currying

问题 I am trying to convert a regular function into curry function but getting Execution was interrupted Below is the code where I am currying a function and doing an unsafeBitCast to call a function with one parameter and call it later with the second parameter. func curry<T>(f: (T, T) -> T) -> T -> T -> T { return { a in typealias Function = (T) -> (T -> T) let fn = unsafeBitCast(f, Function.self) return curry(fn(a)) } } func curry<T>(f: T -> T) -> T -> T { return { f($0) } // Throws Runtime

问题 I'm trying to grasp the concept behind curry functions. Below is the code: class MyHelloWorldClass { func helloWithName(name: String) -> String { return "hello, \(name)" } } I can create a variable that points to the class’s helloWithName function: let helloWithNameFunc = MyHelloWorldClass.helloWithName // MyHelloWorldClass -> (String) -> String My new helloWithNameFunc is of type MyHelloWorldClass -> (String) -> String , a function that takes in an instance of my class and returns another

问题 I was running the below Scala code in Worksheet: package src.com.sudipta.week2.coursera import scala.math.abs import scala.annotation.tailrec object FixedPoint { println("Welcome to the Scala worksheet") //> Welcome to the Scala worksheet val tolerance = 0.0001 //> tolerance : Double = 1.0E-4 def isCloseEnough(x: Double, y: Double): Boolean = { abs((x - y) / x) / x < tolerance } //> isCloseEnough: (x: Double, y: Double)Boolean def fixedPoint(f: Double => Double)(firstGuess: Double): Double =

问题 I recently started learning Scala and came across currying. From an answer in this post, this code snippet def sum(a: Int)(b: Int) = a + b expands out to this def sum(a: Int): Int => Int = b => a + b Then I saw a snippet from scala-lang, which shows it's possible to write something like this to emulate a while loop def whileLoop (cond : => Boolean) (body : => Unit) : Unit = { if (cond) { body whileLoop (cond) (body) } } Out of curiosity, I tried to expand this out, and got this def whileLoop2

问题 I was thinking about how to go about currying a method with varargs, and I realized that I don't even have an intuition for how one would go about doing it. Ideally, it would be something that would let you start using it whenever you liked, and then end it with an iterable. def concat(strs: String*) = strs.mkString val curriedConcat = concat.curry curriedConcat("OK")("hello", "world")("welcome")(Seq(): _*) Is there support for this in scala? I couldn't figure out how to do anything more than

问题 Let's suppose I have a def that takes multiple type parameters: def foo[A, B, C](b: B, c: C)(implicit ev: Writer[A]) However, the intended usage is that type parameters B and C should be inferred (based on the passed-in arguments). And the caller should only need to really specify A explicitly (e.g. to have an appropriate implicit chosen by the compiler). Unfortunately, Scala only allows all or none of the type parameters to be specified by the caller. In a sense, I want the type parameters

问题 I'm reading "Eloquent JavaScript". Chapter 3 introduces "Closure" concept and gives you a couple of examples. One of these is next one: function multiplier(factor) { return function(number) { return number * factor; }; } var twice = multiplier(2); console.log(twice(5)); // → 10 I think I understood the concept. If first I execute console.log(twice) , since variable number is undefined, what I get is [Function] . What I don't understand is how twice(5) works. Why local variable number is

问题 Recently I read about function composition in a Javascript book, and then on a website I saw someone reference it as currying. Are they the same concept? 回答1: @Omarjmh's answer is good but the compose example is overwhelmingly complex for a learner, in my opinion Are they the same concept? No. First, currying is translating a function that takes multiple arguments into a sequence of functions, each accepting one argument. // not curried const add = (x,y) => x + y; add(2,3); // => 5 // curried

问题 What happens when I do the following? (define ((func x) y) (if (zero? y) ((func x) 1) 12)) I understand that I can do this: (define curried (func 5)) And now I can use curried. What I'm curious about is in the definition of the function. Does the line ((func x) 1) create a new lambda with x as the argument, and then invoke it on 1? Or is it smarter than that and it just re-uses the existing one. (For example, if I do (curried 0) , the ((func x) 1) line would be equivalent to (curried 1) -

问题 I'd like to compose functions in a certain way. Please consider these 2 functions in pseudocode (not F#) F1 = x + y F2 = F1 * 10 // note I did not specify arguments for F1, 'reverse curry' for lack of a better word What I would like for F# to do is figure out that since let F1 x y = x + y //val F1 : int -> int -> int the code let F2 = F1 * 10 would give me the same signature as F1: val F2 : int -> int -> int , and calling F2 2 3 would result in 50: (2 + 3) * 10. That would be rather clever...