473,549 Members | 2,791 Online
Bytes | Software Development & Data Engineering Community
+ Post

Home Posts Topics Members FAQ

Floating point arithmetic.

Hi there.
I am cross posting this on comp.lang.c as well: sorry for same.
The problem I am facing is as follows:
For example:
double a= 0.15;
double b=2.4;
const double VERYTINY =1.e-10;
I know b/a = 16 and hence the remainder is zero; but I am not
able to find any suitable thing to encode it into in c.
for example (fmod(b,a)>VERY TINY) returns true!
Now for this particular instance, (fmodf(b,a)>VER YTINY)
does return false.
But now if
a=0.15;
b=4.5;
then fmodf and fmod both don't help...

any suggestions on this?
I was pointed to a reference on floating point arithmetic, where they talk
of ulps etc, but is there a small function or fix to deal with this problem
available somewhere?

thanks,
amit.
Jul 22 '05 #1
14 2607

"Amit Bhatia" <bh*****@nospam .com> wrote:
The problem I am facing is as follows:
For example:
Example of what? It makes more sense to state the
thing you want to give an example of, before giving
examples of it.
double a= 0.15;
double b=2.4;
const double VERYTINY =1.e-10;
I know b/a = 16 and hence the remainder is zero;
but I am not able to find any suitable thing to
encode it into in c.
Define "encode".

Define "it".

If you want to code b/a in C, the code is:

b/a
for example
Example of what? Please give some clue as to the thing
you're trying to give an example OF, before giving
examples!
(fmod(b,a)>VERY TINY) returns true!
It returns false on my system. However, it seems to me
you're pushing the granularity of type double to the max.
Even if you WERE off by one part per 1e10, so what?
That's one part per ten billion. Who cares?

But if you really DO care, try acquiring a compiler that
supports type "long double". Or, perhaps, use a computer
with a 64-bit processor and compiler that defines "double"
to be 64 bits. Even if 4 or so of those bits were used
for sign and exponent, you'd still have a granularity of
about 1 part per quintillion (1e18). More precision than
you'll EVER need.
Now for this particular instance, (fmodf(b,a)>VER YTINY)
does return false.
What is fmodf? That's not a part of the C or C++ std.
libraries. I don't have it on my compiler (DJGPP) either.
But now if
a=0.15;
b=4.5;
then fmodf and fmod both don't help...
I get b/a = 30 and (fmod(b,a)>VERY TINY) = false.

What do YOU get?
any suggestions on this?
Suggestions on what, precisely?
I was pointed to a reference on floating point
arithmetic, where they talk of ulps etc,
What's "ulps"?
is there a small function or fix to deal with this problem
available somewhere?


What problem are you trying to fix?

--
Cheers,
Robbie Hatley
Tustin, CA, USA
lo**********@pa cbell.net
http://home.pacbell.net/earnur/

Jul 22 '05 #2
On Sun, 11 Jul 2004 01:17:09 -0700, Robbie Hatley <lonewolfintj at pacbell
dot net> wrote:
I was pointed to a reference on floating point
arithmetic, where they talk of ulps etc,


What's "ulps"?


ULP stands for unit in the last place. Its a measure of rounding error and
a common enough term in floating point arithmetic.

john
Jul 22 '05 #3
On Sat, 10 Jul 2004 23:08:01 -0500, Amit Bhatia <bh*****@nospam .com> wrote:
Hi there.
I am cross posting this on comp.lang.c as well: sorry for same.
The problem I am facing is as follows:
For example:
double a= 0.15;
double b=2.4;
const double VERYTINY =1.e-10;
I know b/a = 16 and hence the remainder is zero; but I am not
able to find any suitable thing to encode it into in c.
for example (fmod(b,a)>VERY TINY) returns true!
Now for this particular instance, (fmodf(b,a)>VER YTINY)
does return false.
But now if
a=0.15;
b=4.5;
then fmodf and fmod both don't help...

any suggestions on this?
Maybe use a bigger value for VERYTINY.
Maybe use integral or rational aritmetic.
Maybe use fixed point arithmetic.
I was pointed to a reference on floating point arithmetic, where they
talk
of ulps etc, but is there a small function or fix to deal with this
problem
available somewhere?


There is not easy answer to the problem of floating point rounding errors.
The simplest thing is not to write code that depends on any particular
exactness of floating point arithmetic. Obviously this is not always
possible but it is possible more often than people think.

So the asnwer to your question really depends on what kind of problem you
are actually trying to solve.

john
Jul 22 '05 #4
On Sat, 10 Jul 2004 23:08:01 -0500 in comp.lang.c++, Amit Bhatia
<bh*****@nospam .com> wrote,
Hi there.
I am cross posting this on comp.lang.c as well: sorry for same.
In fact you did so a few days ago and got some good tips.
The problem I am facing is as follows:
For example:
double a= 0.15;
double b=2.4;
const double VERYTINY =1.e-10;
b = two and two fifths. Fifths can not be represented exactly in base
two floating point format, so this number cannot be stored exactly.
a is 16 times that, so of course has a similar approximation.

And so on. Never expect any form of exactitude from any floating point
operation, and your life will be less frustrating.
I know b/a = 16 and hence the remainder is zero; but I am not
able to find any suitable thing to encode it into in c.


It may be close to zero, or it may be just less than 0.15.

If you persist in this folly then I guess you need to test for both
possibilities.
Jul 22 '05 #5
Amit Bhatia <bh*****@nospam .com> wrote in message news:<cc******* ***@news.ks.uiu c.edu>...
Hi there.
I am cross posting this on comp.lang.c as well: sorry for same.
The problem I am facing is as follows:
For example:
double a= 0.15;
double b=2.4;
const double VERYTINY =1.e-10;
I know b/a = 16 and hence the remainder is zero; but I am not
able to find any suitable thing to encode it into in c.
for example (fmod(b,a)>VERY TINY) returns true!
Now for this particular instance, (fmodf(b,a)>VER YTINY)
does return false.
But now if
a=0.15;
b=4.5;
then fmodf and fmod both don't help...

any suggestions on this?
I was pointed to a reference on floating point arithmetic, where they talk
of ulps etc, but is there a small function or fix to deal with this problem
available somewhere?

thanks,
amit.

int main() {
double a = 0.15;
double b = 2.4;
const double VERYTINY =1.e-10;
printf("%d", (fmod(b,a) > VERYTINY));
return 0;
}

This gives me false!
Jul 22 '05 #6
Amit Bhatia <bh*****@nospam .com> wrote in message news:<cc******* ***@news.ks.uiu c.edu>...
Hi there.
I am cross posting this on comp.lang.c as well: sorry for same.
The problem I am facing is as follows:
For example:
double a= 0.15;
double b=2.4;
const double VERYTINY =1.e-10;
I know b/a = 16 and hence the remainder is zero; but I am not
able to find any suitable thing to encode it into in c.
for example (fmod(b,a)>VERY TINY) returns true!
Now for this particular instance, (fmodf(b,a)>VER YTINY)
does return false.
But now if
a=0.15;
b=4.5;
then fmodf and fmod both don't help...

any suggestions on this?
I was pointed to a reference on floating point arithmetic, where they talk
of ulps etc, but is there a small function or fix to deal with this problem
available somewhere?

thanks,
amit.

Try the following loop

cout.precision( 20);
cout << b << endl << a << endl;
while(b>a)
b -= a;
cout << b << endl << a << endl;

and see the value of a and b

Your problem is that the method of not comparing floating
point values but comparing their difference to a certain limit
does not work for modulo arithmetic. The reason is simple.

Assume x(mod y) = 0
That is, x = yp

In your case, x and y are floating point values,
while p is an integer. Since p is an integer, you
can say x is divisible by y.

But a computer cannot store floating point numbers exactly.
so a floating point number x is actually stored as (x+dx)
or as (x-dx). In your case, 4.5 can be exactly stored, but
neither can 2.4 or 0.15 . Now consider what happens when you
divide (x+dx) or (x-dx) by y. (x,y,p > 0)

(x+dx)(mod y) = dx
This is ok, your comparison will work in this case, as your
VERYTINY is > dx

However, (x-dx)(mod y) = (yp-dx)(mod y)
= (y(p-1) + y - dx)(mod y) = y-dx
This y-dx is definitely mmuch greater than your VERYTINY
and therefore you do not get the desired result.

I'll give you an example with integers.

Consider dividing 57 by 19. 57(mod 19) = 0
58(mod 19) = 1 ( x+dx case)
56(mod 19) = 18 (x-dx case)

So you should write
( fmod(b,a)<VERYT INY || (a-fmod(b,a))<VERY TINY )

-Arijit
Jul 22 '05 #7

When using floating-point, usually we don't test for equality, instead do
the kind of test you do (fmod(b, a) > VERYTINY)...
But if we take a look at the case in particular, the test doesn't fail on my
machine, and I suspect the > VERYTINY doesn't fail in yours, except for
negative numbers. That is,

2.4 can only be finitely represented in binary as 2.3999... that is
10.0110011(0011 )
0.15 as 0.14(9) (or 0.00100110011(0 011))

in my machine, perhaps yours too, fmod(2.4, 0.15) is exactly 0. This is not
surprising if we examine the mantissa bits of both numbers, they are equal.

but 4.5 does have a finite binary representation as 100.1

but fmod(4.5, 0.15) is only 1.66533e-16, and so is smaller than 1e-10... and
I don't know why the test should fail....
Since the case of the denominator being represented exactly implies that a
multiple is represented exactly too

So, given best possible representations of the input values the test
(abs(fmod(b, a)) < VERYTINY)) should work.

One exception arises when input values are themselves approximations, and,
in particular, the dividend is slightly less than it would take to be an
exact multiple. In that case, the remainder will be less than, but very
close to the denominator. So, we must test the result also against it being
very close to the denominator.

So, the full test, that should work for all cases would be (assuming we want
to know about approximate multiples):

(abs(fmod(b,a)) < VERYTINY) || (abs(fmod(b,a) - a) < VERYTINY)

Now, the value you choose for VERYTINY, depends on the application and how
many operations you have done on the input values. Of course, because the
result of fmod is in absolute value always less than a, if you know your
inputs are positive numbers, you can drop both abs function calls.
Miguel Ramos
Jul 22 '05 #8
What is fmodf? That's not a part of the C or C++ std.
libraries. I don't have it on my compiler (DJGPP) either.


actually fmodf does exist, it is part of the C99 standard, and most modern
compilers have it.
but I think we shouldn't be very harsh, it's obvious he meant modf.
Jul 22 '05 #9

"Miguel" <mi****@exstare .com> wrote in message news:40******** *************** @news.telepac.p t...
What is fmodf? That's not a part of the C or C++ std.
libraries. I don't have it on my compiler (DJGPP) either.
actually fmodf does exist, it is part of the C99 standard,


"fmodf" is a not a keyword in C or C++, nor is it
a part of their standard libraries, according to my
reading. I found reference to "fmodf" in the
documentation for my compiler (DJGPP), where it is
listed as a "non-ANSI extention to fmod". So unless
fmodf has been added to the standard libraries very
recently, it's not standard.
and most modern compilers have it.
Most compilers have lots of added non-ANSI functions.
Useful, but not standard.
it's obvious he meant modf.


Probably.

--
Cheers,
Robbie Hatley
Tustin, CA, USA
lo**********@pa cbell.net
http://home.pacbell.net/earnur/
Jul 22 '05 #10

This thread has been closed and replies have been disabled. Please start a new discussion.

Similar topics

4
7830
by: Roger Leigh | last post by:
Hello, I'm writing a fixed-precision floating point class, based on the ideas in the example fixed_pt class in the "Practical C++ Programming" book by Steve Oualline (O' Reilly). This uses a long int to store the value, and the precision (number of decimal points) is variable (it's a templated class): template <size_t _decimal_places =...
3
2119
by: Mantorok Redgormor | last post by:
What have some of you guys read to have a solid understanding of how floating-point numbers are represented or handled by the processor and what the difference between single and double precision is? I found this: http://docs.sun.com/source/806-3568/ncg_goldberg.html Not sure if this is what I should be reading? Maybe a more authoritive...
687
23004
by: cody | last post by:
no this is no trollposting and please don't get it wrong but iam very curious why people still use C instead of other languages especially C++. i heard people say C++ is slower than C but i can't believe that. in pieces of the application where speed really matters you can still use "normal" functions or even static methods which is...
24
2209
by: j0mbolar | last post by:
C supports single precision floating point and double precision floating point but does it support fixed floating point? i've read that fixed floating point is more accurate than single precision floating point when dealing with dollars and cents.
10
2247
by: Shawn | last post by:
Hello all, I apologize as I am sure this has probably been dealth with before... but I am doing an exercise from "Practical C Programming" and I have been unable to get it to work perfectly due to problems with floating point arithmetic and I am looking for a way to solve it. See the code below... Given a certain amount of change (below...
7
3380
by: Vinoth | last post by:
I'm working in an ARM (ARM9) system which does not have Floating point co-processor or Floating point libraries. But it does support long long int (64 bits). Can you provide some link that would discuss about ways to emulate floating point calculations with just long int or long long int. For eg., if i've a formula X=(1-b)*Y + b*Z in floating...
1
2393
by: Satpreet | last post by:
I'm looking to simulate the behavior of a digital hardware arithmetic block in a C/C++ program. I was just wondering if there are any libraries (with datatypes and overloaded operators etc.) available for arithmetic operations on Fixed/Block-Floating Point variables ?? Alternatively could anyone throw me a pointer to sample code if...
32
4063
by: ma740988 | last post by:
template <class T> inline bool isEqual( const T& a, const T& b, const T epsilon = std::numeric_limits<T>::epsilon() ) { const T diff = a - b; return ( diff <= epsilon ) && ( diff >= -epsilon ); } int main() { std::deque<double> pt ;
70
3547
by: Robert Gamble | last post by:
9899:1999 5.1.2.3 Example 4 reads: "EXAMPLE 4 Implementations employing wide registers have to take care to honor appropriate semantics. Values are independent of whether they are represented in a register or in memory. For example, an implicit spilling of a register is not permitted to alter the value. Also, an explicit store and load is...
0
7554
marktang
by: marktang | last post by:
ONU (Optical Network Unit) is one of the key components for providing high-speed Internet services. Its primary function is to act as an endpoint device located at the user's premises. However, people are often confused as to whether an ONU can Work As a Router. In this blog post, we’ll explore What is ONU, What Is Router, ONU & Router’s main...
0
7480
by: Hystou | last post by:
Most computers default to English, but sometimes we require a different language, especially when relocating. Forgot to request a specific language before your computer shipped? No problem! You can effortlessly switch the default language on Windows 10 without reinstalling. I'll walk you through it. First, let's disable language...
0
7751
Oralloy
by: Oralloy | last post by:
Hello folks, I am unable to find appropriate documentation on the type promotion of bit-fields when using the generalised comparison operator "<=>". The problem is that using the GNU compilers, it seems that the internal comparison operator "<=>" tries to promote arguments from unsigned to signed. This is as boiled down as I can make it. ...
1
7511
by: Hystou | last post by:
Overview: Windows 11 and 10 have less user interface control over operating system update behaviour than previous versions of Windows. In Windows 11 and 10, there is no way to turn off the Windows Update option using the Control Panel or Settings app; it automatically checks for updates and installs any it finds, whether you like it or not. For...
0
7840
tracyyun
by: tracyyun | last post by:
Dear forum friends, With the development of smart home technology, a variety of wireless communication protocols have appeared on the market, such as Zigbee, Z-Wave, Wi-Fi, Bluetooth, etc. Each protocol has its own unique characteristics and advantages, but as a user who is planning to build a smart home system, I am a bit confused by the...
0
6088
agi2029
by: agi2029 | last post by:
Let's talk about the concept of autonomous AI software engineers and no-code agents. These AIs are designed to manage the entire lifecycle of a software development project—planning, coding, testing, and deployment—without human intervention. Imagine an AI that can take a project description, break it down, write the code, debug it, and then...
1
5396
isladogs
by: isladogs | last post by:
The next Access Europe User Group meeting will be on Wednesday 1 May 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM). In this session, we are pleased to welcome a new presenter, Adolph Dupré who will be discussing some powerful techniques for using class modules. He will explain when you may want to use classes...
0
3509
by: adsilva | last post by:
A Windows Forms form does not have the event Unload, like VB6. What one acts like?
1
1087
muto222
by: muto222 | last post by:
How can i add a mobile payment intergratation into php mysql website.

By using Bytes.com and it's services, you agree to our Privacy Policy and Terms of Use.

To disable or enable advertisements and analytics tracking please visit the manage ads & tracking page.