c++,casting,int,double,precision

Almost all widely available hardware uses IETF754 floating point numbers, although C++ does not require it. Assuming IETF754 floating point numbers and a direct mapping of ::std::sqrt to the IETF754 floating point square root operation, you are assured of the following: 16 and 4 can both be represented exactly in...

c++,matlab,math,precision,scientific-computing

Converting to double in C++ inside the computation solves the difference problem. We reach a very satisfactory match after converting to double.

matlab,floating-point,fortran,precision

Matlab is using double precision by default, you are using single precision floats! They are limited to 7-8 digits... If you use double precision as well, you will get the same precision: program test write(*,*) 'single precision:', tanh(1.0) write(*,*) 'double precision:', tanh(1.0d0) end program Output: single precision: 0.761594176 double precision:...

You can use sprintf: def time_conversion(minutes) hour = minutes / 60 min_side = minutes % 60 sprintf("%d:%02d", hour, min_side) end ...

java,android,math,precision,pow

omg... I speend a few days on this. I have tried many algorithms and none of them don't worked... And then I wrote in my function: pow(new BigDecimal(3), new BigDecimal("8.73")) // <-- this is String in 2 argument (good) NOT: pow(new BigDecimal(3), new BigDecimal(8.73)) // <-- this is double in...

java,double,precision,double-precision

I'm using a decimal floating point arithmetic with a precision of three decimal digits and (roughly) with the same features as the typical binary floating point arithmetic. Say you have 123.0 and 4.56. These numbers are represented by a mantissa (0<=m<1) and an exponent: 0.123*10^3 and 0.456*10^1, which I'll write...

perl,printf,precision,long-double

bignum will overload all operators in the current scope to use arbitrary precision integers and floating point operations. use bignum; my $f = 123456789.123456789; print "$f\n"; # 123456789.123456789 print $f + $f, "\n"; # 246913578.246913578 Behind the scenes, bignum turns all numeric constants into Math::BigInt and Math::BigNum objects as appropriate....

performance,precision,arbitrary-precision

It is a trade-off between performance and features/safety. I cannot think of any reason why I would prefer using overflowing integers other then performance. Also, I could easily emulate overflowing semantics with non-overflowing types if I ever needed to. Also, overflowing a signed int is a very rare occurrence in...

This should work for you: (Just used php predefined constants to compare it) $id = "21321312412435453453453454"; if($id > PHP_INT_MAX) echo "too big"; else echo "okay"; Output: too big EDIT: For negative you just can define the constant like this: define('PHP_INT_MIN', ~PHP_INT_MAX); You can read about PHP_INT_MAX even in the manual:...

opengl-es,precision,framebuffer,depth-buffer

I had 2 issues in my case: 1. Large distance between nearZ and farZ 2. I tried use gl_FragCoord.z which is return low precision. It's was resolved with rendering to framebuffer which is have just attached texture for depth component (without color buffer!) and then render this depth-texture to another...

c,math,casting,double,precision

There is indeed no difference between the two, but compilers are used to take some freedom when it comes down to floating point computations. For example compilers are free to use higher precision for intermediate results of computations but higher still means different so the results may vary. Some compilers...

With typical FP is binary64, double x = 0.3; results in x with the value more like 0.29999999999999998890... so code has an difference from the beginning. Scale x by 10 to stay with exact math - or use a decimal64 double int main(void) { double x = 3.0; double binary...

This problem can not really be avoided since any computer will always suffer from machine precision: since decimal numbers are stored internally as a binary number, you will always have a finite precision. This means there is a minimum step size between any 2 consecutive numbers. You can find this...

javascript,integer,dart,precision

Yes, this is because JS doesn't have 64 int. I'm not sure what you mean by the rest of your question. If you store bigger values as double, you loose precision.

c++,c,precision,floating-point-precision,double-precision

From the standard: There are three ï¬‚oating point types: float, double, and long double. The type double provides at least as much precision as float, and the type long double provides at least as much precision as double. The set of values of the type float is a subset of...

Well, if you must (although I don't understand why you would want the 'wrong' number) #include <iomanip> #include <iostream> #include <sstream> using namespace std; void concat_print(double val, int precision) { stringstream stream; stream << setprecision(precision + 1) << val; // for (auto i = 0; i < stream.str().size() - 1;...

postgresql,double,precision,numeric,to-char

The reason is that for the purpose of the equality comparison, the type with the higher resolution is cast to the type of the lower resolution. I.e.: in the example the numeric is cast to double precision. Demo: SELECT * , num = dp AS no_cast , num::float8 = dp...

Ok, this was a bit tricky. In the line int omega[100] = {}; inside omegaN(int n) you fix the size of the array omega to 100. However, you perform an out-of-bound access when omegaN's input parameter n >= 100. You end up overwriting some memory you shouldn't (undefined behaviour), so...

c++,floating-point,precision,floating-accuracy,numerical-methods

Will it be more stable (or faster, or at all different) in general to implement a simple function to perform log(n) iterated multiplies if the exponent is known to be an integer? The result of exponentiation by squaring for integer exponents is in general less accurate than pow, but...

matlab,matrix,precision,significant-digits

maybe something simple like this? m = [10 15.675; 13.5 34.987; 20 55.5]; file = fopen('file.txt', 'w'); for ii = 1:size(m, 1) fprintf(file, '%0.1f %0.2f\n', m(ii, 1), m(ii, 2)); end I've edited to add the '\n'...

double,cross-platform,precision,floating-accuracy,deterministic

The C# standard, ECMA-334, says in section 11.1.6 Floating point types "Floating-point operations can be performed with higher precision than the result type of the operation.". Given that sort of rule, you cannot be sure of getting exactly the same answer everywhere. "can" means some implementations may use higher precision...

python,list,floating-point,decimal,precision

This apparent change in precision is because the repr representation of a float may contain more digits than the str representation of that same float. objects printed directly use str, and objects within a collection such as a list use repr. Ex: >>> repr(1/3.0) '0.3333333333333333' >>> str(1/3.0) '0.333333333333' >>> print...

All you need is epsilon. This is a "small number" which is small enough so you're no longer interested in. You could use: double epsilon = 1E-50; and whenever one of your factors gets smaller than epislon you take action (for example treat it like 0.0)...

opengl-es,webgl,shader,precision

No default precision exists for fp types in fragment shaders in OpenGL ES 2.0. In vertex shaders, if you do not explicitly set the default precision for floating-point types, it defaults to highp. However, if the fragment shader were to default to highp as well, that would cause issues since...

c#,double,type-conversion,precision,ieee-754

Being able to store the maximum value of UInt32 without losing information doesn't necessarily mean you'll be able to store all values in UInt32 without losing information - after all, there are plenty of long values which can be stored in a double even though some smaller values can't. (262...

ruby,precision,equality,bigdecimal

I reported the bug to the Ruby core team. However, this is not a bug as you can see in the rejection response. BigDecimal, though offers arbitrary precision, cannot represent numbers like 1/3 precisely. Thus during some arithmetic you can encounter imprecisions. You can use Rational in Ruby for exact...

c,sse,precision,matrix-multiplication,rounding-error

I am summarizing the discussion in order to close this question as answered. So according to the article (What Every Computer Scientist Should Know About Floating-Point Arithmetic) in link, floating point always results in a rounding error which is a direct consequence of the approximate representation nature of the floating...

double,cross-platform,rounding,precision,deterministic

I've actually been thinking about the whole matter in the wrong way. Instead of trying to minimize errors with greater accuracy (and more overhead), I should just use a type that stores and operates deterministically. I used the answer here: Fixed point math in c#? which helped me create a...

c++,c,floating-point,precision,floating-point-precision

No, it's not standard - neither the type nor the header. That's why the type has a double underscore (reserved name). Apparently, quadmath.h provides a quadmath_snprintf method. In C++ you would have used <<, of course.

matlab,math,precision,floating-point-precision,largenumber

Ah, large numbers. The value 3.3896e+038 is higher than the maximum integer that can be represented by a double without loss of accuracy. That maximum number is 2^53 or >> flintmax('double') ans = 9.0072e+15 So you are losing accuracy and you cannot reverse the computation. Doing the computations with uint64...

not so familiar with pandas but if you convert to a numpy array it works, try np.asarray(valores.iloc[:,5], dtype=np.float).mean() ...

Since money needs an exact representation don't use data types that are only approximate like double which is a floating-point. You can use a fixed-point numeric data type for that like numeric(15,2) 15 is the precision (total length of value including decimal places) 2 is the number of digits after...

browser,svg,coordinates,precision,specifications

The SVG specification says double precision. No UA implements this though because they are all constrained by platform capabilities such as direct3d which are generally single precision.

ios,objective-c,64bit,md5,precision

You simply need to cast [inputData length] to either int or uint32_t as you suspect. If you want to check for a number that is too large you can use the UINT32_MAXto check. Here is an example on an NSAssert you could use. NSAssert([inputData length] > UINT32_MAX, @"too big!"); ...

javascript,floating-point,precision

Remember that, with floating points, a number you see on-screen is not necessarily exactly the number the computer is modelling. For example, using node: > 1359.6 1359.6 > (1359.6).toFixed(20) '1359.59999999999990905053' Node shows you 1359.6, but when you ask for more precision, you can see that the real number is not...

You could work with logarithms, and handle the result later as you need. foreach (var word in fileDictionary) { dictionaryTotal.TryGetValue(word.Key, out percent); temp = percent; A += Math.Log10(temp); B += Math.Log10(1 - temp); } You could then operate the resulting logarithms, instead of the resulting numbers....

math,svg,precision,arbitrary-precision

Of course it is. Floating point math deals with relative, not absolute, precision. If you created a regular polygon at the origin, with radius 1e-7, then zoomed it to 1e7X size, you would expect to see a regular polygon with the same size and precision as an unzoomed circle with...

You are advised to use Inf,-Inf. Also it seems, interestingly, that integrate() is converting -Inf to Inf when used as upper limit. : > integrate(p, -Inf, -1*.Machine$double.xmax) 0 with absolute error < 0 > integrate(p, -Inf, -2*.Machine$double.xmax) 1.772454 with absolute error < 4.3e-06 This didn't result in what we would...

You get an ArithmeticException as described in BigDecimal.divide(BigDecimal) - "if the exact quotient does not have a terminating decimal expansion" because to properly represent the result of 1/3 would take infinite memory. By using BigDecimal.divide(BigDecimal,int,RoundingMode) you explicitly set what precision you'll tolerate, meaning any division can be approximated in-memory. For...

c++,c,performance,floating-point,precision

From GCC's man page: -ffast-math Sets -fno-math-errno, -funsafe-math-optimizations, -ffinite-math-only, -fno-rounding-math, -fno-signaling-nans and -fcx-limited-range. This option causes the preprocessor macro __FAST_MATH__ to be defined. This option is not turned on by any -O option besides -Ofast since it can result in incorrect output for programs that depend on an exact implementation...

php,rounding,precision,floating-point-precision

There is no need to replace round with number_format. You could use round with the flag to round down. Choose what is best for the application. Are you rounding a float? Use round Are you wanting to format a number and group by thousands? Use number_format <?php $number = '0.5600';...

c++,c,precision,floating-point-precision,gmp

From the manual page: Function: void mpf_init (mpf_t x) Initialize x to 0. Normally, a variable should be initialized once only or at least be cleared, using mpf_clear, between initializations. The precision of x is undefined unless a default precision has already been established by a call to mpf_set_default_prec. There's...

javascript,php,decimal,precision,rounding

Maybe this? <?php $num = '49.82'; $new_num = $num; $hundredth = substr($num, -1, 1); switch($hundredth) { case '0': break; case '1': $new_num = ($new_num - 0.01); break; case '2': $new_num = ($new_num - 0.02); break; case '3': $new_num = ($new_num - 0.03); break; case '4': $new_num = ($new_num - 0.04);...

System.out.println( String.format( "%.2f", totalCost ) );

How about using substring instead? It's not really a number anyway. int pos = x.indexOf('.'); String y = x.substring(pos - 12, pos + 3); If you are really crazy enough to try processing this as numbers (collissions and differences in rounding are going to break your neck sooner or later!),...

java,precision,biginteger,bigdecimal,largenumber

Using BigDecimal and doing calculation in double is the same as not using BigDecimal at all. Use BigDecimal to do calculations: BigDecimal bigNumber = new BigDecimal(1000).pow(10); System.out.println(bigNumber); ...

c#,entity-framework,sql-server-2008,precision,savechanges

I tried everything. I tried increasing the Precision and Range.. But the issue still persisted. Funny thing is It automatically got solved once I restarted my Visual Studio. Strange. But some times some things just need a restart to fix themselves....

The techniques mentioned below use eps and fix. These might be less "uglier" than rounding for your case. Let's look at these step by step with some codes and comments. %%// Original indices org_indices = [15 30 45 60] %%// Scaled indices scaled_indices = org_indices/3165 %%// Retrieved indices that are...

perl,split,floating-point,precision,dot

IF you're going to treat it as a string, do so all the way through. You wouldn't assign a string without quoting, right? my $a = "123456789.123456789"; ...

fortran,precision,gfortran,quad

In general, things like "dabs" were from the days before type-generic intrinsics. In particular, there is no corresponding "qabs/absq" or whatever you might want to call it, but rather only the type-generic "abs" which is then resolved to the correct library symbol at compile time. Secondly, your choice of "abs"...

c++,floating-point,precision,floating-accuracy,numerical-methods

You are correct. The precision from 1.0 to 2.0 will be uniform across the surface, like you are using fixed point. The precision from -0.5 to 0.5 will be highest around the center point, and lower near the edges (but still quite good). The precision from 0.0 to 1.0 will...

Use a std::string instead of number literal. Any value of the format aaa.bbb is treated as a double (limited precision) and with f suffix as float (limited precision again). In your case it is probably treated as double. const std::string number = "3.141592653589793238462643383279502884197169399375105820974944592307816406286"; mpz_class pi2(number); ...

javascript,math,precision,arbitrary-precision

There are three short answers: By using an external library that supports math to that precision (then effectively using it) By doing calculations as integers (for example, multiply by 1,000,000,000, do the calculations, then divide by the same amount) By building the zoom function in to the code and re-drawing...

python,fortran,precision,double-precision,f2py

You should set output in the Fortran code using double precision, output = 1.3d0 In Fortran 1.3 is a single precision constant....

c++,unix,precision,double-precision,scientific-notation

The number you are getting is related to the machine epsilon for the double data type. A double is 64 bits long, with 1 bit for the sign, 11 bits for the exponent, and 52 bits for the mantissa fraction. A double's value is given by 1.mmmmm... * (2^exp) With...

Use DecimalFormat as explained here. DecimalFormat df = new DecimalFormat("#0.######"); df.format(0.912385); ...

c++,solaris,precision,double-precision,scientific-notation

You could declare norm as a long double for some more precision. long double wiki Although there are some compiler specific issues to be aware of. Some compilers make long double synonymous with double. Another way to go about solving this precision problem is to work with numbers in the...

c,precision,double-precision,fmod

I'd solve this by simply not using floating point variables in that way: for (int i = 0; i <= 500; i += 5) { double d = i / 100.0; // in case you need to use it. if ((i % 25) == 0) print 'X'; } They're usually...

c,precision,arbitrary-precision,exponentiation

You can precompute and tabulate all 256-bit values of the function for x in [65280, 65535] (i.e. 255 x 256 + i); you will lookup the table by the 8 least significant bits of the argument. That will take 8KB of storage. For lower values of the argument, shift the...

You are adding 5 to 2128 using double. For the addition of 5 to have any effect it would need to use, at least, 127 bits to represent the significant. It uses 52 if I recall correctly. You'd need to use a much bigger representation. Since you only do integer...

delphi,floating-point,cross-platform,precision,deterministic

I think there is no simple answer. Similar task was discussed here. In general, there are two standards for presentation of floating point numbers: IEEE 754-1985 and EEE 754-2008. All modern (and quite old actually) CPUs follow the standards and it guarantees some things: Binary presentation of same standard floating...

There's more precision in that number than you're displaying. You just have to ask for it: cout << std::setprecision(40) << pi << endl; That gives me 3.141592653543740176758092275122180581093 when run on Codepad. As double should have way more than enough precision for basic calculations. If you need to compute millions of...

Are you confused about the unboxing conversion? From type Integer to type int From type Double to type double If r is a reference of type Integer, then unboxing conversion converts r into r.intValue() If r is a reference of type Double, then unboxing conversion converts r into r.doubleValue() And...

machine-learning,precision,recall,precision-recall

First of all, what you have written is not the F1-score. That is Precision! To compute the F1-score, set precision=TP/(TP+FP) and recall=TP/(TP+FN). Their harmonic mean is the F1-score. So, F1=2*(P*R)/(P+R). See this for further details. You can compute these values for each class and see how well you are doing...

floating-point,precision,numerical-methods

Exponentiation magnifies relative error and, by extension, ulp error. Consider this illustrative example: float x = 0x1.fffffep6; printf ("x=%a %15.8e exp2(x)=%a %15.8e\n", x, x, exp2f (x), exp2f(x)); x = nextafterf (x, 0.0f); printf ("x=%a %15.8e exp2(x)=%a %15.8e\n", x, x, exp2f (x), exp2f(x)); This will print something like x=0x1.fffffep+6 1.27999992e+02 exp2(x)=0x1.ffff4ep+127...

At a guess: it's the number of significant digits: there is also a digit in front of the decimal point for the second and third example (the 0 in the first example is not significant). Note that the fourth bullet in the documentation says: The decimal module incorporates a notion...

sql-server,entity-framework,precision,data-type-conversion,sqldatatypes

(@p__linq__0 decimal(5,0)) is the Precision of the parameter value your provided. For example, decimal number = 12345m; var result = context.MyTables.Where(x => x.Number == number).ToList(); Then the query will be like this - exec sp_executesql N'SELECT [Extent1].[Id] AS [Id], [Extent1].[Number] AS [Number] FROM [dbo].[MyTable] AS [Extent1] WHERE [Extent1].[Number] = @p__linq__0',N'@p__linq__0...

c,math,floating-point,precision,ieee-754

-1.#IND is Microsoft's way of outputting an indeterminate value, specifically NaN. One of the ways this can happen is with 0 / 0 but I would check all operations to see where the issue lies: double secant_method(double(*f)(double), double a, double b){ double c; printf("DBG =====\n"); for (int i = 0;...

objective-c,floating-point,double,precision,rounding

Use NSDecimalNumber for greater precision. See NSRoundingMode for the different rounding modes (I think you want NSRoundPlain) NSDecimalNumberHandler *rounder = [NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:NSRoundPlain scale:precision raiseOnExactness:NO raiseOnOverflow:NO raiseOnUnderflow:NO raiseOnDivideByZero:NO]; NSDecimalNumber *number; NSDecimaNumber *rounded = [number decimalNumberByRoundingAccordingToBehavior:rounder]; See the docs for more info on the BOOL parameters in the handler. It might seem...

fortran,precision,hdf5,double-precision

The type double precision and the corresponding kind kind(1.d0) are perfectly well defined by the standard. But they are also not exactly fixed. Indeed there were many computers in history and use different kind of native formats for their floating point numbers and the standard must allow this! So, a...

The cleanest way, for recent libraries (the ones that ship with GHC 7.8 or later), is to use finiteBitSize from Data.Bits. This is exactly the function you requested. With earlier versions, you can use bitSize, also from Data.Bits, a non-total version of the same thing. Specifically, bitSize will throw an...