Given a double value like 1.00500000274996E-8, how do I convert it to it's non scientific format with a maximum number of digits after the decimal point - in this case with 8 digits it would be 1.00500000?
The conversion should not pad with zeros, so 2007 would come out as 2007, and 2012.33 and 2012.33.
I've tried lots of combinations using Format, FormatFloat, FloatToStrF but can't quite seem to hit the jackpot. Many thanks for any help.
Edit: I should clarify that I am trying to convert it to a string representation, without the Exponent (E) part.
FormatFloat('0.######################', 1.00500000274996E-8) should do the trick.
Output is: 0,0000000100500000274996
It will not output more digits than absolutely necessary.
See John Herbster's Exact Float to String Routines in CodeCentral. Perhaps not exactly what youre after but might be good starting point... CC item's description:
This module includes
(a) functions for converting a floating binary point number to its *exact* decimal representation in an AnsiString;
(b) functions for parsing the floating point types into sign, exponent, and mantissa; and
(c) function for analyzing a extended float number into its type (zero, normal, infinity, etc.)
Its intended use is for trouble shooting problems with floating point numbers.
His DecimalRounding routines might be of intrest too.
Related
I'm getting a rounding oddity in Delphi 2010, where some numbers are rounding down in roundto, but up in formatfloat.
I'm entirely aware of binary representation of decimal numbers sometimes giving misleading results, but in that case I would expect formatfloat and roundto to give the same result.
I've also seen advice that this is the sort of thing "Currency" should be used for, but as you can see below, Currency and Double give the same results.
program testrounding;
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils,Math;
var d:Double;
c:Currency;
begin
d:=534.50;
c:=534.50;
writeln('Format: ' +formatfloat('0',d));
writeln('Roundto: '+formatfloat('0',roundto(d,0)));
writeln('C Format: ' +formatfloat('0',c));
writeln('C Roundto: '+formatfloat('0',roundto(c,0)));
readln;
end.
The results are as follows:
Format: 535
Roundto: 534
C Format: 535
C Roundto: 534
I've looked at Why is the result of RoundTo(87.285, -2) => 87.28 and the suggested remedies do not seem to apply.
First of all, we can remove Currency from the question, because the two functions that you use don't have Currency overloads. The value is converted to an IEEE754 floating point value and then follows the same path as your Double code.
Let's look at RoundTo first of all. It is quick to check, using the debugger, or an additional Writeln that RoundTo(d,0) = 534. Why is that?
Well, the documentation for RoundTo says:
Rounds a floating-point value to a specified digit or power of ten using "Banker's rounding".
Indeed in the implementation of RoundTo we see that the rounding mode is temporarily switched to TRoundingMode.rmNearest before being restored to its original value. The rounding mode only applies when the value is exactly half way between two integers. Which is precisely the case we have here.
So Banker's rounding applies. Which means that when the value is exactly half way between two integers, the rounding algorithm chooses the adjacent even integer.
So it makes sense that RoundTo(534.5,0) = 534, and equally you can check that RoundTo(535.5,0) = 536.
Understanding FormatFloat is quite a different matter. Quite frankly its behaviour is somewhat opaque. It performs an ad hoc rounding in code that differs for different platforms. For instance it is assembler on 32 bit Windows, but Pascal on 64 bit Windows. The overall approach appears to be to take the mantissa of the floating point value, convert it to an integer, convert that to text digits, and then perform the rounding based on those text digits. No respect is paid to the current rounding mode when the rounding is performed, and the algorithm appears to implement the round half away from zero policy. However, even that is not implemented robustly for all possible floating point values. It works correctly for your value, but for values with more digits in the mantissa the algorithm breaks down.
In fact it is fairly well known that the Delphi RTL routines for converting between floating point values and text are fundamentally broken by design. There are no routines in the Delphi RTL that can correctly convert from text to float, or from float to text. In fact, I have recently implemented my own conversion routines, that do this correctly, based on existing open source code used by other language runtimes. One of these days I will get around to publishing this code for use by others.
I'm not sure what your exact needs are, but if you are wishing to exert some control over rounding, then you can do so if you take charge of the rounding. Whilst RoundTo always uses Banker's rounding, you can instead use Round which uses the current rounding mode. This will allow you to perform the round using the rounding algorithm of your choice (by calling SetRoundMode), and then you can convert the rounded value to text. That's the key. Keep the value in an arithmetic type, perform the rounding, and only convert to text at the very last moment, after the correct rounding has been applied.
In this case, the value 534.5 is exactly representable in Double precision.
Looking into source code, reveals that the FormatFloat function rounds upwards if the last pending digit is 5 or more.
RoundTo uses the Banker's rounding, and rounds to nearest even number (534) in this case.
A simple question that turned out to be quite complex:
How do I turn a float to a String in GForth? The desired behavior would look something like this:
1.2345e fToString \ takes 1.2345e from the float stack and pushes (addr n) onto the data stack
After a lot of digging, one of my colleagues found it:
f>str-rdp ( rf +nr +nd +np -- c-addr nr )
https://www.complang.tuwien.ac.at/forth/gforth/Docs-html-history/0.6.2/Formatted-numeric-output.html
Convert rf into a string at c-addr nr. The conversion rules and the
meanings of nr +nd np are the same as for f.rdp.
And from f.rdp:
f.rdp ( rf +nr +nd +np – )
https://www.complang.tuwien.ac.at/forth/gforth/Docs-html/Simple-numeric-output.html
Print float rf formatted. The total width of the output is nr. For
fixed-point notation, the number of digits after the decimal point is
+nd and the minimum number of significant digits is np. Set-precision has no effect on f.rdp. Fixed-point notation is used if the number of
siginicant digits would be at least np and if the number of digits
before the decimal point would fit. If fixed-point notation is not
used, exponential notation is used, and if that does not fit,
asterisks are printed. We recommend using nr>=7 to avoid the risk of
numbers not fitting at all. We recommend nr>=np+5 to avoid cases where
f.rdp switches to exponential notation because fixed-point notation
would have too few significant digits, yet exponential notation offers
fewer significant digits. We recommend nr>=nd+2, if you want to have
fixed-point notation for some numbers. We recommend np>nr, if you want
to have exponential notation for all numbers.
In humanly readable terms, these functions require a number on the float-stack and three numbers on the data stack.
The first number-parameter tells it how long the string should be, the second one how many decimals you would like and the third tells it the minimum number of decimals (which roughly translates to precision). A lot of implicit math is performed to determine the final String format that is produced, so some tinkering is almost required to make it behave the way you want.
Testing it out (we don't want to rebuild f., but to produce a format that will be accepted as floating-point number by forth to EVALUATE it again, so the 1.2345E0 notation is on purpose):
PI 18 17 17 f>str-rdp type \ 3.14159265358979E0 ok
PI 18 17 17 f.rdp \ 3.14159265358979E0 ok
PI f. \ 3.14159265358979 ok
I couldn't find the exact word for this, so I looked into Gforth sources.
Apparently, you could go with represent word that prints the most significant numbers into supplied buffer, but that's not exactly the final output. represent returns validity and sign flags, as well as the position of decimal point. That word then is used in all variants of floating point printing words (f., fp. fe.).
Probably the easiest way would be to substitute emit with your word (emit is a deferred word), saving data where you need it, use one of available floating pint printing words, and then restoring emit back to original value.
I'd like to hear the preferred solution too...
As the title of the question states I'm looking to take the following string in hexadecimal base:
b9c84ee012f4faa7a1e2115d5ca15893db816a2c4df45bb8ceda76aa90c1e096456663f2cc5e6748662470648dd663ebc80e151d4d940c98a0aa5401aca64663c13264b8123bcee4db98f53e8c5d0391a7078ae72e7520da1926aa31d18b2c68c8e88a65a5c221219ace37ae25feb54b7bd4a096b53b66edba053f4e42e64b63
And convert it to its decimal equivalent string:
130460875511427281888098224554274438589599458108116621315331564625526207150503189508290869993616666570545720782519885681004493227707439823316135825978491918446215631462717116534949960283082518139523879868865346440610923729433468564872249430429294675444577680464924109881111890440473667357213574597524163283811
I've looked to use this code, found at this link:
unsigned result = 0;
NSScanner *scanner = [NSScanner scannerWithString:hexString];
[scanner setScanLocation:1]; // bypass '#' character
[scanner scanHexInt:&result];
NSLog(#" %u",result);
However, I keep getting the following result: 4294967295. Any ideas on how I can solve this problem?
This sounds like a homework/quiz question, and SO isn't to get code written, so here are some hints in hope they help.
Your number is BIG, far larger than any standard integer size, so you are not going to be able to do this with long long or even NSDecimal.
Now you could go and source an "infinite" precision arithmetic package, but really what you need to do isn't that hard (but if you are going to be doing more than this then such using a package would make sense).
Now think back to your school days, how were you taught to do base conversion? The standard method is long division and reminders.
Example: start with BAD in hex and convert to decimal:
BAD ÷ A = 12A remainder 9
12A ÷ A = 1D remainder 8
1D ÷ A = 2 remainder 9
2 ÷ A = 0 remainder 2
now read the remainder back, last first, to give 2989 decimal.
Long division is a digit at a time process, starting with the most significant digit, and carrying the remainder as you move to the next digit. Sounds like a loop.
Your initial number is a string, the most significant digit is first. Sounds like a loop.
Processing characters one at a time from an NSString is, well, painful. So first convert your NSString to a standard C string. If you copy this into a C-array you can then overwrite it each time you "divide". You'll probably find the standard C functions strlen() and strcpy() helpful.
Of course you have characters in your string, not integer values. Include ctype.h in your code and use the digittoint() function to convert each character in your number to its numeric equivalent.
The standard library doesn't have the inverse of digittoint(), so to convert an integer back to its character equivalent you need to write your own code, think indexing into a suitable constant string...
Write a C function, something like int divide(char *hexstring) which does one long division of hexstring, writing the result into hexstring and returning the remainder. (If you wish to write more general code, useful for testing, write something like int divide(char *buf, int base, int divisor) - so you can convert hex to decimal and then back again to check you get the back to where you started.)
Now you can loop calling your divide and accumulating the remainders (as characters) into another string.
How big should your result string be? Well a number written in decimal typically has more digits than when written in hex (e.g. 2989 v. BAD above). If you're being general then hex uses the fewest digits and binary uses the most. A single hex digit equates to 4 binary digits, so a working buffer 4 times the input size will always be long enough. Don't forget to allow for the terminating NUL in C strings in your buffer.
And as hinted above, for testing make your code general, convert your hex string to a decimal one, then convert that back to a hex one and check the result is the same as the input.
If this sounds complicated don't despair, it only takes around 30 lines of well spaced code.
If you get stuck coding it ask a new question showing your code, explain what goes wrong, and somebody will undoubtedly help you out.
HTH
Your result is the maximum of unsinged int 32 bit, the type you are using. As far as I can see, in the NSScanner documentation long long is the biggest supported type.
I want to convert a double to a string and only display needed decimals.
So I cannot use
d := 123.4
s := Format('%.2f', [d]);
As it display as the result is 123.40 when I want 123.4.
Here is a table of samples and expected result
|Double|Result as string|
-------------------------
|5 |5 |
|5.1 |5.1 |
|5.12 |5.12 |
|5.123 |5.123 |
You can use the %g format string:
General: The argument must be a floating-point value. The value is converted to the shortest possible decimal string using fixed or
scientific format. The number of significant digits in the resulting
string is given by the precision specifier in the format string; a
default precision of 15 is assumed if no precision specifier is
present. Trailing zeros are removed from the resulting string, and a
decimal point appears only if necessary. The resulting string uses the
fixed-point format if the number of digits to the left of the decimal
point in the value is less than or equal to the specified precision,
and if the value is greater than or equal to 0.00001. Otherwise the
resulting string uses scientific format.
This is not as simple as you think. It all boils down to representability.
Let's consider a simple example of 0.1. That value is not exactly representable in double. This is because double is a binary representation rather than a decimal representation.
A double value is stored in the form s*2^e, where s and e are the significand and exponent respectively, both integers.
Back to 0.1. That value cannot be exactly represented as a binary floating point value. No combination of significand and exponent exist that represent it. Instead the closest representable value will be used:
0.10000 00000 00000 00555 11151 23125 78270 21181 58340 45410 15625
If this comes as a shock I suggest the following references:
Is floating point math broken?
http://download.oracle.com/docs/cd/E19957-01/806-3568/ncg_goldberg.html
http://floating-point-gui.de/
So, what to do? An obvious option is to switch to a decimal rather than binary representation. In Delphi that typically means using the Currency type. Depending on your application that might be a good choice, or it might be a terrible choice. If you wish to perform scientific or engineering calculations efficiently, for instance, then a decimal type is not appropriate.
Another option would be to look at how Python handles this. The repr function is meant, where possible, to yield a string with the property that eval(repr(x)) == x. In older versions of Python repr produced very long strings of the form 1.1000000000000001 when in fact 1.1 would suffice. Python adopted an algorithm that finds the shortest decimal expression that represents the floating point value. You could adopt the same approach. The snag is that the algorithm is very complex.
I need to parse floating-point literals in C code using OCaml.
OCaml's float type is 64 bit. I have the string of the literal, its numeric value rounded to 64 bits and its kind (float, double or long double).
The problem are literals with a numeric value bigger than 64 bit:
long double literals
float literals with 'f'-suffix for which double rounding errors would occur if they wouldn't have the suffix.
OCaml's arbitrary-precision module can parse rational numbers from strings like "123/123", but not "123.123", "123e123", "0x1.23p-1" like they might appear in C.
Background: I do value analysis of C programs using CIL.
Double literals of any size and float literals with a numeric value that fits into 64 bit are always correctly represented. By rounding from double- to single-precision I can also reproduce double rounding errors.
I wrote my answer in the form of a blog post
To summarize some of the points here: you could interface strtold() and strtof() from OCaml. For the former, you would have to consider how you are going to store the result it produces, since there only is a point if long double is larger than double on your host architecture. There remains the problem that these functions are buggy in one of the most widely used C library. Very slightly buggy, but buggy for exactly the examples that are going to be of interest if you are doing this to study double rounding.
Another way is to write your own function, starting from another post in the blog you refer to.
Finally, the phrase "Even getting single-precision floats right requires me to parse literals with values bigger than 64 bit" that you use in the comments is still a strange way to put it. The intermediate format(s) in which you can parse the representation of a single-precision float before you round it to single-precision have to be lossless, otherwise there will be double rounding. Double rounding may be more or less difficult to exhibit depending on the precision of the lossy intermediate format, but using 80 bits or 128 bits binary floating-point formats is not going to remove the problem, just make it more subtle. In the simple algorithm that I recommend, the intermediate format is a fraction of two multiprecision integers.
I don't see the question in this question :)
Assuming that you need an ocaml parser for "C float literals" - the answer is - write one yourself, it is not very hard and you will have strict control on the implementation details and what "C float literal" actually means.