overloading

问题 I want to understand the code from this answer type Mult = Mult with static member inline ($) (Mult, v1: 'a list) = fun (v2: 'b list) -> v1 |> List.collect (fun x -> v2 |> List.map (fun y -> (x, y))) : list<'a * 'b> static member inline ($) (Mult, v1:'a ) = fun (v2:'a) -> v1 * v2 :'a let inline (*) v1 v2 = (Mult $ v1) v2 F# can resolve overloaded members. (Because it doesn't support currying of members). So, I supposed, it should work for methods as well But it doesn't: type Mult = Mult with

问题 I'm receiving a double definition error on the following two methods: def apply[T](state: T, onRender: T => Graphic, onMouseEvent: (MouseEvent, T) => T): GraphicPanel = apply(state, onRender, onMouseEvent = Some(onMouseEvent)) and def apply[T](state: T, onRender: T => Graphic, onKeyEvent: (KeyEvent, T) => T): GraphicPanel = apply(state, onRender, onKeyEvent = Some(onKeyEvent)) which are both method overloads for the more general apply method with the signature: def apply[T](state: T, onRender

问题 I have a class like this : class Foo { public: Foo() { for(int i = 0; i < 10; ++i) v.push_back(i); }; const vector<double>& V() const {return v;}; protected: vector<double>& V() {return v;}; private: vector<double> v; }; And then a piece of code like this : Foo foo; for(int i = 0; i < (int) foo.V().size(); ++i) cout << foo.V().at(i) << endl; However, the latter raises a compilation error saying the V() call is a protected method while i am just trying to read from it, not modify it. I have

问题 Consider the code below: #include <iostream> #include <vector> void f(std::vector<int> v) {std::cout << __PRETTY_FUNCTION__ << std::endl;} void f(int n) {std::cout << __PRETTY_FUNCTION__ << std::endl;} int main() { f({42}); // the int overload is being picked up } Live on Coliru I was a bit surprised to realize that in this case the int overload is being picked up, i.e. the output of the program is: void f(int) with the warning warning: braces around scalar initializer [-Wbraced-scalar-init]

问题 I am trying to add a class object with a number, but I'm confused on how to go about adding a class object with two numbers. For example, this is my hypothetical add class method: class A: def __add__(self, b): return something I know how to add this so far: object = A() print(object + 1) But, what if I want to add it like this? object = A() print(object + 1 + 2) Should I use *args for the add class method? 回答1: No, you can't use multiple arguments. Python executes each + operator separately,

问题 While writing a CRTP template that enables classes to provide overloads for operator+ based on template arguments, I found that an in-class friend operator doesn't seem to participate in overload resolution if none of it's arguments is of the type of the class it was defined in. Boiled down: enum class FooValueT{ zero, one, two }; class Foo{ FooValueT val_; public: Foo(FooValueT x) : val_(x){}; Foo& operator+=(Foo other){ val_ = (FooValueT)((int)val_ + (int)other.val_); return *this; } /

问题 Suppose I have two unrelated classes A and B . I also have a class Bla that uses boost::shared_ptr like this: class Bla { public: void foo(boost::shared_ptr<const A>); void foo(boost::shared_ptr<const B>); } Notice the const . That's the important part which the original version of this question lacked. This compiles, and the following code works: Bla bla; boost::shared_ptr<A> a; bla.foo(a); However, if I switch from using boost::shared_ptr to using std::shared_ptr in the above examples, I

问题 Here is an example playground: protocol P { associatedtype T func getValue() -> T } class Foo: P { func getValue() -> String { return "hello" } } class Bar { func test<T: P>(_ o: T) { print("Generic", o.getValue()) } func test(_ o: Any) { print("Any") } } let foo = Foo() let bar = Bar() bar.test(foo) This outputs: Any . If I remove the Any version of test , the generic method is called. Class Foo conforms to protocol P , why does Swift not pick the generic method since it is more specific? Is

问题 I asked a question earlier but it turns out my problem was not properly modeled by my example. So here is my actual problem: I have class A , and class B inheriting from A , I have two functions foo(A&) and foo(B&) , I have a list of A* pointers, containing instances of A and B . How do I get to call foo(A&) for instances of A and foo(B&) for instances of B ? Constraints: I can modify A and B implementation, but not foo 's implementation. See below an example: #include <iostream> #include

问题 Related to Why does casting a function to a function type that is identical except for return type fail?, I would like to understand, in a fuller way, the distinction between a function's type and a function's signature . For example, the type of a function must typically be considered when dealing with function pointers, and the type of the function includes the return type of that function. However, as noted in Mike Seymour's answer to the above-linked question, the signature of a function