fold

I added the following code to my .vimrc: " save and restore folds when a file is closed and re-opened autocmd BufWinLeave *.* mkview autocmd BufWinEnter *.* silent loadview HTML and CSS documents save and restore their folds but code folding is not being saved in my .vimrc Any suggestions? EDIT: The following code solves the problem: au BufWinLeave ?* mkview au BufWinEnter ?* silent loadview but if I write it, the MRU files disappear from my list (and I have to open MRU twice in order to see my list of recent files why?) The problem is that your original autocmd lines are set to match the

Looking over the C++17 paper on folds, (and on cppreference ), I'm confused as to why the choice was made to only work with operators? At first glance it seems like it would make it easier to expand (... + args) by just shoving a + token between the elements of args , but I'm unconvinced this is a great decision. Why can't a binary lambda expression work just as well and follow the same expansion as the latter above? It's jarring to me that a fold syntax would be added to a language without support for arbitrary callables, so does the syntax allow a way to use them that I'm just not seeing?

How can I write a function which takes a tuple of functions of type ai -> b -> ai and returns a function which takes a tuple of elements of type ai , one element of type b , and combines each of the elements into a new tuple of ai : That is the signature should be like f :: (a1 -> b -> a1, a2 -> b -> a2, ... , an -> b -> an) -> (a1, a2, ... , an) -> b -> (a1, a2, ... , an) Such that: f (min, max, (+), (*)) (1,2,3,4) 5 = (1, 5, 8, 20) The point of this is so I can write: foldlMult' t = foldl' (f t) And then do something like: foldlMult' (min, max, (+), (*)) (head x, head x, 0, 0) x to do

Consider the following list of Boolean values in Scala List(true, false, false, true) How would you using either foldRight or foldLeft emulate the function of performing a logical AND on all of the values within the list? val l = List(true, false, false, true) val andAll = l.foldLeft(true)(_ && _) Instead of using foldLeft/Right , you can also use forall(identity) for the logical AND, or exists(identity) for the logical OR. edit: The benefit of these functions is the early exit. If forall hits a false or exists a true , they will immediately return. Without initial value as in foldLeft , List

Our internal git-lab wiki works with Markdown. I made several summaries of articles and want to post them in our wiki, in such a way that if I click on the header, it should unfold and the text should become visible, basically like in this example Does Markdown have this expand/collapse/fold function? Short Answer: No, Markdown does not offer a feature like that directly, but with some work you might be able to build something that works. For a feature like that to work you would need some CSS and/or JavaScript to control the animations, etc. While you might be able to get such a feature to

Why can you reverse a list with the foldl? reverse' :: [a] -> [a] reverse' xs = foldl (\acc x-> x : acc) [] xs But this one gives me a compile error. reverse' :: [a] -> [a] reverse' xs = foldr (\acc x-> x : acc) [] xs Error Couldn't match expected type `a' with actual type `[a]' `a' is a rigid type variable bound by the type signature for reverse' :: [a] -> [a] at foldl.hs:33:13 Relevant bindings include x :: [a] (bound at foldl.hs:34:27) acc :: [a] (bound at foldl.hs:34:23) xs :: [a] (bound at foldl.hs:34:10) reverse' :: [a] -> [a] (bound at foldl.hs:34:1) In the first argument of `(:)',

While going through Functional Programming in Scala , I came across this question: Can you right foldLeft in terms of foldRight? How about the other way around? In solution provided by the authors they have provided an implementation as follows: def foldRightViaFoldLeft_1[A,B](l: List[A], z: B)(f: (A,B) => B): B = foldLeft(l, (b:B) => b)((g,a) => b => g(f(a,b)))(z) def foldLeftViaFoldRight[A,B](l: List[A], z: B)(f: (B,A) => B): B = foldRight(l, (b:B) => b)((a,g) => b => g(f(b,a)))(z) Can somebody help me trace through this solution and make me understand how this actually gets the foldl