For “more precise”: this recipe in the Python Cookbook has summation algorithms which keep the full precision (by keeping track of the subtotals). Code is in Python but even if you don’t know Python it’s clear enough to adapt to any other language. All the details are given in this paper.
If the implementation is IEEE 754 compliant there exist special cases for different bit-combinations, like documented here. For your case: When all bits (sign, exponent, mantissa) are zero the floating point value represents also zero, as defined by IEEE 754. More specifically, that value is a “positive zero”, also written as +0.
This question has a bureaucratic part and an algorithmic part. A floating point number is stored internally as (2e × m), where e is an exponent (itself in binary) and m is a mantissa. The bureaucratic part of the question is how to access this data, but R. seems more interested in the algorithmic part of the question, … Read more
Try to do it like this: AVG(Cast(e.employee_level as Float)) as avg_level Also i found this topic where you can find some another approach but i not used its and don’t know exactly whether works or not.
Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea: static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348; static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696; // Floating-point modulo // The result (the remainder) has same sign as the divisor. // Similar to matlab’s mod(); Not similar to fmod() – Mod(-3,4)= 1 fmod(-3,4)= -3 template<typename T> T Mod(T … Read more
Be extremely careful using any of the other suggestions. It all depends on context. I have spent a long time tracing bugs in a system that presumed a==b if |a-b|<epsilon. The underlying problems were: The implicit presumption in an algorithm that if a==b and b==c then a==c. Using the same epsilon for lines measured in … Read more
The explanation is in the comments in the code. Java has double values for both 0.0 and -0.0, as well as “not a number” (NaN). You can’t use simple == operator for these values. Take a peek into the doubleToLongBits() source and at the Javadoc for the Double.equals() method: Note that in most cases, for … Read more
Well, here’s a quick explanation: An int and float usually take up “one-word” in memory. Today, with the shift to 64bit systems this may mean that your word is 64 bits, or 8 bytes, allowing the representation of a huge span of numbers. Or, it could still be a 32bit system meaning each word in … Read more
But what if the equation used to calculate the number is the same? can I assume the outcome would be the same too? No, not necessarily. In particular, in some situations the JIT is permitted to use a more accurate intermediate representation – e.g. 80 bits when your original data is 64 bits – whereas … Read more
The precision is fixed, which is exactly 53 binary digits for double-precision (or 52 if we exclude the implicit leading 1). This comes out to about 15 decimal digits. The OP asked me to elaborate on why having exactly 53 binary digits means “about” 15 decimal digits. To understand this intuitively, let’s consider a less-precise … Read more