Deviation of double

[Christian Meier]

Hallo NG

My problem is the deviation of the floating point datatypes. 90.625 is not
exactly 90.625. It is 90.624999999.... on my system. Now I tried to find a
solution for my problem but I couldn't think of one. So I tried to write a
work-arround which should solve the problem for the most numbers.
What do you think of this: (I know it is not a perfect solution....)

double dValue = ...;

if (dValue >= 10000000000000LL) {
nInValue += 0.01;
} else if (dValue >= 1000000000000LL) {
nInValue += 0.001;
} else if (dValue >= 100000000000LL) {
nInValue += 0.0001;
} else if (dValue >= 10000000000LL) {
nInValue += 0.00001;
} else if (dValue >= 1000000000) {
nInValue += 0.000001;
} else if (dValue >= 100000000) {
nInValue += 0.0000001;
} else if (dValue >= 10000000) {
nInValue += 0.00000001;
} else if (dValue >= 1000000) {
nInValue += 0.000000001;
} else if (dValue >= 100000) {
nInValue += 0.0000000001;
} else if (dValue >= 10000) {
nInValue += 0.00000000001;
} else if (dValue >= 1000) {
nInValue += 0.000000000001;
} else if (dValue >= 100) {
nInValue += 0.0000000000001;
} else if (dValue >= 10) {
nInValue += 0.00000000000001;
} else if (dValue < 0) {
nInValue -= 0.000000000000001;
} // if

Greets Chris

 

[Peter van Merkerk]

Hallo NG

My problem is the deviation of the floating point datatypes. 90.625 is not
exactly 90.625. It is 90.624999999.... on my system. Now I tried to find a
solution for my problem but I couldn't think of one. So I tried to write a
work-arround which should solve the problem for the most numbers.
What do you think of this: (I know it is not a perfect solution....)

double dValue = ...;

if (dValue >= 10000000000000LL) {
nInValue += 0.01;
} else if (dValue >= 1000000000000LL) {
nInValue += 0.001;
} else if (dValue >= 100000000000LL) {
nInValue += 0.0001;
} else if (dValue >= 10000000000LL) {
nInValue += 0.00001;
} else if (dValue >= 1000000000) {
nInValue += 0.000001;
} else if (dValue >= 100000000) {
nInValue += 0.0000001;
} else if (dValue >= 10000000) {
nInValue += 0.00000001;
} else if (dValue >= 1000000) {
nInValue += 0.000000001;
} else if (dValue >= 100000) {
nInValue += 0.0000000001;
} else if (dValue >= 10000) {
nInValue += 0.00000000001;
} else if (dValue >= 1000) {
nInValue += 0.000000000001;
} else if (dValue >= 100) {
nInValue += 0.0000000000001;
} else if (dValue >= 10) {
nInValue += 0.00000000000001;
} else if (dValue < 0) {
nInValue -= 0.000000000000001;
} // if

I'm not sure what you are trying to achieve by doing this, and what about
negative numbers? The problem is that many numbers cannot be exactly
represented in floating point format, just like 1/3 cannot be exactly
represented with just decimal numbers. Fiddling with floating point values
will only help in very specific cases at best. If the inevitable rounding
errors of floating point numbers is not acceptable for your application,
maybe a fixed point format is more appropriate for your application.

 

[Karl Heinz Buchegger]

Hallo NG

My problem is the deviation of the floating point datatypes. 90.625 is not
exactly 90.625. It is 90.624999999.... on my system. Now I tried to find a
solution for my problem but I couldn't think of one.

Forgive me. What was your problem?

So I tried to write a
work-arround which should solve the problem for the most numbers.
What do you think of this: (I know it is not a perfect solution....)

I doesn't look like solution to anything.

 

[bartek]

Hallo NG

My problem is the deviation of the floating point datatypes. 90.625 is
not exactly 90.625. It is 90.624999999.... on my system. Now I tried
to find a solution for my problem but I couldn't think of one. So I
tried to write a work-arround which should solve the problem for the
most numbers. What do you think of this: (I know it is not a perfect
solution....)

(...)

You can't "work around" the floating point precision limitation. You can
only use more bits to have a better approximation. An exact representation
would require an infinite amount of bits, wouldn't it?
That's how computers work, sorry.

You can use a specialised extended-precision arithmetics library, though,
like the GNU MP http://www.swox.com/gmp/ for example.

 

[Victor Bazarov]

[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is 725/8.
If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the numbers).

BTW, to the OP: I have a hard time believing that 90 5/8 is not
represented precisely on your machine. It's quite possible that you
_expect_ it to be represented precisely, like after multiplying 0.90625
with 100, but you don't have your 0.90625 precisely to begin with. That's
the problem, I believe.

V

 

[Christian Meier]

Forgive me. What was your problem?


I doesn't look like solution to anything.

Sorry, wasn't a really good description of my problem. I will give more
infos: I have a function with two parameters. One of type double, the other
one is an int.
string func (double dValue, int iDigits);
This function must return a string which represents the double value.
iDigits is the number of digits after the point. For this I use sprintf. And
now I got a problem. When passing 90.625 to the function (iDigits = 2) then
the function returned 90.62 instead of 90.63. I found out that the double is
stored as 90.624999999... on my system. So I need a solution for this. The
function has to return 90.63. So I thought I can add a small value to dValue
before rounding. And the double has a limit of 15 decimal digits on my
system. So I tried to write a work-arround as described in my first post.
Can you now understand my problem?

Regards,
Chris

 

[Christian Meier]

[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is 725/8.
If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the numbers).

BTW, to the OP: I have a hard time believing that 90 5/8 is not
represented precisely on your machine. It's quite possible that you
_expect_ it to be represented precisely, like after multiplying 0.90625
with 100, but you don't have your 0.90625 precisely to begin with. That's
the problem, I believe.

V

It does..... these lines:

double d = 90.625;
char c[10];
sprintf(c, "%.2f", h);
std::cout << c << std::endl;

return the following output:
90.62

Greetings Chris

 

[Karl Heinz Buchegger]

[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is 725/8.
If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the numbers).

Hmm. I wonder how you would represent PI in such a system

 

[Victor Bazarov]

[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is 725/8.
If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the numbers).

Hmm. I wonder how you would represent PI in such a system

22/7 of course (Gods know a better value). On a serious note, however, I
suspect that the OP starts with integral values. If that's so, then the
result should be representable as a fractional value, if all operations
are arithmetic.

V

 

[Victor Bazarov]

[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is 725/8.
If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the
numbers).

BTW, to the OP: I have a hard time believing that 90 5/8 is not
represented precisely on your machine. It's quite possible that you
_expect_ it to be represented precisely, like after multiplying 0.90625
with 100, but you don't have your 0.90625 precisely to begin with. That's
the problem, I believe.

V

It does..... these lines:

double d = 90.625;
char c[10];
sprintf(c, "%.2f", h);
std::cout << c << std::endl;

return the following output:
90.62

They do????? Declare/define/initialise 'd', then print 'h' and expect
something sensible? Come on...

I took your code, fixed it, put it in a write surrounding and got 90.63.
Try again, this time the right version:

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
double d = 90.625;
char c[10];
sprintf(c, "%.2f", d);
cout << c << endl;

return 0;
}

V

 

[Victor Bazarov]

Sorry, wasn't a really good description of my problem. I will give more
infos: I have a function with two parameters. One of type double, the other
one is an int.
string func (double dValue, int iDigits);
This function must return a string which represents the double value.
iDigits is the number of digits after the point. For this I use sprintf. And
now I got a problem. When passing 90.625 to the function (iDigits = 2) then
the function returned 90.62 instead of 90.63. I found out that the double is
stored as 90.624999999... on my system. So I need a solution for this. The
function has to return 90.63. So I thought I can add a small value to dValue
before rounding. And the double has a limit of 15 decimal digits on my
system. So I tried to write a work-arround as described in my first post.
Can you now understand my problem?

I believe you're confused about rounding. 90.6249999999999 simply _must_
round to 90.62 if only two decimal digits are required. That's how
rounding is defined. You cannot assume that "since it's so close to
90.625, I should begin my rounding from 90.625 (instead of the actual value)".

If your function "has to return 90.63", add this statement:

return 90.63;

to your function. If, OTOH, you want it to work correctly, analyse the
problem further. Why do you get 90.6249999999999 instead of [expected]
90.625? Why are you expecting 90.625? What values do you start with?
What operations are you performing on those values? And so on...

Victor

 

[Karl Heinz Buchegger]

Sorry, wasn't a really good description of my problem. I will give more
infos: I have a function with two parameters. One of type double, the other
one is an int.
string func (double dValue, int iDigits);
This function must return a string which represents the double value.
iDigits is the number of digits after the point. For this I use sprintf. And
now I got a problem. When passing 90.625 to the function (iDigits = 2) then
the function returned 90.62 instead of 90.63. I found out that the double is
stored as 90.624999999... on my system. So I need a solution for this. The
function has to return 90.63. So I thought I can add a small value to dValue
before rounding. And the double has a limit of 15 decimal digits on my
system. So I tried to write a work-arround as described in my first post.
Can you now understand my problem?

I see.
Hmm. There is indeed not much you can do other then adding some
small constant depending on the number of digits you want
after comma.

But note: This doesn't solve your problem, it just hides it
for that number.
There will always be some border where one or the other digit
shows up. If your numbers are close to that border you will
see exactly that problem. In your case that border is the
digit 5. By adding some small constant you simply shift
that border a little bit higher or lower, but the border
is still there.

 

[Christian Meier]

bartek wrote:

[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is 725/8.
If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the
numbers).

BTW, to the OP: I have a hard time believing that 90 5/8 is not
represented precisely on your machine. It's quite possible that you
_expect_ it to be represented precisely, like after multiplying 0.90625
with 100, but you don't have your 0.90625 precisely to begin with. That's
the problem, I believe.

V

It does..... these lines:

double d = 90.625;
char c[10];
sprintf(c, "%.2f", h);
std::cout << c << std::endl;

return the following output:
90.62

They do????? Declare/define/initialise 'd', then print 'h' and expect
something sensible? Come on...

I took your code, fixed it, put it in a write surrounding and got 90.63.
Try again, this time the right version:

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
double d = 90.625;
char c[10];
sprintf(c, "%.2f", d);
cout << c << endl;

return 0;
}

V

Believe me!!! Ok, I'll prove it.... The following is my bash output:
~/> cat main.cpp
#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
double d = 90.625;
char c[10];
sprintf(c, "%.2f", d);
cout << c << endl;

return 0;
}
~/> g++ -o prog main.cpp
~/> prog
90.62

Do you need more infos that you can believe me?

Greets Chris

 

[Christian Meier]

I see.
Hmm. There is indeed not much you can do other then adding some
small constant depending on the number of digits you want
after comma.

But note: This doesn't solve your problem, it just hides it
for that number.
There will always be some border where one or the other digit
shows up. If your numbers are close to that border you will
see exactly that problem. In your case that border is the
digit 5. By adding some small constant you simply shift
that border a little bit higher or lower, but the border
is still there.

Thanks for your help. Yes, you're right. It is not really a solution. And I
must not estimate the the deviation is only 0.00001.... And I know that I
really get trouble when someone is calling the function with the argument
90.6249999999. Because then it should be rounded to 90.62. But as I said...
it is just a work-arround for my current problem. It is not proper but I
works now for the most cases.

Greetings Chris

 

[Victor Bazarov]

bartek wrote:


[...]
You can't "work around" the floating point precision limitation.

Yes, he can (and often should). He could use fractions. 90.625 is

725/8.

If all arithmetic operations are done in fractions, it's possible to
achieve exact solution (using large enough integers to represent the

numbers).


BTW, to the OP: I have a hard time believing that 90 5/8 is not
represented precisely on your machine. It's quite possible that you
_expect_ it to be represented precisely, like after multiplying 0.90625
with 100, but you don't have your 0.90625 precisely to begin with.

That's

the problem, I believe.

V


It does..... these lines:

double d = 90.625;
char c[10];
sprintf(c, "%.2f", h);
std::cout << c << std::endl;

return the following output:
90.62

They do????? Declare/define/initialise 'd', then print 'h' and expect
something sensible? Come on...

I took your code, fixed it, put it in a write surrounding and got 90.63.
Try again, this time the right version:

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
double d = 90.625;
char c[10];
sprintf(c, "%.2f", d);
cout << c << endl;

return 0;
}

V

Believe me!!! Ok, I'll prove it.... The following is my bash output:
~/> cat main.cpp
#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
double d = 90.625;
char c[10];
sprintf(c, "%.2f", d);
cout << c << endl;

return 0;
}
~/> g++ -o prog main.cpp
~/> prog
90.62

Do you need more infos that you can believe me?

Sigh...

So your compiler and my compiler create code that uses different rules for
rounding... Mine rounds .5 up, yours probably towards the even. Use that
rule wisely. See my other posts. Heed your own words that you need a
real solution. What you call "a work-around" is not really a work-around,
it's painting over a rusty spot, it's hiding the problem. For all we care
you could simply do

double myfunctionthatneedstoreturn90point63(double whatever)
{
// need to round 'whatever' but for now...
return 90.63;
}

it's just as good a solution as your current "adding a small value".
Believe me, you'd be better off if you never do anything like that and
instead solve the real problem using real methods.

If you need to round 0.5 _always_up_, then you need to do it manually. It
is not as difficult a task (and we often simply rely on the CPU to do the
right thing) if you have to roll your own. Look on the web for a
solution. Somehow I am convinced that there are several in public domain.

Victor

 

[tom_usenet]

Sorry, wasn't a really good description of my problem. I will give more
infos: I have a function with two parameters. One of type double, the other
one is an int.
string func (double dValue, int iDigits);
This function must return a string which represents the double value.
iDigits is the number of digits after the point. For this I use sprintf. And
now I got a problem. When passing 90.625 to the function (iDigits = 2) then
the function returned 90.62 instead of 90.63. I found out that the double is
stored as 90.624999999... on my system. So I need a solution for this.

Yes, the literal 90.625 is not the precise number 90.625. If you're
having a problem with 90.625, just use 90.626.

Read the FAQs on this point: http://www.eskimo.com/~scs/C-faq/s14.html

The

function has to return 90.63. So I thought I can add a small value to dValue
before rounding. And the double has a limit of 15 decimal digits on my
system. So I tried to write a work-arround as described in my first post.
Can you now understand my problem?

Either you're getting these numbers from literals in the code or from
input of some kind. In the former case, change the literals. In the
latter case, use a rational class to store the numbers, such as
boost::rational (see www.boost.org). That way you don't lose precision
(as long as the numerators and denominators lie in the allowed range).

Tom

 

[Kurt Watzka]

Hallo NG

My problem is the deviation of the floating point datatypes. 90.625 is not
exactly 90.625. It is 90.624999999.... on my system.

Are you sure? 725 / 8 has an exact representation in a floating point system
if FLT_RADIX (in <cfloat>) is 2, 4 or 8, and if FLT_DIG is greater than 5.

Even for

float a = 1.0f / 8.0f, s = 0.0f;
for (int i = 0; i < 725; ++i)
s += a;
assert(s == 90.625f);

your chances are quite good that the assertion will hold for reasonable
floating point implementations. If you want to know which numbers have
exact representations, or how floating point arithmetic is defined on
your system, looking at <cfloat> cannot be avoided.

If you are comparing 90.625f to the result of floating point operations
that exceeded the possibilities of your floating point representation,
you cannot expect exact equality. A good strategy to deal with the
limitations of floating point representations is to think about
these limitations while you perform flaoting point operations, not
to assume that "all floating point results are inexact". If you use
floating point results in further calculation, you have _no_ guarantee
that 14 _decimal_ digits of your result are exact, even if you use
double on an implementation with DBL_DIG greater than 14.

Kurt Watzka