 decimal fractions with 2 digits.
So the answer is: while exact equality can in theory make sense in narrow cases, it is best avoided.
The only case where I ever use == (or !=) for floats is in the following:
if (x != x)
{
// Here x is guaranteed to be Not a Number
}
and I must admit I am guilty of using Not A Number as a magic floating point constant (using numeric_limits<double>::quiet_NaN() in C++).
There is no point in comparing floating point numbers for strict equality. Floating point numbers have been designed with predictable relative accuracy limits. You are responsible for knowing what precision to expect from them and your algorithms.
It's probably ok if you're never going to calculate the value before you compare it. If you are testing if a floating point number is exactly pi, or -1, or 1 and you know that's the limited values being passed in...
I also used it a few times when rewriting few algorithms to multithreaded versions. I used a test that compared results for single- and multithreaded version to be sure, that both of them give exactly the same result.
Let's say you have a function that scales an array of floats by a constant factor:
void scale(float factor, float *vector, int extent) {
int i;
for (i = 0; i < extent; ++i) {
vector[i] *= factor;
}
}
I'll assume that your floating point implementation can represent 1.0 and 0.0 exactly, and that 0.0 is represented by all 0 bits.
If factor is exactly 1.0 then this function is a no-op, and you can return without doing any work. If factor is exactly 0.0 then this can be implemented with a call to memset, which will likely be faster than performing the floating point multiplications individually.
The reference implementation of BLAS functions at netlib uses such techniques extensively.
In my opinion, comparing for equality (or some equivalence) is a requirement in most situations: standard C++ containers or algorithms with an implied equality comparison functor, like std::unordered_set for example, requires that this comparator be an equivalence relation (see C++ named requirements: UnorderedAssociativeContainer).
Unfortunately, comparing with an epsilon as in abs(a - b) < epsilon does not yield an equivalence relation since it loses transitivity. This is most probably undefined behavior, specifically two 'almost equal' floating point numbers could yield different hashes; this can put the unordered_set in an invalid state.
Personally, I would use == for floating points most of the time, unless any kind of FPU computation would be involved on any operands. With containers and container algorithms, where only read/writes are involved, == (or any equivalence relation) is the safest.
abs(a - b) < epsilon is more or less a convergence criteria similar to a limit. I find this relation useful if I need to verify that a mathematical identity holds between two computations (for example PV = nRT, or distance = time * speed).
In short, use == if and only if no floating point computation occur;
never use abs(a-b) < e as an equality predicate;
Yes. 1/x will be valid unless x==0. You don't need an imprecise test here. 1/0.00000001 is perfectly fine. I can't think of any other case - you can't even check tan(x) for x==PI/2
The other posts show where it is appropriate. I think using bit-exact compares to avoid needless calculation is also okay..
Example:
float someFunction (float argument)
{
// I really want bit-exact comparison here!
if (argument != lastargument)
{
lastargument = argument;
cachedValue = very_expensive_calculation (argument);
}
return cachedValue;
}
I would say that comparing floats for equality would be OK if a false-negative answer is acceptable.
Assume for example, that you have a program that prints out floating points values to the screen and that if the floating point value happens to be exactly equal to M_PI, then you would like it to print out "pi" instead. If the value happens to deviate a tiny bit from the exact double representation of M_PI, it will print out a double value instead, which is equally valid, but a little less readable to the user.
I have a drawing program that fundamentally uses a floating point for its coordinate system since the user is allowed to work at any granularity/zoom. The thing they are drawing contains lines that can be bent at points created by them. When they drag one point on top of another they're merged.
In order to do "proper" floating point comparison I'd have to come up with some range within which to consider the points the same. Since the user can zoom in to infinity and work within that range and since I couldn't get anyone to commit to some sort of range, we just use '==' to see if the points are the same. Occasionally there'll be an issue where points that are supposed to be exactly the same are off by .000000000001 or something (especially around 0,0) but usually it works just fine. It's supposed to be hard to merge points without the snap turned on anyway...or at least that's how the original version worked.
It throws of the testing group occasionally but that's their problem :p
So anyway, there's an example of a possibly reasonable time to use '=='. The thing to note is that the decision is less about technical accuracy than about client wishes (or lack thereof) and convenience. It's not something that needs to be all that accurate anyway. So what if two points won't merge when you expect them to? It's not the end of the world and won't effect 'calculations'.