You could experiment with delta values in the order of 10-15 but you will notice that some calculations give a larger rounding error. Furthermore, the more operations you make the larger will be the accumulated rounding error. One particularly bad case is if you subtract two almost equal numbers, for example 1.0000000001 – 1.0 and … Read more
From The Floating-Point Guide: This is a bad way to do it because a fixed epsilon chosen because it “looks small” could actually be way too large when the numbers being compared are very small as well. The comparison would return “true” for numbers that are quite different. And when the numbers are very large, … Read more
What you actually want is new BigDecimal(“0.1”) .add(new BigDecimal(“0.1”)) .add(new BigDecimal(“0.1”)); The new BigDecimal(double) constructor gets all the imprecision of the double, so by the time you’ve said 0.1, you’ve already introduced the rounding error. Using the String constructor avoids the rounding error associated with going via the double.
All numbers in JavaScript are 64-bit floating point numbers. Ref: http://www.hunlock.com/blogs/The_Complete_Javascript_Number_Reference http://www.crockford.com/javascript/survey.html
The problem In binary 2.64 is 10.10100011110101110000101000111101 recurring, in other words not exactly representable in binary, hence the small error. Java is being kind to you with d3 but as soon as actual calculations are involved it has to fall back on the real representation. Binary Calculator Further more: 2.64= 10.10100011110101110000101000111101 4.64=100.1010001111010111000010100011110 Now, even though … Read more
The specific reason in your case is that the real number 0.94 cannot be represented exactly in a double precision floating point. When you type 0.94, the actual number stored is 0.939999999999999946709294817992486059665679931640625.
See the docs for num.toStringAsFixed(). String toStringAsFixed(int fractionDigits) Returns a decimal-point string-representation of this. Converts this to a double before computing the string representation. If the absolute value of this is greater or equal to 10^21 then this methods returns an exponential representation computed by this.toStringAsExponential(). Examples: 1000000000000000000000.toStringAsExponential(3); // 1.000e+21 Otherwise the result is the … Read more
If you want to know the true answer, you should read What Every Computer Scientist Should Know About Floating-Point Arithmetic. In short, although double allows for higher precision in its representation, for certain calculations it would produce larger errors. The “right” choice is: use as much precision as you need but not more and choose … Read more
There is no general solution for comparing floating-point numbers that contain errors from previous operations. The code that must be used is application-specific. So, to get a proper answer, you must describe your situation more specifically. For example, if you are sorting numbers in a list or other data structure, you should not use any … Read more
This happens because in your statement if(f == 0.7) the 0.7 is treated as a double. Try 0.7f to ensure the value is treated as a float: if(f == 0.7f) But as Michael suggested in the comments below you should never test for exact equality of floating-point values.