division

I'm making a function in Haskell that halves only the evens in a list and I am experiencing a problem. When I run the complier it complains that you can't perform division of an int and that I need a fractional int type declaration. I have tried changing the type declaration to float, but that just generated another error. I have included the function's code below and was hoping for any form of help. halfEvens :: [Int] -> [Int] halfEvens [] = [] halfEvens (x:xs) | odd x = halfEvens xs | otherwise = x/2:halfEvens xs Thank you for reading. Joey Adams Use div , which performs integer division:

Possible Duplicates: Is shifting bits faster than multiplying and dividing in Java? .NET? Quick Java Optimization Question Many years ago in college, I learned that bit-shifting right by one accomplishes the same thing as dividing by two, but is generally significantly faster. I'm not sure how Java has come along in that regards since the 9-10 years ago I learned about that. Does the Java compiler automatically converts a divide-by-two into a bit-shift operation, or should I manually perform the bit-shift operation in the code myself? Unless you're working in a shop and a codebase where bit

This question already has an answer here: Why do the division (/) operators behave differently in VB.NET and C#? 5 answers Anyone care to explain why these two pieces of code exhibit different results? VB.NET v4.0 Dim p As Integer = 16 Dim i As Integer = 10 Dim y As Integer = p / i //Result: 2 C# v4.0 int p = 16; int i = 10; int y = p / i; //Result: 1 Christian When you look at the IL-code that those two snippets produce, you'll realize that VB.NET first converts the integer values to doubles, applies the division and then rounds the result before it's converted back to int32 and stored in y.

I want to implement unsigned a integer division by an arbitrary power of two, rounding up , efficiently. So what I want, mathematically, is ceiling(p/q) 0 . In C, the strawman implementation, which doesn't take advantage of the restricted domain of q could something like the following function 1 : /** q must be a power of 2, although this version works for any q */ uint64_t divide(uint64_t p, uint64_t q) { uint64_t res = p / q; return p % q == 0 ? res : res + 1; } ... of course, I don't actually want to use division or mod at the machine level, since that takes many cycles even on modern

Trying to figure out how to write a recursive predicate divide_by(X, D, I, R) that takes as input a positive integer X and a divisor D, and returns the answer as the whole number part I and the remainder part R, however, I can't seem to get my head around Prolog. How would I go about doing this? false There are predefined evaluable functors for this. (div)/2 and (mod)/2 always rounding down. Recommended by LIA-1, Knuth etc. (//)/2 and (rem)/2 rounding toward zero (actually, it's implementation defined, but all current implementations do it like that). You can ask this via current_prolog_flag

I have the following variables: int first = 0; int end = 0; Declare in the public class. Within a method: double diff = end / first; double finaldiff = 1 - diff; The end variable on System.out.println is 527 , the first is 480 . Why is the answer for diff coming out as 1 ? It should be 1.097916667 , I thought using a double would enable me to calculate into decimals? Dividing two int s will get you an int , which is then implicitly converted to double . Cast one to a double before the divison: double diff = (double)end / first; 来源： https://stackoverflow.com/questions/10376322/java-division