I have a class that internally represents some quantity in fixed point as 32-bit integer with somewhat arbitrary denominator (it is neither power of 2 nor power of 10).
For communicating with other applications the quantity is converted to plain old double on output and back on input. As code inside the class it looks like:
int32_t quantity;
double GetValue() { return double(quantity) / DENOMINATOR; }
void SetValue(double x) { quantity = x * DENOMINATOR; }
Now I need to ensure that if I output some value as double and read it back, I will always get the same value back. I.e. that
x.SetValue(x.GetValue());
will never change x.quantity (x is arbitrary instance of the class containing the above code).
The double representation has more digits of precision, so it should be possible. But it will almost certainly not be the case with the simplistic code above.
What rounding do I need to use and
How can I find the critical would-be corner cases to test that the rounding is indeed correct?
Any 32 bits will be represented exactly when you convert to a double, but when you divide then multiply by an arbitrary value you will get a similar value but not exactly the same. You should lose at most one bit per operations, which means your double will be almost the same, prior to casting back to an int.
However, since int casts are truncations, you will get the wrong result when very minor errors turn 2.000 into 1.999, thus what you need to do is a simple rounding task prior to casting back.
You can use std::lround() for this if you have C++11, else you can write you own rounding function.
You probably don't care about fairness much here, so the common int(doubleVal+0.5) will work for positives. If as seems likely, you have negatives, try this:
int round(double d) { return d<0?d-0.5:d+0.5; }
The problem you describe is the same problem which exists with converting between binary and decimal representation just with different bases. At least it exists if you want to have the double representation to be a good approximation of the original value (otherwise you could just multiply the 32 bit value you have with your fixed denominator and store the result in a double).
Assuming you want the double representation be a good approximation of your actual value the conversions are nontrivial! The conversion from your internal representation to double can be done using Dragon4 ("How to print floating point numbers accurately", Steele & White) or Grisu ("How to print floating point numbers quickly and accurately", Loitsch; I'm not sure if this algorithm is independent from the base, though). The reverse can be done using Bellerophon ("How to read floating point numbers accurately", Clinger). These algorithms aren't entirely trivial, though...
I have a program in C++ where I divide two numbers, and I need to know if the answer is an integer or not. What I am using is:
if(fmod(answer,1) == 0)
I also tried this:
if(floor(answer)==answer)
The problem is that answer usually is a 5 digit number, but with many decimals. For example, answer can be: 58696.000000000000000025658 and the program considers that an integer.
Is there any way I can make this work?
I am dividing double a/double b= double answer
(sometimes there are more than 30 decimals)
Thanks!
EDIT:
a and b are numbers in the thousands (about 100,000) which are then raised to powers of 2 and 3, added together and divided (according to a complicated formula). So I am plugging in various a and b values and looking at the answer. I will only keep the a and b values that make the answer an integer. An example of what I got for one of the answers was: 218624 which my program above considered to be an integer, but it really was: 218624.00000000000000000056982 So I need a code that can distinguish integers with more than 20-30 decimals.
You can use std::modf in cmath.h:
double integral;
if(std::modf(answer, &integral) == 0.0)
The integral part of answer is stored in fraction and the return value of std::modf is the fractional part of answer with the same sign as answer.
The usual solution is to check if the number is within a very short distance of an integer, like this:
bool isInteger(double a){
double b=round(a),epsilon=1e-9; //some small range of error
return (a<=b+epsilon && a>=b-epsilon);
}
This is needed because floating point numbers have limited precision, and numbers that indeed are integers may not be represented perfectly. For example, the following would fail if we do a direct comparison:
double d=sqrt(2); //square root of 2
double answer=2.0/(d*d); //2 divided by 2
Here, answer actually holds the value 0.99999..., so we cannot compare that to an integer, and we cannot check if the fractional part is close to 0.
In general, since the floating point representation of a number can be either a bit smaller or a bit bigger than the actual number, it is not good to check if the fractional part is close to 0. It may be a number like 0.99999999 or 0.000001 (or even their negatives), these are all possible results of a precision loss. That's also why I'm checking both sides (+epsilon and -epsilon). You should adjust that epsilon variable to fit your needs.
Also, keep in mind that the precision of a double is close to 15 digits. You may also use a long double, which may give you some extra digits of precision (or not, it is up to the compiler), but even that only gets you around 18 digits. If you need more precision than that, you will need to use an external library, like GMP.
Floating point numbers are stored in memory using a very different bit format than integers. Because of this, comparing them for equality is not likely to work effectively. Instead, you need to test if the difference is smaller than some epsilon:
const double EPSILON = 0.00000000000000000001; // adjust for whatever precision is useful for you
double remainder = std::fmod(numer, denom);
if(std::fabs(0.0 - remainder) < EPSILON)
{
//...
}
Alternatively, if you want to include values that are close to integers (based on your desired precision), you can modify the if condition slightly (since the remainder returned by std::fmod will be in the range [0, 1)):
if (std::fabs(std::round(d) - d) < EPSILON)
{
// ...
}
You can see the test for this here.
Floating point numbers are generally somewhat precise to about 12-15 digits (as a double), but as they are stored as a mantissa (fraction) and a exponent, rational numbers (integers or common fractions) are not likely to be stored as such. For example,
double d = 2.0; // d might actually be 1.99999999999999995
Because of this, you need to compare the difference of what you expect to some very small number that encompasses the precision you desire (we will call this value, epsilon):
double d = 2.0;
bool test = std::fabs(2 - d) < epsilon; // will return true
So when you are trying to compare the remainder from std::fmod, you need to check it against the difference from 0.0 (not for actual equality to 0.0), which is what is done above.
Also, the std::fabs call prevents you from having to do 2 checks by asserting that the value will always be positive.
If you desire a precision that is greater than 15-18 decimal places, you cannot use double or long double; you will need to use a high precision floating point library.
In my previous question Comparing a double and int, without casting or conversion, I found out how the the difference between two doubles was tripping the comparison.
I came accross the method setprecision(), which will help display all the numbers after decimal.
So, the difference of 6.15 and 3.15 was found to be : 3.00000000000000044408920985006
Now, when it gets compared with 3, it returns a result saying it is greater than 3.
How do I force it to take only a limited number of digits?
When I used 6.1 and 3.1, the difference was : 2.99999999999999955591079014994
How should I make the precision so that we know that it is actually equal to 3, and not less than.
Hopefully you should be knowing that floating/double cannot be exactly represented in binary and truncation happens because of the recurring decimal. Your comparison with a float/double with an integer will always fail.
Even your using setprecision will not work because its a method of iomanip to set the precision of the display and not the value being stored.
The portable way of comparing doubles is not to use the '==' operator but to do something like
bool Compare(double a,double b) {
std::fabs(a - b) < std::numeric_limits<double>::epsilon();
}
You can use this to compare double with float and or integer. You can also write a similar compare function for float
bool Compare(float a,float b) {
std::fabs(a - b) < std::numeric_limits<float>::epsilon();
}
In a comment on your other question, you were already pointed towards this great paper on floating-point numbers. It's well worth a read.
With reference to your particular question, a standard way is to define a tolerance with which comparison between doubles is to be made. For example if you have two doubles a and b, and wish to determine whether a is larger than b within a tolerance of eps (another double), you might do something like:
if (a - b > eps) {
// a is greater than b
} else {
// a is not greater than b
}
Alternatively, if you want to know that a is equal to b within the tolerance specified by eps, you might do something of this sort:
if (std::abs(a - b) <= eps) {
// a and b are equal within the set tolerance
} else {
// a and b are not equal within the set tolerance
}
As pointed out by others, C++ comes with some helpful functions out of the box for performing these sorts of comparisons. Look at std::abs, std::numeric_limits, and this nice post on SO.
Here's a comparison function that determines if two numbers are within one LSB of each other.
bool Compare(double a, double b)
{
return (a <= std::nextafter(b, abs(1.1*b))) && (b <= std::nextafter(a, abs(1.1*a)));
}
std::nextafter is new to C++11, but versions are available in earlier compilers. See Generate next largest or smallest representable floating point number without bit twiddling
setprecision lets you select how many digits you spit out to a stream. It does not decide the number of digits to be considered. For rounding purposes use one of the rounding functions from <cmath>.
The no of digits of precision in float and double actually depends on the size of them respectively.That is why float generally has less precision than double.
You can use
std::cout<<std::setprecision(desired no);
float a=(desired no);
Now you have sucessfully set the precision of float.The same can be done to other appropriate data type including double as well.
Warning do not set precision greater than what a data type has to offer.Double has 15 to 18 digits of precision while float has only 6 to 9 digits of precision.
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'.
Say I want a function that takes two floats (x and y), and I want to compare them using not their float representation but rather their bitwise representation as a 32-bit unsigned int. That is, a number like -495.5 has bit representation 0b11000011111001011100000000000000 or 0xC3E5C000 as a float, and I have an unsigned int with the same bit representation (corresponding to a decimal value 3286614016, which I don't care about). Is there any easy way for me to perform an operation like <= on these floats using only the information contained in their respective unsigned int counterparts?
You must do a signed compare unless you ensure that all the original values were positive. You must use an integer type that is the same size as the original floating point type. Each chip may have a different internal format, so comparing values from different chips as integers is most likely to give misleading results.
Most float formats look something like this: sxxxmmmm
s is a sign bit
xxx is an exponent
mmmm is the mantissa
The value represented will then be something like: 1mmm << (xxx-k)
1mmm because there is an implied leading 1 bit unless the value is zero.
If xxx < k then it will be a right shift. k is near but not equal to half the largest value that could be expressed by xxx. It is adjusted for the size of the mantissa.
All to say that, disregarding NaN, comparing floating point values as signed integers of the same size will yield meaningful results. They are designed that way so that floating point comparisons are no more costly than integer comparisons. There are compiler optimizations to turn off NaN checks so that the comparisons are straight integer comparisons if the floating point format of the chip supports it.
As an integer, NaN is greater than infinity is greater than finite values. If you try an unsigned compare, all the negative values will be larger than the positive values, just like signed integers cast to unsigned.
If you truly truly don't care about what the conversion yields, it isn't too hard. But the results are extremely non-portable, and you almost certainly won't get an ordering that at all resembles what you'd get by comparing the floats directly.
typedef unsigned int TypeWithSameSizeAsFloat; //Fix this for your platform
bool compare1(float one, float two)
union Convert {
float f;
TypeWithSameSizeAsFloat i;
}
Convert lhs, rhs;
lhs.f = one;
rhs.f = two;
return lhs.i < rhs.i;
}
bool compare2(float one, float two) {
return reinterpret_cast<TypeWithSameSizeAsFloat&>(one)
< reinterpret_cast<TypeWithSameSizeAsFloat&>(two);
}
Just understand the caveats, and chose your second type carefully. Its a near worthless excersize at any rate.
In a word, no. IEEE 754 might allow some kinds of hacks like this, but they do not work all the time and handle all cases, and some platforms do not use that floating point standard (such as doubles on x87 having 80 bit precision internally).
If you're doing this for performance reasons I suggest you strongly reconsider -- if it's faster to use the integer comparison the compiler will probably do it for you, and if it is not, you pay for a float to int conversion multiple times, when a simple comparison may be possible without moving the floats out of registers.
Maybe I'm misreading the question, but I suppose you could do this:
bool compare(float a, float b)
{
return *((unsigned int*)&a) < *((unsigned int*)&b);
}
But this assumes all kinds of things and also warrants the question of why you'd want to compare the bitwise representations of two floats.