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Floating point rounding error

 P: n/a 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 Jun 16 '07 #1
15 Replies

 P: n/a 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. Jun 16 '07 #2

 P: n/a On Jun 16, 4:32 pm, Richard Heathfield

 P: n/a Mu***************@yahoo.com said: I know binary arithmetic with integrals. I sometimes wondered and never bothered myself as to how decimals were represented as binaries. The representation of floating-point numbers is implementation-defined. 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) = (three-quarters) = 0.75 0.011 would be (quarter + eighth) = (three-eighths) = 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/806-3568/ncg_goldberg.html (Title: "What Every Computer Scientist Should Know About Floating-Point Arithmetic") -- Richard Heathfield "Usenet is a strange place" - dmr 29/7/1999 http://www.cpax.org.uk email: rjh at the above domain, - www. Jun 16 '07 #4

 P: n/a 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 Jun 16 '07 #5

 P: n/a 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*****@acm-dot-org.invalid Jun 16 '07 #6

 P: n/a On Jun 16, 4:53 pm, Richard Heathfield I know binary arithmetic with integrals. I sometimes wondered and never bothered myself as to how decimals were represented as binaries. The representation of floating-point numbers is implementation-defined. 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) = (three-quarters) = 0.75 0.011 would be (quarter + eighth) = (three-eighths) = 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/806-3568/ncg_goldberg.html (Title: "What Every Computer Scientist Should Know About Floating-Point 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 754-1985 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. Jun 16 '07 #7

 P: n/a I recommend that you readhttp://docs.sun.com/source/806-3568/ncg_goldberg.html > (Title: "What Every Computer Scientist Should Know About Floating-Point 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. Jun 16 '07 #8

 P: n/a Mu***************@yahoo.com wrote: On Jun 16, 4:32 pm, Richard Heathfield Mukesh_Singh_N...@yahoo.com said: >>Why does floating point have a rounding error? Consider 1234.12345678It's easy enough to deal with 1234. Here are the bits: 10011010010So 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), andso on.0.1 = 1/2 = 0.5 - too large0.01 = 1/4 = 0.25 - too large0.001 = 1/8 = 0.125 - too large0.0001 = 1/16 = 0.0625 - too small0.00011 = 3/32 = 0.09375 - too small0.000111 = 7/64 = 0.109375 - too small0.0001111 = 15/128 = 0.1171875 - too small0.00011111 = 31/256 = 0.12109375 - too small0.000111111 = 63/512 = 0.123046875 - too small0.0001111111 = 127/1024 = 0.1240234375 - too large0.00011111101 = 253/2048 = 0.12353515625 - too large0.000111111001 = 505/4096 = 0.123291015625 - too small0.0001111110011 = 1011/8192 = 0.1234130859375 - too small0.00011111100111 = 2023/16384 = 0.12347412109375 - too large0.000111111001101 = 4045/32768 = 0.123443603515625 - too small0.0001111110011011 = 8091/65536 = 0.1234588623046875 - too largeSo far, we've used 16 bits on this. Keep on calculatin', and find outhow many bits you need if you're to get an *exact* representation of0.12345678. 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.ukemail: 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 non-integral 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 1.2345678000000000e-01 I hope it didn't wrap on you. -- Joe Wright "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Jun 16 '07 #9

 P: n/a 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). -- cbfalconer at maineline dot net -- Posted via a free Usenet account from http://www.teranews.com Jun 16 '07 #10

 P: n/a 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. -- cbfalconer at maineline dot net -- Posted via a free Usenet account from http://www.teranews.com Jun 16 '07 #11

 P: n/a CBFalconer wrote: Mu***************@yahoo.com wrote: >Why does floating point have a rounding error? How to work aroundit? For example, the following: flaot f = 1234.12345678F; printf("%2f\n", f) //prints 1234.123413and 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 --- Jun 16 '07 #12

 P: n/a 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 thosewho are worthy. To prove your worth, you must undertake aQuest and complete it successfully. Your Quest, Mukesh, isto discover what fraction one day is of one week, and expressthe answer as a decimal number, using as many decimal placesas are needed for perfect accuracy. When you have done thisI will know you are indeed worthy, and I will reveal to youthe 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*****@acm-dot-org.invalid Jun 16 '07 #13

 P: n/a 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 Mukesh_Singh_N...@yahoo.com said: >The representation of floating-point numbers is implementation-defined.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 columnto its left.So we might reasonably think of the columns beyond the binary point asrepresenting a half, a quarter, an eighth, etc.So 0.11 would be (half + quarter) = (three-quarters) = 0.750.011 would be (quarter + eighth) = (three-eighths) = 0.375and so on. Is this rule of representing decimals in binary applicable only to754? 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"? Jun 20 '07 #14

 P: n/a On Sat, 16 Jun 2007 08:05:28 -0400, Eric Sosman - formerly david.thompson1 || achar(64) || worldnet.att.net Jul 1 '07 #15

 P: n/a David Thompson Right. This calendar was in use 13 years or so. Yours, -- Jean-Marc Jul 2 '07 #16

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