xslt,rounding,xslt-2.0,xpath-2.0,rounding-error

How about: <xsl:value-of select="format-number(Price * Quantity, '0.00')"/> Note also that I get the same - expected - results with all of these: <xsl:value-of select="round(Price * Quantity * 100) div 100" /> and: <xsl:value-of select="round(xs:decimal(Price) * xs:decimal(Quantity) * 100) div 100" /> and: <xsl:value-of select="round(xs:decimal(Price) * xs:decimal(Quantity) * 10 * 10)...

For what it's worth, this would be more straight forward to do with a std::vector and a for loop. std::vector<double> createArray(double lower, double upper, double step) { std::vector<double> values; for (double value = lower; value <= upper; value += step) { values.push_back(value); } return values; } Working example This will...

For the second result, the impact is quite small in absolute terms: 1.22464679915e-16 is a very small number and very close to zero, about 0.00000000000000012246 If you simply want to avoid that small error having a large impact in how your results are displayed, use a formatting string, e.g: for...

calculated-columns,decimalformat,rounding-error,calculated,calculated-measure

The problem you are having is because SQL is doing integer division, which will only return whole numbers. To get a decimal return value you must have at least one value as a decimal. Try this: (CAST([N] as decimal(12,6))/[D]) * 1000 Adjust decimal(12,6) based on the precision you are expecting....

java,floating-point,double,rounding,rounding-error

If you know what precision you want you can round it to say 6 decimal places. eg. public static double round6(double x) { return Math.round(x * 1e6) / 1e6; } round6(0.7f) == 0.7 The fact the x passed in was a float doesn't matter provided this precision is suitable....

javascript,trigonometry,rounding-error

Like you said, i/8 has an exact binary representation. Therefore (i/8) * ((2*Math.PI) / (2*Math.PI)) will not suffer from floating point rounding error, but (i/8*2*Math.PI) / (2*Math.PI) will. This is due to the fact that the intermediate result (i/8*2*Math.PI) has no exact floating point representation. Having said that, why do...

Divakar gave me the hint towards the proper answer. it was solved by using t(n) = floor((eps(t(n-1))+t(n-1))/dt)*dt+dt; But most likely an alternative solution would be the suggestion of Mendi Barel, to convert the time to miliseconds so that all numbers are integers. Thank you all for the help and this...

While your code is very messy and should be improved overall, the solution to your problem should be simple: average=average/(len(splitedline)-1) should be: average /= float(len(splitedline) - 1) average = int(round(average)) By default in Python 2.x / with two integers does flooring division. You must explicitly make one of the parameters...

c++,opencv,visual-c++,rounding-error

The rounding behavior of the SSE instructions is configurable via the floating point environment (specifically, the MXCSR register). There are several IEEE rounding modes. The default rounding mode is round-to-nearest, ties-to-even, so if the value is exactly in the middle of two representable values, the result is rounded to the...

matlab,rounding-error,symbolic-computation

There is the High Precision Float class. After adding it to the path you can simply write k = hpf(10); before evaluating your expression. The result will be 370371188237525.0106290979251578332118698122510380699168308638036 With the Symbolic Math Toolbox I would write syms k expr = 1/1034591578977116160000*prod(1:19)*(29576428208904825-17729494921579950*k - 20479697577410832*k^2 + 13867226524449248*k^3 - 836937224095392*k^4 - 869194297188672*k^5...

This is because float is based on floating point notation. In rude words it tries to represent your decimal number as a sum of fractions of power 2. It means it will try to sum 1/2^n1 + 1/2^n2 + 1/2^n3 .... 1/2^nm until gets closes or exact value that you...

c#,excel,rounding,rounding-error,number-rounding

First: The problem is hard. Because 4.53/2 = 2.265. This rounds to 2.27. However the tiniest rounding error in the calculation resulting in a smaller result (2.264999999....) will lead to a rounding to 2.26. This is what is happening here. To solve this problem you need to have a floating...

c++,date,datetime,boost,rounding-error

That kind of depends on your circumstances. The general problem is that 1 / (24. * 60. * 60.) is not exactly representable as a binary float (because 86400 is not a power of two). The DATE you get is very nearly exact, but there will be a rounding error....

javascript,math,rounding-error

Possibly the most common way to do this is simply multiply both numbers by the same multiplier to make them have no decimals, and then do the operation, then divide again. Here's a crude implementation: function getDigits(n){ return n.toString().substring(n.toString().indexOf('.')+1).length; } function xNums(n1,n2){ var highRes=(n1*Math.pow(10,getDigits(n1))*(n2*Math.pow(10,getDigits(n2)))); return highRes/Math.pow(10,getDigits(n1))/Math.pow(10,getDigits(n2)); } Then, run xNums(31435517643980,(1...

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...