Use the “R” numeric format string: double d = 0.00034101243963859839; string s = d.ToString(“R”); //… double d2 = double.Parse(s); if(d == d2) { //– Success } The R stands for “round-trip”. From the linked document: This format is supported only for the Single and Double types. The round-trip specifier guarantees that a numeric value converted … Read more
Matlab’s symbolic math toolbox contains a function called double. This function converts the result of the solve function – which is a sym – to a double: double(a) ans = 5.98921078320145 0.834834535131742
Floating point numbers (in this case doubles) cannot represent decimal values exactly. I would recommend reading David Goldberg’s What every computer scientist should know about floating-point arithmetic This applies to all languages that deal with floating point numbers whether it be C#, Java, JavaScript, … This is essential reading for any developer working with floating … Read more
When you create a double, the value 0.3 cannot be represented exactly. You can create a BigDecimal from a string without the intermediate double, as in new BigDecimal(“0.3”) A floating point number is represented as a binary fraction and an exponent. Therefore there are some number that cannot be represented exactly. There is an analogous … Read more
You can use Alignment Component for this purpose. Like this: Console.WriteLine($”value: {d,-17} {s}”); The optional alignment component is a signed integer indicating the preferred formatted field width. If the value of alignment is less than the length of the formatted string, alignment is ignored and the length of the formatted string is used as the … Read more
doubles are not exact. It is because there are infinite possible real numbers and only finite number of bits to represent these numbers. have a look at: what every programmer should know about floating point arithmetic
%d stands for decimal and it expects an argument of type int (or some smaller signed integer type that then gets promoted). Floating-point types float and double both get passed the same way (promoted to double) and both of them use %f. In C99 you can also use %lf to signify the larger size of … Read more
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
When you are using a binary operator, both arguments should be of a same type and the result will be in their type too. When you want to divide (int)/(long) it turns into (long)/(long) and the result is (long). you shouldmake it (double)/(long) or (int)/(double) to get a double result. Since double is greater that … Read more
If you want to be really careful you can test whether it is within some epsilon of zero with something like double epsilon = 0.0000001; if ( f <= ( 0 – epsilon ) ) { .. } else if ( f >= ( 0 + epsilon ) ) { .. } else { /* … Read more