infix-operator

问题 I am recently learning about Prolog and I find the three types used for defining infix operators confusing. What are the differences between xfx, xfy and yfx when specifying the type of an operator? I have googled about the problem and haven't found anything useful. I tried typing the following codes in Prolog: :- op(500,yfx,is_alive). is_alive(A,B) :- display([A,B]). :- op(500,xfy,is_alive2). is_alive2(A,B) :- display([A,B]). :- op(500,xfx,is_alive3). is_alive3(A,B) :- display([A,B]). and

问题 I am recently learning about Prolog and I find the three types used for defining infix operators confusing. What are the differences between xfx, xfy and yfx when specifying the type of an operator? I have googled about the problem and haven't found anything useful. I tried typing the following codes in Prolog: :- op(500,yfx,is_alive). is_alive(A,B) :- display([A,B]). :- op(500,xfy,is_alive2). is_alive2(A,B) :- display([A,B]). :- op(500,xfx,is_alive3). is_alive3(A,B) :- display([A,B]). and

问题 It is easy to introduce new infix operators in C++ // User-defined infix operator framework template <typename LeftOperand, typename Operation> struct LeftHelper { const LeftOperand& leftOperand; const Operation& operation; LeftHelper(const LeftOperand& leftOperand, const Operation& operation) : leftOperand(leftOperand), operation(operation) {} }; template <typename LeftOperand, typename Operation > auto operator < (const LeftOperand& leftOperand, Operation& operation) { return LeftHelper

问题 Is it possible to define my own infix function/operator in CoffeeScript (or in pure JavaScript)? e.g. I want to call a foo b or a `foo` b instead of a.foo b or, when foo is global function, foo a, b Is there any way to do this? 回答1: ES6 enables a very Haskell/Lambda calculus way of doing things. Given a multiplication function: const multiply = a => b => (a * b) You can define a doubling function using partial application (you leave out one parameter): const double = multiply (2) And you can

问题 Is it possible to define my own infix function/operator in CoffeeScript (or in pure JavaScript)? e.g. I want to call a foo b or a `foo` b instead of a.foo b or, when foo is global function, foo a, b Is there any way to do this? 回答1: ES6 enables a very Haskell/Lambda calculus way of doing things. Given a multiplication function: const multiply = a => b => (a * b) You can define a doubling function using partial application (you leave out one parameter): const double = multiply (2) And you can

问题 Is it possible to define my own infix function/operator in CoffeeScript (or in pure JavaScript)? e.g. I want to call a foo b or a `foo` b instead of a.foo b or, when foo is global function, foo a, b Is there any way to do this? 回答1: ES6 enables a very Haskell/Lambda calculus way of doing things. Given a multiplication function: const multiply = a => b => (a * b) You can define a doubling function using partial application (you leave out one parameter): const double = multiply (2) And you can

问题 Is it possible to define my own infix function/operator in CoffeeScript (or in pure JavaScript)? e.g. I want to call a foo b or a `foo` b instead of a.foo b or, when foo is global function, foo a, b Is there any way to do this? 回答1: ES6 enables a very Haskell/Lambda calculus way of doing things. Given a multiplication function: const multiply = a => b => (a * b) You can define a doubling function using partial application (you leave out one parameter): const double = multiply (2) And you can

问题 Methods taking a single argument can be written as an infix operators in Scal. I.e. adding *(other:C) = foo(this, other) to class C, will allow us to write c1 * c2 instead of foo(c1,c2). But is there a way to define infix operators on existing classes that you cannot modify? E.g. if I wanted to write c1 + c2 instead of xor(c1,c2) , where c1,c2:Array[Byte] , I obviously cannot modify the Array-Class. I found this and tried implicit class Bytearray(a1:Array[Byte]) extends Anyval { def +(a2

问题 Why is f <$> g <$> x equivalent to (f . g) <$> x although <$> is not right-associative? (This kind of equivalence is valid in a popular idiom with plain $ , but currently $ is right-associative!) <*> has the same associativity and precedence as <$> , but behaves differently! Example: Prelude Control.Applicative> (show . show) <$> Just 3 Just "\"3\"" Prelude Control.Applicative> show <$> show <$> Just 3 Just "\"3\"" Prelude Control.Applicative> pure show <*> pure show <*> Just 3 <interactive>

问题 I'm trying to understand the implementation of List s in Scala. In particular I'm trying to get my head around how you can write match expressions using an infix operator, for example: a match { case Nil => "An empty list" case x :: Nil => "A list without a tail" case x :: xs => "A list with a tail" } How is the match expression allowed to be x :: xs rather than List(x, xs) ? 回答1: Jay Conrad's answer is almost right. The important thing is that somewhere there is an object named :: which