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

Converting from float to double preserves the value. For this reason, in the first snippet, d contains exactly the approximation to float precision of 112391/100000. The rational 112391/100000 is stored in the float format as 9428040 / 223. If you carry out this division, the result is exactly 1.12390995025634765625: the...

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

From the standard: There are three ﬂoating 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...

c++,math,floating-point-precision

That's impossible. There's no way to represent the exact value of pi in a finite floating-point data type. In general, you can't expect any operation on floating-point values to be exactly equal to the corresponding mathematical operation on real values. The usual approach is to compare with a tolerance, in...

swift,modulo,floating-point-precision,modulus,integer-division

52 goes into 66 1 time evenly with 14 left over. When you divide 66 / 52, you get 1.2692, or 52 / 52 + 14 / 52. So, if you're only after the decimals here, they can be acquired like this (66 % 52) / 52

c,gcc,floating-point,floating-point-precision

I'm surprised this does not work because I explicitly define the associativity of my operations when it matters. For example in the following code I but parentheses where it matters. This is exactly what -fassociative-math does: it ignores the ordering defined by your program (which is just as defined...

javascript,floating-point,floating-point-precision,floating-point-conversion

As pointed out by Mark Dickinson in a comment on the question, the ECMA-262 ECMAScript Language Specification requires the use of IEEE 754 64-bit binary floating point to represent the Number Type. The relevant rounding rules are "Choose the member of this set that is closest in value to x....

c++,floating-point,memcpy,floating-point-precision,floating-point-conversion

Your assumption is wrong. Look at this: float f = -.022119; std::cout << std::setprecision(20) << f << std::endl; Prints: -0.022119000554084778...

mysql,floating-point-precision

When you run the query: SELECT * FROM some_table WHERE id = 123 You are relying on the user interface to format the floating point numbers. The interface you are using is using two characters rather than more. After all, there is no information on the "right" number to show....

php,floating-point-precision,bcmath

probably the "scale" is different on the two environments. Try to set the scale with the bcscale function before doing your operations, For example: bcscale(3); $SR = "0"; $SPR = "149"; $SR = bcadd($SR, $SPR); echo "$SR"; Or simply use the third parameter in bcadd to set the scale: $SR...

sql,floating-point,floating-accuracy,floating-point-precision,kognitio-wx2

But you are using the full precision of the 32-bit floating point number. Remember that these are binary floating point values, not decimal. When you view it in decimal, you get a bunch of trailing zeros, and it looks very nice and clean. But if you were to view those...

bash,awk,floating-point,floating-point-precision,gawk

I get the expected output from awk version 3.1.5. I get your output from awk version 3.1.7. You can force awk to convert a string to a number by adding zero to it. So try this awk script instead: printf '4\n3e-20\n4.5e-320\n3\n1e-10\n' | awk '$1+0 > 1e-15' ...

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

When you pass a floating-point constant to printf, it is passed as a double. So it is not the same as passing a float variable with "the same" value to printf. Change this constant value to 0.7f (or 1.7f), and you'll get the same results. Alternatively, change float a to...

bash,floating-point,floating-point-precision,floating-point-conversion

What about a=`echo "5+50*3/20 + (19*2)/7" | bc -l` a_rounded=`printf "%.3f" $a` echo "a = $a" echo "a_rounded = $a_rounded" which outputs a = 17.92857142857142857142 a_rounded = 17.929 ?...

excel,floating-point,floating-point-precision

If you find the closest doubles to 170! and 169!, they are double oneseventy = 5818033100654137.0 * 256; double onesixtynine = 8761273375102700.0; times the same power of two. The closest double to the quotient of these is exactly 170.0. Also, Excel may compute 170! by multiplying 169! by 170. William...

c,floating-accuracy,floating-point-precision,approximation

Truncation vs. rounding. Due to subtle rounding effect of FP arithmetic, the product nol * noc may be slightly less than an integer value. Conversion to int results in fractional truncation. Suggest rounding before conversion to int. #include <math.h> int result = (int) roundf(nol * noc); ...

You can use number_format in case you need comma as decimal separator, see example below: $price = 26.20; $increase = 0.00; $decrease = 0.00; $result = ($price + $increase - $decrease); echo number_format($result , 2, ',', ''); ...

matlab,floating-point,double,floating-point-precision

It's usually not a good idea to test the equality of floating-point numbers. The behavior of binary floating-point numbers can differ drastically from what you may expect from base-10 decimals. Consider the example: >> isequal(0.1, 0.3/3) ans = 0 Ultimately, you have 53 bits of precision. This means that integers...

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.

webgl,floating-point-precision

After many tries and examining how float bits works, i've realized that when my values of float are between -1 to 1 the precision is very high. around +/-1.e-45. So what i'm basically doing is offsetting my mapping to origin - my middle point will be (0,0). and so on...

javascript,math,floating-point,floating-point-precision

It doesn't work. What you see there is not exactly 0.02, but a number that is close enough (to 15 significant decimal digits) to look like it. It just happens that multiplying an operand by 1000, then dividing the result by 1000, results in rounding errors that yield an apparently...

xcode,swift,floating-point-precision

The Swift Floating-Point Number documentation states: Note Double has a precision of at least 15 decimal digits, whereas the precision of Float can be as little as 6 decimal digits. The appropriate floating-point type to use depends on the nature and range of values you need to work with in...

mysql,sql-server,data-type-conversion,floating-point-precision

I found an answer. Instead of specifying decimal values in mysql you can just change the code to cast(value as DECIMAL)

java,datetime,floating-point-precision,date-calculation,date-difference

To expand on Sotirios' comment: The integer literals 365, 24, 60, and 1000 all have type int. Therefore, multiplication will be performed using the int type. Since the mathematical result is 31536000000, and the largest possible int is 2147483648, the result overflows and the result will wrap around. The result...

c#,floating-point,floating-point-precision,numerical-stability

Because the multiplication to get from [0.0f .. 1.0f] to [0 .. UInt32.MaxValue] can itself be approximative, the order of operations that most obviously has the property you desire is multiply, then clamp, then round. The maximum value to clamp to is the float immediately below 232, that is, 4294967040.0f....

d,floating-accuracy,floating-point-precision

This is probably because 0.6 can't be represented purely in a floating point number. You write 0.6, but that's not exactly what you get - you get something like 0.599999999. When you divide that by 0.1, you get something like 5.99999999, which converts to an integer of 5 (by rounding...

ios,swift,floating-point-precision

You should calculate by an integer to avoid the floating point precision issue. Therefore, convert the float to an integer at first. Is what you want the following code? func gcd(var m: Int, var n: Int) -> Int { if m < n { (m, n) = (n, m) }...

c++,binary,decimal,floating-point-precision,significant-digits

what is the most significant decimal digits precision that can be converted to binary and back to decimal without loss of significance? The most significant decimal digits precision that can be converted to binary and back to decimal without loss of significance (for single-precision floating-point numbers or 24-bits) is...

php,forms,floating-point,floating-point-precision

Form inputs are always a string, so the if condition will always return the error message you specified. Make use of filter_var() instead. if(filter_var($_POST['gpa'], FILTER_VALIDATE_FLOAT) === false) { echo "GPA must be in #.## format."; } If you want to validate the format you need to use Regular Expressions. ([0-9]{1,})\.([0-9]{2,2})...

c++,floating-point,language-lawyer,floating-point-precision,minimum

If std::numeric_limits<F>::is_iec559 is true, then the guarantees of the IEEE 754 standard apply to floating point type F. Otherwise (and anyway), minimum permitted values of symbols such as DBL_DIG are specified by the C standard, which, undisputably for the library, “is incorporated into [the C++] International Standard by reference”, as...

matlab,floating-point,integer,floating-point-precision

let x be the rounded number to the closest integer. if abs(x - number) < threshold, then this number is an integer....

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

why by using an union object and there defining the same memory as int32 and float32 i get different solution? The only reason the float/int union even makes sense is by virtue of the fact that both float and int share a storage size of 32-bits. What you are missing...

python,list,formatting,floating-point-precision,floating-point-conversion

Use string formatting to get the desired number of decimal places. >>> nums = [1883.95, 1878.3299999999999, 1869.4300000000001, 1863.4000000000001] >>> ['{:.2f}'.format(x) for x in nums] ['1883.95', '1878.33', '1869.43', '1863.40'] The format string {:.2f} means "print a fixed-point number (f) with two places after the decimal point (.2)". str.format will automatically round...

python,numpy,random,floating-point-precision

Q: is it possible to specify a dtype for random numbers when I create them. A: No it isn't. randn accepts the shape only as randn(d0, d1, ..., dn) Simply try this: x = np.random.randn(10, 10).astype('f') Or define a new function like np.random.randn2 = lambda *args, **kwarg: np.random.randn(*args).astype(kwarg.get('dtype', np.float64))...

python,floating-point,decimal,floating-point-precision,floating

One way of doing it is to set prec much higher than you need, then use round. Python 3: >>> getcontext().prec=100 >>> print(round(Decimal(748327402479023).sqrt(), 50)) 27355573.51764029457632865944595074348085555311409538175920 >>> print(round(Decimal(7483).sqrt(), 50)) 86.50433515148243713481567854198645604944135142208905 For Python 2, do: >>> print Decimal(748327402479023).sqrt().quantize(Decimal('1E-50')) 27355573.51764029457632865944595074348085555311409538175920 The value for prec depends on how...

haskell,floating-point,formatting,floating-point-precision,scientific-notation

After some research, I manage to get what I want. The function to get the engineering format work in a few steps : 1. Dissociate the exponent from the mantissa It's necessary to get the exponent apart of the mantissa. The function decodeFloat (provided by base) decode a floating point...

c,floating-point,printf,floating-point-precision,format-string

Yes, you're in luck. You need to make use of the precision field in the format string. In that, you can provide a .* notation and supply the corresponding integer argument holding the value of the precision. You can use the following pattern to make this happen, example with printf()....

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

Simply use an int, Start at 100 decrement down to 0 and divide the value by 100.0 for (int i=100; i>=0; --i) { float f = i/100.0f; ... } ...

language-agnostic,floating-point,floating-point-precision,ieee-754

From the Wikipedia entry on IEEE-754: The number representations described above are called normalized, meaning that the implicit leading binary digit is a 1. To reduce the loss of precision when an underflow occurs, IEEE 754 includes the ability to represent fractions smaller than are possible in the normalized representation,...

sql,floating-point,double,floating-point-precision

Big thanks to @Paqogomez: The ToString() fixed the issue. Here is the working code: Using cmdc = New SqlCommand("Insert into " & TextBox2.Text & " Select CallType,ChargeCode,Destination,TariffUsed,(case when chargecode in ('" + ComboBox2.Text + "') then " + NumericUpDown1.Value.ToString() + " else Peak*2 end) as Peak,(case when chargecode in ('"...

c++,c,floating-point-precision

The general, full answer is that floating-point numbers are compared according to the IEEE 754 specification. To answer your question specifically, most of the time two floating-point numbers are compared bitwise, with a few exceptional cases: Positive and negative zero are considered equal NaN is considered unequal to everything, even...

c++,fstream,truncate,floating-point-precision,digits

The following works fine on my system (Win7, VS2012): #include <fstream> #include <iostream> int main (void) { std::ifstream file ("test.txt") ; long double d = 0 ; file >> d ; std::cout.precision (20) ; std::cout << d << "\n" ; return 0 ; } The text file: 2.7239385667867091 The output:...

php,math,floating-point-precision

The output you are expecting is actually the result of $v[$i] = $d_negatif[$i] / ($d_negatif[$i] + $d_positif[$i]); not $v[$i] = $d_positif[$i] / ($d_negatif[$i] + $d_positif[$i]); ...

std::setprecision by default sets the total number of significant digits that are shown (ie. including the ones before the decimal point). To set the number of digits shown after the decimal point, use std::fixed : std::cout << std::fixed << std::setprecision(2) << 24.0; will display : 24.00 ...

floating-point,intel,floating-point-precision,floating-point-conversion

Problem was mxcsr register's DAZ bit. One machine A had it set so it was treating denormals as 0 . Where as on machine 2 DAZ was reset and hence it was normalizing it and treating denormals as different values , which resulted in comparison result as not equal. But...

c++,math,c++11,floating-point-precision

The problem here is withstd::cout. I fixed it using std::setprecision(50), as explained in How do I print a double value with full precision using cout? That shows me values like this: 1.6180339887498948482072100296669248109537875279784 To make it flexible, I gave the user the option to enter the desired level of precision: void...

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...

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

Some decimal numbers can only be represented with a certain precision on a system. In the case of normal floating point numbers that precision depends on the format of the number (usually IEEE 754). Hardware differences and base constraints can introduce small rounding errors. There are also other factors which...

limit,floating-accuracy,floating-point-precision,maxima

(1) use horner to rearrange the expression so that it can be evaluated more accurately. (%i1) display2d : false; (%o1) false (%i2) horner (x^5-6*x^4+14*x^3-20*x^2+24*x-16, x); (%o2) x*(x*(x*((x-6)*x+14)-20)+24)-16 (%i3) subst (x=1.999993580023622, %); (%o3) -1.77635683940025E-15 (2) use bigfloat (variable precision) arithmetic. (%i4) subst (x=1.999993580023622b0, x^5-6*x^4+14*x^3-20*x^2+24*x-16); (%o4) -1.332267629550188b-15 (%i5) fpprec : 50 $...

java,c#,floating-point-precision

The cause of not getting 15 digits is the subtraction of two similar numbers. z / Math.sin(lat0) is about 6347068.978968251. n0 * (1 - eSq) is about 6346068.978912728. With 7 digits before the decimal point, a change in the 9th decimal place in the subtraction result corresponds to a change...

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

Can I get away with setting the rounding mode of the FP environment appropriately and performing a simple cast (e.g. (double) N)? Yes, as long as the compiler implements the IEEE 754 (most of them do at least roughly). Conversion from one floating-point format to the other is one...

python,floating-point-precision

TL;DR The last digit of str(float) or repr(float) can be "wrong" in that it seems that the decimal representation is not correctly rounded. >>> 0.100000000000000040123456 0.10000000000000003 But that value is still closer to the original than 0.1000000000000000 (with 1 digit less). In the case of math.pi, the decimal approximation of...

For anything requiring larger range than available with integers, and where limited accuracy of number representation isn't important enough to use longer floats. In terms of accuracy, nothing beats integer or fixed point, at the price of their limited range. Say if i wanted cosmological distances in a unit which...

c++,compiler-optimization,floating-point-precision,quadruple-precision

Compiler optimizations work like this: the compiler recognizes certain patterns in your code, and replaces them by equivalent, but faster versions. Without knowing exactly what your code looks like and what optimizations the compiler performs, we can't say what the compiler is missing. It's likely that several optimizations that the...

python,floating-point,decimal,rounding,floating-point-precision

You could use the quantize() method: >>> import decimal >>> decimal.getcontext().prec = 20 >>> a = decimal.Decimal(321.12345) >>> a Decimal('321.12344999999999117790139280259609222412109375') >>> TWO_PLACES = decimal.Decimal("0.01") >>> a.quantize(TWO_PLACES) Decimal('321.12') ...

c,floating-point-precision,double-precision

A float is never more accurate than a double since the former must be a subset of the latter, by the C standard: 6.2.5/6: "The set of values of the type float is a subset of the set of values of the type double; the set of values of the...

See http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html for an in-depth explanation. In short, floating point numbers are approximations to the real numbers, and they have a limit on digits they can hold. With float, this limit is quite small, with doubles, it's more, but still not perfect. Try printf("%20.12lf\n",(double)2000008/(double)3); and you'll see a better, but...

ios,swift,floating-point,floating-point-precision,floating-point-conversion

This is due to the way the floating-point format works. A Float is a 32-bit floating-point number, stored in the IEEE 754 format, which is basically scientific notation, with some bits allocated to the value, and some to the exponent (in base 2), as this diagram from the single-precision floating-point...

binary,floating-point,decimal,floating-point-precision,significant-digits

The precision is fixed, which is exactly 53 binary digits for double-precision (or 52 if we exclude the implicit leading 1). This comes out to about 15 decimal digits. The OP asked me to elaborate on why having exactly 53 binary digits means "about" 15 decimal digits. To understand this...

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

The reason is that unsigned long long will store exact integers whereas double stores a mantissa (with limited 52-bit precision) and an exponent. This allows double to store very large numbers (around 10308) but not exactly. You have about 15 (almost 16) valid decimal digits in a double, and the...

c#,mysql,linq,nhibernate,floating-point-precision

x.Amount is being converted to a low precision minimum type from "LINQ-to-SQL" conversion, because your collection is IQueryable. There are several workarounds, the easiest of which is to change the type of your collection to IList, or call ToList() on your collection, forcing the linq query to run as LINQ-to-Objects....

ieee-754,floating-point-precision

The standard double has 52 bit mantissa, so yes, it is capable to hold and exactly reproduce a 32 bit integer. Another problem is the requirement that they have to be beetween 0 and 1. The reciprocal is not the way to do that! Counterexample: 1/3 is not exactly representable...

c,floating-point-precision,c11,floating-point-conversion

All that you need to know to which limits you may go and still have integer precision should be available to you through the macros defined in <float.h>. There you have the exact description of the floating point types, FLT_RADIX for the radix, FLT_MANT_DIG for the number of the digits,...

c,if-statement,compiler-errors,floating-point,floating-point-precision

Because 0.5 has an exact representation in IEEE-754 binary formats (like binary32 and binary64). 0.5 is a negative power of two. 0.6 on the other hand is not a power of two and it cannot be represented exactly in float or double.

java,vector,floating-point,normalization,floating-point-precision

As indicated above, this is a common issue, one you're going to have to deal with if you're going to use floating point binary arithmetic. The problem mostly crops up when you want to compare two floating point binary numbers for equality. Since the operations applied to arrive at the...

c,printf,floating-point-precision

floating-suffix: one of f l F L (C99 §6.4.4.2 ¶1) An unsuffixed floating constant has type double. If suffixed by the letter f or F, it has type float. If suffixed by the letter l or L, it has type long double. (ibidem, ¶4)...

java,floating-accuracy,floating-point-precision

Simple answer is - yes. This example definitely returns false: public boolean alwaysFalse(){ float a=Float.MIN_VALUE; float b=Float.MAX_VALUE; float c = a / b; return a == c * b; } Updated More general answer is that there are two cases when false happens in your method: 1) when significand if...

python,numpy,floating-point-precision

As others have commented, what constitutes 'almost zero' really does depend on your particular application, and how large you expect the rounding errors to be. If you must use a hard threshold, a sensible value might be the machine epsilon, which is defined as the upper bound on the relative...

c,floating-point,floating-point-precision

This is happening because the float and double data types store numbers in base 2. Most base-10 numbers can’t be stored exactly. Rounding errors add up much more quickly when using floats. Outside of embedded applications with limited memory, it’s generally better, or at least easier, to use doubles for...

c#,trigonometry,floating-point-precision

Did you notice how much the two values are close of 0, and close to each other ? The floating point (im)precision and the implementation of each methods may probably perfectly explain that. Those methods are not perfect, for example they are relying on an approximation of Pi (of course,...

php,mysql,floating-point,floating-point-precision,decimal-point

Understand that float is not a precise decimal number but a value that's close to it. The way it's stored is very efficient at the cost of not being exact. Without knowing your exact needs, one possible solution is to store the numbers as floats and round them to one...

javascript,floating-point-precision

Floating point rounding errors are at fault here. 2620.825 * 100 is 262082.49999999997, so when you round it, the result is 262082. Here is a more robust approach that corrects for this: function round(rnum, rlength) { var shifted = rnum * Math.pow(10, rlength), rounded = Math.round(shifted), delta = Math.abs(shifted -...

python,math,numpy,floating-point-precision

This is just a questions how the numbers are displayed: >>> result[0] 1.1475018493845227e+32 and: >>> math.pow(A[0],2.5) 1.1475018493845227e+32 Both ways lead to same value: >>> result[0] == math.pow(A[0],2.5) True result has its own method __repr__() that shows numbers differently than the standard Python __repr__() for float....

c,floating-point,floating-accuracy,floating-point-precision

I imagine speed is of some concern, or else you could just try the floating point-based estimate and adjust if it turned out to be too small. In that case, one can sacrifice tightness of the estimate for speed. In the following, let dst_base be 2^w, src_base be b, and...

floating-point,computer-science,computer-architecture,floating-point-precision

According to the paper referenced in the question, it is possible to calculate the dot product of a pair of length N vectors with a single rounding operation at the end, getting the closest representable result to the dot product. In practice, current computers round intermediate results, which often results...

ruby,floating-point-precision,fixnum

This is because Ruby by default uses Double-precision floating-point format. You can read about issues related to it here. However here's a short and crisp answer: Because internally, computers use a format (binary floating-point) that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all. When the code...

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...

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++,floating-accuracy,floating-point-precision

It is indeed a floating-point issue: a typical 64-bit double only gives 53 bits, or about 15 decimal digits, of precision. You might (or might not) get more precision from long double. Or, for integers up to about 1019, you could use uint64_t. Otherwise, there's no standard type with more...

algorithm,floating-point-precision,moving-average

You could recalculate a fresh sum over every SMA_LENGTH values to stop the errors accumulating: Queue values; double sum = 0.0; double freshsum = 0.0; int count = 0; double CalcSMA(double next) { values.push(next); sum -= values.pop(); sum += next; freshsum += next; if(++count == SMA_LENGTH) { sum = freshsum;...

python,arrays,numpy,floating-point,floating-point-precision

The type of your diff-array is the type of H1 and H2. Since you are only adding many 1s you can convert diff to bool: print diff.astype(bool).sum() or much more simple print (H1 == H2).sum() But since floating point values are not exact, one might test for very small differences:...