Why does floating point have a rounding error? How to work around it?
For example, the following:
flaot f = 1234.12345678F;
printf("%2f\n", f) //prints 1234.123413
and
printf("%8.9f\n", f) //prints 1234.123413086 15 7747 Mu***************@yahoo.com said:
Why does floating point have a rounding error?
Consider 1234.12345678
It's easy enough to deal with 1234. Here are the bits: 10011010010
So let's try to deal with 0.12345678, using binary notation.
So 0.1 (binary) is 0.5 (decimal), 0.01 (binary) is 0.25 (decimal), and
so on.
0.1 = 1/2 = 0.5  too large
0.01 = 1/4 = 0.25  too large
0.001 = 1/8 = 0.125  too large
0.0001 = 1/16 = 0.0625  too small
0.00011 = 3/32 = 0.09375  too small
0.000111 = 7/64 = 0.109375  too small
0.0001111 = 15/128 = 0.1171875  too small
0.00011111 = 31/256 = 0.12109375  too small
0.000111111 = 63/512 = 0.123046875  too small
0.0001111111 = 127/1024 = 0.1240234375  too large
0.00011111101 = 253/2048 = 0.12353515625  too large
0.000111111001 = 505/4096 = 0.123291015625  too small
0.0001111110011 = 1011/8192 = 0.1234130859375  too small
0.00011111100111 = 2023/16384 = 0.12347412109375  too large
0.000111111001101 = 4045/32768 = 0.123443603515625  too small
0.0001111110011011 = 8091/65536 = 0.1234588623046875  too large
So far, we've used 16 bits on this. Keep on calculatin', and find out
how many bits you need if you're to get an *exact* representation of
0.12345678. You might well be surprised by the result.
How to work around it?
That depends on what you want to achieve.

Richard Heathfield
"Usenet is a strange place"  dmr 29/7/1999 http://www.cpax.org.uk
email: rjh at the above domain,  www.
On Jun 16, 4:32 pm, Richard Heathfield <r...@see.sig.invalidwrote:
Mukesh_Singh_N...@yahoo.com said:
Why does floating point have a rounding error?
Consider 1234.12345678
It's easy enough to deal with 1234. Here are the bits: 10011010010
So let's try to deal with 0.12345678, using binary notation.
So 0.1 (binary) is 0.5 (decimal), 0.01 (binary) is 0.25 (decimal), and
so on.
0.1 = 1/2 = 0.5  too large
0.01 = 1/4 = 0.25  too large
0.001 = 1/8 = 0.125  too large
0.0001 = 1/16 = 0.0625  too small
0.00011 = 3/32 = 0.09375  too small
0.000111 = 7/64 = 0.109375  too small
0.0001111 = 15/128 = 0.1171875  too small
0.00011111 = 31/256 = 0.12109375  too small
0.000111111 = 63/512 = 0.123046875  too small
0.0001111111 = 127/1024 = 0.1240234375  too large
0.00011111101 = 253/2048 = 0.12353515625  too large
0.000111111001 = 505/4096 = 0.123291015625  too small
0.0001111110011 = 1011/8192 = 0.1234130859375  too small
0.00011111100111 = 2023/16384 = 0.12347412109375  too large
0.000111111001101 = 4045/32768 = 0.123443603515625  too small
0.0001111110011011 = 8091/65536 = 0.1234588623046875  too large
So far, we've used 16 bits on this. Keep on calculatin', and find out
how many bits you need if you're to get an *exact* representation of
0.12345678. You might well be surprised by the result.
How to work around it?
That depends on what you want to achieve.

Richard Heathfield
"Usenet is a strange place"  dmr 29/7/1999http://www.cpax.org.uk
email: rjh at the above domain,  www.
Thank you for replying with a very elaborate example, Richard. I would
disappoint you if I told you I am intrigued by the representation of
nonintegral decimal numbers in their binary form.
I know binary arithmetic with integrals. I sometimes wondered and
never bothered myself as to how decimals were represented as binaries.
I want to understand your example.
I can see a pattern in the representation.
0.1 is half.
0.01 is a right shift and you further halve it.
0.001 two right shifts further halving it and so on.
You lost me at 0.011. Can I please request you to explain. Mu***************@yahoo.com said:
<snip>
I know binary arithmetic with integrals. I sometimes wondered and
never bothered myself as to how decimals were represented as binaries.
The representation of floatingpoint numbers is implementationdefined.
IEEE 754 is common but not universal.
I want to understand your example.
I can see a pattern in the representation.
0.1 is half.
0.01 is a right shift and you further halve it.
0.001 two right shifts further halving it and so on.
You lost me at 0.011. Can I please request you to explain.
You understand binary integers. Extend the concept.
13 in binary is 1101 (1 * eight + 1 * four + 0 * two + 1 * one).
So each column represents a multiplier half as big as that of the column
to its left.
So we might reasonably think of the columns beyond the binary point as
representing a half, a quarter, an eighth, etc.
So 0.11 would be (half + quarter) = (threequarters) = 0.75
0.011 would be (quarter + eighth) = (threeeighths) = 0.375
and so on.
That isn't necessarily how they're stored internally, of course, but it
does give you a good idea of the number of bits you need for storing a
particular value to a particular precision. I recommend that you read http://docs.sun.com/source/8063568/ncg_goldberg.html
(Title: "What Every Computer Scientist Should Know About FloatingPoint
Arithmetic")

Richard Heathfield
"Usenet is a strange place"  dmr 29/7/1999 http://www.cpax.org.uk
email: rjh at the above domain,  www. Mu***************@yahoo.com wrote, On 16/06/07 11:49:
Why does floating point have a rounding error?
Calculate 2/3 as a decimal number. Come back when you have understood
the answer to your question or when you have finished writing it without
any rounding or truncation. I'll even get you started 0.6667 (I've
rounded it here).
How to work around it?
Use sufficient care and analysis for the problem at hand. This is a
general problem, not a C specific one, so comp.programming, and there is
no one correct solution for all situations.

Flash Gordon Mu***************@yahoo.com wrote:
Why does floating point have a rounding error?
Knowledge of the Great Mysteries is reserved for those
who are worthy. To prove your worth, you must undertake a
Quest and complete it successfully. Your Quest, Mukesh, is
to discover what fraction one day is of one week, and express
the answer as a decimal number, using as many decimal places
as are needed for perfect accuracy. When you have done this
I will know you are indeed worthy, and I will reveal to you
the secret origin of rounding errors.

Eric Sosman es*****@acmdotorg.invalid
On Jun 16, 4:53 pm, Richard Heathfield <r...@see.sig.invalidwrote:
Mukesh_Singh_N...@yahoo.com said:
<snip>
I know binary arithmetic with integrals. I sometimes wondered and
never bothered myself as to how decimals were represented as binaries.
The representation of floatingpoint numbers is implementationdefined.
IEEE 754 is common but not universal.
I want to understand your example.
I can see a pattern in the representation.
0.1 is half.
0.01 is a right shift and you further halve it.
0.001 two right shifts further halving it and so on.
You lost me at 0.011. Can I please request you to explain.
You understand binary integers. Extend the concept.
13 in binary is 1101 (1 * eight + 1 * four + 0 * two + 1 * one).
So each column represents a multiplier half as big as that of the column
to its left.
So we might reasonably think of the columns beyond the binary point as
representing a half, a quarter, an eighth, etc.
So 0.11 would be (half + quarter) = (threequarters) = 0.75
0.011 would be (quarter + eighth) = (threeeighths) = 0.375
and so on.
That isn't necessarily how they're stored internally, of course, but it
does give you a good idea of the number of bits you need for storing a
particular value to a particular precision. I recommend that you readhttp://docs.sun.com/source/8063568/ncg_goldberg.html
(Title: "What Every Computer Scientist Should Know About FloatingPoint
Arithmetic")

Richard Heathfield
"Usenet is a strange place"  dmr 29/7/1999http://www.cpax.org.uk
email: rjh at the above domain,  www.
Thank you so very much, Richard. You just taught me something
*fantabulous*. I just learnt something terrific, something I could not
have learnt reading a thousand words. Actually, I think I've just
understood the IEEE 7541985 implementation in a nutshell.
Is this rule of representing decimals in binary applicable only to
754?
And then I revisited your previous table wherein you try to reach a
precision for 1234.12345678 by heuristic computation. It suddenly
removed a big block in my head.
Thank you, everyday.
I recommend that you readhttp://docs.sun.com/source/8063568/ncg_goldberg.html
>
(Title: "What Every Computer Scientist Should Know About FloatingPoint
Arithmetic")

Richard Heathfield
"Usenet is a strange place"  dmr 29/7/1999http://www.cpax.org.uk
email: rjh at the above domain,  www.
Thank you. I certainly will. Mu***************@yahoo.com wrote:
On Jun 16, 4:32 pm, Richard Heathfield <r...@see.sig.invalidwrote:
> Mukesh_Singh_N...@yahoo.com said:
>>Why does floating point have a rounding error?
Consider 1234.12345678
It's easy enough to deal with 1234. Here are the bits: 10011010010
So let's try to deal with 0.12345678, using binary notation.
So 0.1 (binary) is 0.5 (decimal), 0.01 (binary) is 0.25 (decimal), and so on. 0.1 = 1/2 = 0.5  too large 0.01 = 1/4 = 0.25  too large 0.001 = 1/8 = 0.125  too large 0.0001 = 1/16 = 0.0625  too small 0.00011 = 3/32 = 0.09375  too small 0.000111 = 7/64 = 0.109375  too small 0.0001111 = 15/128 = 0.1171875  too small 0.00011111 = 31/256 = 0.12109375  too small 0.000111111 = 63/512 = 0.123046875  too small 0.0001111111 = 127/1024 = 0.1240234375  too large 0.00011111101 = 253/2048 = 0.12353515625  too large 0.000111111001 = 505/4096 = 0.123291015625  too small 0.0001111110011 = 1011/8192 = 0.1234130859375  too small 0.00011111100111 = 2023/16384 = 0.12347412109375  too large 0.000111111001101 = 4045/32768 = 0.123443603515625  too small 0.0001111110011011 = 8091/65536 = 0.1234588623046875  too large
So far, we've used 16 bits on this. Keep on calculatin', and find out how many bits you need if you're to get an *exact* representation of 0.12345678. You might well be surprised by the result.
>>How to work around it?
That depends on what you want to achieve.
 Richard Heathfield "Usenet is a strange place"  dmr 29/7/1999http://www.cpax.org.uk email: rjh at the above domain,  www.
Thank you for replying with a very elaborate example, Richard. I would
disappoint you if I told you I am intrigued by the representation of
nonintegral decimal numbers in their binary form.
I know binary arithmetic with integrals. I sometimes wondered and
never bothered myself as to how decimals were represented as binaries.
I want to understand your example.
I can see a pattern in the representation.
0.1 is half.
0.01 is a right shift and you further halve it.
0.001 two right shifts further halving it and so on.
You lost me at 0.011. Can I please request you to explain.
Here's something to chew on..
00111111 10111111 10011010 11011101 00010000 10010001 11001000 10010101
Exp = 1019 (3)
111 11111101
Man = .11111 10011010 11011101 00010000 10010001 11001000 10010101
1.2345678000000000e01
I hope it didn't wrap on you.

Joe Wright
"Everything should be made as simple as possible, but not simpler."
 Albert Einstein 
Eric Sosman wrote:
Mu***************@yahoo.com wrote:
>Why does floating point have a rounding error?
Knowledge of the Great Mysteries is reserved for those
who are worthy. To prove your worth, you must undertake a
Quest and complete it successfully. Your Quest, Mukesh, is
to discover what fraction one day is of one week, and express
the answer as a decimal number, using as many decimal places
as are needed for perfect accuracy. When you have done this
I will know you are indeed worthy, and I will reveal to you
the secret origin of rounding errors.
Easy. Just use a septal base (which is not decimal), get 0.1.
:) (Very useful for writing lock combinations).

<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net

Posted via a free Usenet account from http://www.teranews.com Mu***************@yahoo.com wrote:
>
Why does floating point have a rounding error? How to work around
it? For example, the following:
flaot f = 1234.12345678F;
printf("%2f\n", f) //prints 1234.123413
and
printf("%8.9f\n", f) //prints 1234.123413086
It doesn't have a rounding error. It has a precision limit.

<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net

Posted via a free Usenet account from http://www.teranews.com
CBFalconer wrote:
Mu***************@yahoo.com wrote:
>Why does floating point have a rounding error? How to work around it? For example, the following:
flaot f = 1234.12345678F; printf("%2f\n", f) //prints 1234.123413 and printf("%8.9f\n", f) //prints 1234.123413086
It doesn't have a rounding error. It has a precision limit.
Indeed. 1.23412341e+03 is all there is in 32 bits.

Joe Wright
"Everything should be made as simple as possible, but not simpler."
 Albert Einstein 
CBFalconer wrote:
Eric Sosman wrote:
>Mu***************@yahoo.com wrote:
>>Why does floating point have a rounding error?
Knowledge of the Great Mysteries is reserved for those who are worthy. To prove your worth, you must undertake a Quest and complete it successfully. Your Quest, Mukesh, is to discover what fraction one day is of one week, and express the answer as a decimal number, using as many decimal places as are needed for perfect accuracy. When you have done this I will know you are indeed worthy, and I will reveal to you the secret origin of rounding errors.
Easy. Just use a septal base (which is not decimal), get 0.1.
:) (Very useful for writing lock combinations).
Since "as a decimal number" was clearly specified in the
rules of the Quest, your change of base is not septal but septic.
I'll have to look up the traditional rules for punishment of
failure. "Something lingering, with boiling oil in it, I fancy.
Something of that sort. I think boiling oil occurs in it, but
I'm not sure. I know it's something humorous, but lingering,
with either boiling oil or melted lead."
AND you'll never get to learn about rounding error. Nyaahh!

Eric Sosman es*****@acmdotorg.invalid
Groovy hepcat Mu***************@yahoo.com was jivin' on Sat, 16 Jun
2007 05:07:47 0700 in comp.lang.c.
Re: Floating point rounding error's a cool scene! Dig it!
>On Jun 16, 4:53 pm, Richard Heathfield <r...@see.sig.invalidwrote:
> Mukesh_Singh_N...@yahoo.com said:
>The representation of floatingpoint numbers is implementationdefined. IEEE 754 is common but not universal.
I want to understand your example.
I can see a pattern in the representation.
0.1 is half.
0.01 is a right shift and you further halve it.
0.001 two right shifts further halving it and so on.
You lost me at 0.011. Can I please request you to explain.
You understand binary integers. Extend the concept.
13 in binary is 1101 (1 * eight + 1 * four + 0 * two + 1 * one).
So each column represents a multiplier half as big as that of the column to its left.
So we might reasonably think of the columns beyond the binary point as representing a half, a quarter, an eighth, etc.
So 0.11 would be (half + quarter) = (threequarters) = 0.75
0.011 would be (quarter + eighth) = (threeeighths) = 0.375
and so on.
Is this rule of representing decimals in binary applicable only to 754?
This doesn't represent decimals. It represents values. Values are
neither decimal nor binary, but may be expressed in decimal or binary
or, indeed, any other number system, such as hexadecimal or octal. A
computer stores values expressed internally in binary, but may read in
or write out the values in decimal or other systems.
What Richard showed you is not a complete floating point
representation. It was simply a binary representation of a value. This
is refered to as "fixed point". Floating point representations have a
mantissa part and an exponent part. These parts are (typically)
expressed in binary. For example, instead of representing .75 as .11
in fixed point binary, it might be represented as (just to keep things
simple) an 8 bit binary mantissa, 00000011, and an 8 bit exponent,
11111110 (that's 2 expressed as an 8 bit binary number). This
represents 3 * 2 ** 2 (using ** for exponentiation) or 3 >2. Real
floating point implementations, however, use more bits for the
mantissa and exponent, often with separate sign bits.

Dig the even newer still, yet more improved, sig! http://alphalink.com.au/~phaywood/
"Ain't I'm a dog?"  Ronny Self, Ain't I'm a Dog, written by G. Sherry & W. Walker.
I know it's not "technically correct" English; but since when was rock & roll "technically correct"?
On Sat, 16 Jun 2007 08:05:28 0400, Eric Sosman
<es*****@acmdotorg.invalidwrote: Mu***************@yahoo.com wrote:
Why does floating point have a rounding error?
Knowledge of the Great Mysteries is reserved for those
who are worthy. To prove your worth, you must undertake a
Quest and complete it successfully. Your Quest, Mukesh, is
to discover what fraction one day is of one week, and express
the answer as a decimal number, using as many decimal places
as are needed for perfect accuracy. When you have done this
I will know you are indeed worthy, and I will reveal to you
the secret origin of rounding errors.
<OTDidn't the French Revolution, among its many variously intriguing
and alarming ideas, (try to) change the week to 10 days? </>
 formerly david.thompson1  achar(64)  worldnet.att.net
David Thompson <da************@verizon.netwrites:
On Sat, 16 Jun 2007 08:05:28 0400, Eric Sosman
<es*****@acmdotorg.invalidwrote: Mu***************@yahoo.com wrote:
Why does floating point have a rounding error?
Knowledge of the Great Mysteries is reserved for those
who are worthy. To prove your worth, you must undertake a
Quest and complete it successfully. Your Quest, Mukesh, is
to discover what fraction one day is of one week, and express
the answer as a decimal number, using as many decimal places
as are needed for perfect accuracy. When you have done this
I will know you are indeed worthy, and I will reveal to you
the secret origin of rounding errors.
<OTDidn't the French Revolution, among its many variously intriguing
and alarming ideas, (try to) change the week to 10 days? </>
Right. This calendar was in use 13 years or so.
Yours,

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