function

问题 I define the piecewise function def Li(x): return piecewise(x, [x < 0, x >= 0], [lambda t: sin(t), lambda t: cos(t)]) And when I evaluate Li(1.0) The answer is correct Li(1.0)=array(0.5403023058681398) , But if I write Li(1) the answer is array(0) . I don't understand this behaviour. 回答1: It seems that piecewise() converts the return values to the same type as the input so, when an integer is input an integer conversion is performed on the result, which is then returned. Because sine and

问题 I have gone through the following threads: warning: implicit declaration of function Implicit Declaration of Function in C UNIX Possibly my issue is linked. But while they offer the solution that the function prototype should be declared before the function is used, I wanted to explore what happens when the function name is not matching. In my test, it still works fine. Main C file #include "node.h" int main(){ nd *head=NULL; nd *tail=NULL; create_node(&head, &tail, 10); create_node(&head,

问题 I have gone through the following threads: warning: implicit declaration of function Implicit Declaration of Function in C UNIX Possibly my issue is linked. But while they offer the solution that the function prototype should be declared before the function is used, I wanted to explore what happens when the function name is not matching. In my test, it still works fine. Main C file #include "node.h" int main(){ nd *head=NULL; nd *tail=NULL; create_node(&head, &tail, 10); create_node(&head,

问题 In R, I would like to do something like this: I have a function f1, that has an argument with a default value; k=3. f1 = function(x,k=3){ u=x^2+k u } I then later define a second function, f2 that calls f1. f2 = function(z,s){ s*f1(z) } What's the cleanest way to allow users of f2 to override the default value of k in the inner function f1? One trivial solution is to redefine f2 as: f2 = function(z,s,K){ s*f1(z,K) } However, I feel this might be cumbersome if I'm dealing with a large

问题 When do closures have parameters (or how do closures with parameters work)? I know that use() is used to import variables outside the anonymous function, but what about the parameter(s) of the closure itself? 回答1: An example of closures with parameters is currying: function greeter($greeting) { return function($whom) use ($greeting) { // greeting is the closed over variable return "$greeting $whom"; }; } $hello_greeter = greeter('hello'); echo $hello_greeter('world'); // will print 'hello

问题 When do closures have parameters (or how do closures with parameters work)? I know that use() is used to import variables outside the anonymous function, but what about the parameter(s) of the closure itself? 回答1: An example of closures with parameters is currying: function greeter($greeting) { return function($whom) use ($greeting) { // greeting is the closed over variable return "$greeting $whom"; }; } $hello_greeter = greeter('hello'); echo $hello_greeter('world'); // will print 'hello

问题 Can I do something like this? $arrayFn[1] = function fnName(){ echo "test"; } $arrayFn[1]; 回答1: Anonymous functions cannot have a name: $arrayFn[1] = function (){ echo "test"; }; $arrayFn[1](); // run it! 回答2: $arrayFn[1] = function (){ echo "test"; }; $arrayFn[1](); 来源： https://stackoverflow.com/questions/22443213/store-function-inside-an-array

问题 I have the same question as this one, but in the context of JavaScript. From Wikipedia: [a pure function's] return value is the same for the same arguments It's further claimed there that a pure function is not allowed to have a variation in return value with "mutable reference arguments". In JavaScript, every normal object is passed as a "mutable reference argument". Consider the following example: const f = (arr) => arr.length const x = [] console.log( f(x) ) // 0 x.push(1); console.log( f

问题 I have the same question as this one, but in the context of JavaScript. From Wikipedia: [a pure function's] return value is the same for the same arguments It's further claimed there that a pure function is not allowed to have a variation in return value with "mutable reference arguments". In JavaScript, every normal object is passed as a "mutable reference argument". Consider the following example: const f = (arr) => arr.length const x = [] console.log( f(x) ) // 0 x.push(1); console.log( f

问题 This question already has answers here : Are variables declared with let or const hoisted? (6 answers) Closed 1 year ago . I'm having a bit of trouble understanding why my code works. I'm expecting a reference error, but everything works fine. My code: const functionA = () => { let bResult = functionB(); console.log("Function A " + bResult); }; const functionB = () => { return "Function B"; }; functionA(); I get this output (no errors); λ node test.js Function A Function B As I understand it,