Strictfp ensures that you get exactly the same results from your floating point calculations on every platform. If you don’t use strictfp, the JVM implementation is free to use extra precision where available. From the JLS: Within an FP-strict expression, all intermediate values must be elements of the float value set or the double value … Read more
The biggest/largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double. That is, DBL_MAX or approximately 1.8 × 10308 (if your double is an IEEE 754 64-bit double). It’s an integer. It’s represented exactly. What more do you want? Go on, ask me what … Read more
I think you’ve summarised the advantages quite well. You are however missing one point. The decimal type is only more accurate at representing base 10 numbers (e.g. those used in currency/financial calculations). In general, the double type is going to offer at least as great precision (someone correct me if I’m wrong) and definitely greater … Read more
A comment in the Python source code for float objects acknowledges that: Comparison is pretty much a nightmare This is especially true when comparing a float to an integer, because, unlike floats, integers in Python can be arbitrarily large and are always exact. Trying to cast the integer to a float might lose precision and … Read more
General (language agnostic) reason Since not all numbers can be represented exactly in IEEE floating point arithmetic (the standard that almost all computers use to represent decimal numbers and do math with them), you will not always get what you expected. This is especially true because some values which are simple, finite decimals (such as … Read more
It’s doing integer division. You can use to_f to force things into floating-point mode: 9.to_f / 5 #=> 1.8 9 / 5.to_f #=> 1.8 This also works if your values are variables instead of literals. Converting one value to a float is sufficient to coerce the whole expression to floating point arithmetic.
It is not a matter of how the number is stored, it is a matter of how you are displaying it. When converting it to a string you must round to the desired precision, which in your case is two decimal places. E.g.: NSString* formattedNumber = [NSString stringWithFormat:@”%.02f”, myFloat]; %.02f tells the formatter that you … Read more
If they’re meant to be separate values, try this: var values = “554,20”.split(“,”) var v1 = parseFloat(values[0]) var v2 = parseFloat(values[1]) If they’re meant to be a single value (like in French, where one-half is written 0,5) var value = parseFloat(“554,20”.replace(“,”, “.”));
Maybe this question is stupid, but why does simply changing the order of the elements affects the result? It will change the points at which the values are rounded, based on their magnitude. As an example of the kind of thing that we’re seeing, let’s pretend that instead of binary floating point, we were using … Read more