Disclaimer (as suggested by Toby Speight): Although IEEE 754 representations are quite common, an implementation is permitted to use any other representation that satisfies the requirements of the language. The doubles are represented in the form mantissa * 2^exponent, i.e. some of the bits are used for the non-integer part of the double number. bits … Read more
Just use: (easy as pie) double number = 651.5176515121351; number = Math.round(number * 100); number = number/100; The output will be 651.52
What is the return type of the round() method in the snippet? If this is the Math.round() method, it returns a Long when the input param is Double. So, you will have to cast the return value: int a = (int) Math.round(doubleVar);
Depends on what the native hardware does. If the hardware is (or is like) x86 with legacy x87 math, float and double are both extended (for free) to an internal 80-bit format, so both have the same performance (except for cache footprint / memory bandwidth) If the hardware implements both natively, like most modern ISAs … Read more
Some shortcut is to create a NumberFormatInfo class, set its NumberDecimalSeparator property to “.” and use the class as parameter to ToString() method whenever u need it. using System.Globalization; NumberFormatInfo nfi = new NumberFormatInfo(); nfi.NumberDecimalSeparator = “.”; value.ToString(nfi);
Oh, I love these… these are caused by inaccuracy in the double representation and floating-point arithmetic is full of these. It is often caused by recurring numbers in binary (i.e. base-2 floating-point representation). For example, in decimal 1/3 = 0.3333′ In binary 1/10 is a recurring number, which means it cannot be perfectly represented. Try … Read more
Java doubles are in IEEE-754 format, therefore they have a 52-bit fraction; between any two adjacent powers of two (inclusive of one and exclusive of the next one), there will therefore be 2 to the 52th power different doubles (i.e., 4503599627370496 of them). For example, that’s the number of distinct doubles between 0.5 included and … Read more
Use the static Double.isNaN(double) method, or your Double‘s .isNaN() method. // 1. static method if (Double.isNaN(doubleValue)) { … } // 2. object’s method if (doubleObject.isNaN()) { … } Simply doing: if (var == Double.NaN) { … } is not sufficient due to how the IEEE standard for NaN and floating point numbers is defined.
The value of a float does not change when converted to a double. There is a difference in the displayed numerals because more digits are required to distinguish a double value from its neighbors, which is required by the Java documentation. That is the documentation for toString, which is referred (through several links) from the … Read more
The classes from the Decimal TR are not implemented for all compilers. Some compilers, e.g., gcc, implement the C Decimal TR and provide the corresponding extensions in C++, too. In the past there was an open source implementation for the C++ Decimal TR available but I failed to locate it. If your compiler doesn’t support … Read more