lazy-evaluation

I made really time consuming algorithm which produces a short string as the result. When I try to print it (via putStrLn) it appears on the screen character by character. I did understand why that happened, and I tried to force evaluation of the string before actual printing. myPrint !str = putStrLn str But this help very little. When I ran the program in debug I noticed that the !str forced evaluation only for the first character. Does anyone know why is that, and how to deal with this? (!) translates into seq , which evaluates strictly to Weak Head Normal Form -- that is, it only evaluates

I've been doing some computationally intensive work in F#. Functions like Array.Parallel.map which use the .Net Task Parallel Library have sped up my code exponentially for a really quite minimal effort. However, due to memory concerns, I remade a section of my code so that it can be lazily evaluated inside a sequence expression (this means I have to store and pass less information). When it came time to evaluate I used: // processor and memory intensive task, results are not stored let calculations : seq<Calculation> = seq { ...yield one thing at a time... } // extract results from

In Haskell , I might implement if like this: if' True x y = x if' False x y = y spin 0 = () spin n = spin (n - 1) This behaves how I expect : haskell> if' True (spin 1000000) () -- takes a moment haskell> if' False (spin 1000000) () -- immediate In Racket , I could implement a flawed if like this: (define (if2 cond x y) (if cond x y)) (define (spin n) (if (= n 0) (void) (spin (- n 1)))) This behaves how I expect : racket> (if2 #t (spin 100000000) (void)) -- takes a moment racket> (if2 #f (spin 100000000) (void)) -- takes a moment In Idris , I might implement if like this: if' : Bool -> a -> a

I don't know whether the term "Lazy" Binary Search is valid, but I was going through some old materials and I just wanted to know if anyone can explain the algorithm of a Lazy Binary Search and compare it to a non-lazy Binary Search. Let's say, we have this array of numbers: 2, 11, 13, 21, 44, 50, 69, 88 How to look for the number 11 using a Lazy Binary Search? As far as I am aware "Lazy binary search" is just another name for " Binary search ". Justin was wrong. The common binarySearch algorithm first checks whether the target is equal to current middle entry before proceeding to the left or