function-pointers

问题 I would like to declare a variable of type pointer to function returning pointer to function . Essentially what the following does, but without any typedef s: typedef void (*func)(); typedef func (*funky_func)(); funky_func ptr; I tried the following (void (*)()) (*ptr)(); but it gives an "undeclared identifier" -error for ptr (probably due to completely different parsing). Being not that well-versed in the intricacies of parsing C++, I'd like to know if this is even possible and if yes, how

问题 I would like to declare a variable of type pointer to function returning pointer to function . Essentially what the following does, but without any typedef s: typedef void (*func)(); typedef func (*funky_func)(); funky_func ptr; I tried the following (void (*)()) (*ptr)(); but it gives an "undeclared identifier" -error for ptr (probably due to completely different parsing). Being not that well-versed in the intricacies of parsing C++, I'd like to know if this is even possible and if yes, how

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: Disabling “bad function cast” warning I am attempting to wrap my brain around c++ function pointers. To keep my learning experience basic, I created a test function pointer example. Eventually, I would like to pass all-ready instantiated objects by reference so I can callback the object's method; however, for the sake of learning and understand, I would like to stick to the basics of c++ function pointers. I

问题 This question already has an answer here : Why can't std::function bind to C-style variadic functions? (1 answer) Closed 6 years ago . I have a structure that I pass around my application which contains a bunch of callback functions: typedef struct { std::function<void (void)> f1; std::function<void (int)> f2; std::function<int (float *)> f3; // ... and so on } CallbackTable; I handle state control within the application by binding different functions to the various callbacks, depending upon

问题 The Problem Lets say that we have a simple class to sort a list of integers, class Sorter { public: Sorter() {} ~Sorter() {} enum class Algorithm { Bubble, Heap, Merge, Insertion }; void SetVector(const std::vector<int>& vec) { mVector = vec; } void Sort(Algorithm algo) { void (Sorter:: * pfSort)() = nullptr; switch (algo) { case Sorter::Algorithm::Bubble: pfSort = &Sorter::BubbleSort; break; case Sorter::Algorithm::Heap: pfSort = &Sorter::HeapSort; break; case Sorter::Algorithm::Merge:

问题 Suppose there is a pointer f declared to some function say int foo(int) as : int (*f)(int)=foo; While mentioning about calling foo() function by using this ponter which is passed as argument to some other function. I have come across a statement saying that both y=(*f)(x) and y=f(x) are same in C and calls function foo() ....(x and y are of int type). For arrays I know that if p is pointer to any array a. p[i]=*(p+i)=*(&a[0]+i)=*(a+i)=a[i] . So writing p[i] and *(p+i) are same thing. But I

问题 Suppose there is a pointer f declared to some function say int foo(int) as : int (*f)(int)=foo; While mentioning about calling foo() function by using this ponter which is passed as argument to some other function. I have come across a statement saying that both y=(*f)(x) and y=f(x) are same in C and calls function foo() ....(x and y are of int type). For arrays I know that if p is pointer to any array a. p[i]=*(p+i)=*(&a[0]+i)=*(a+i)=a[i] . So writing p[i] and *(p+i) are same thing. But I

问题 This question already has answers here : C# Generic Operators (5 answers) Closed 5 years ago . I would like to create a procedure class that supports connecting to another procedure like so: a|b|c|d (this should result a procedure takes a's input type, and give d's output type) class Procedure<I,O> { public Func<I,O> func; public Procedure(Func<I, O> func) { this.func = func; } public static Procedure<I, O> operator| (Procedure<I, M> proc1, Procedure<M, O> proc2) { return new Procedure<I,O>(

问题 This question already has answers here : How do function pointers in C work? (11 answers) C function pointer syntax (4 answers) Closed 3 years ago . I've been meaning to ask this question for a while now. What's going on with these functions? Why are the names in parenthesis? void (*think)(gentity_t *self); void (*reached)(gentity_t *self); // movers call this when hitting endpoint void (*blocked)(gentity_t *self, gentity_t *other); void (*touch)(gentity_t *self, gentity_t *other, trace_t

问题 A function is returning an anonymous function. I would like to assign the result to a variable. However the compiler thinks that I am trying to assign the function and not the result of the function. How can I resolve this? program Project9; {$APPTYPE CONSOLE} type TMyEvent = reference to function: string; var v1: TMyEvent; function GetHandler: TMyEvent; begin Result := function: string begin Result := ''; end; end; begin v1 := GetHandler; // <- Incompatible types: 'TMyEvent' and 'Procedure'