You could try something like this: template <typename T, typename R, typename …Args> R proxycall(T & obj, R (T::*mf)(Args…), Args &&… args) { return (obj.*mf)(std::forward<Args>(args)…); } Usage: proxycall(obj, &hello::f); Alternatively, to make the PTMF into a template argument, try specialization: template <typename T, T> struct proxy; template <typename T, typename R, typename …Args, R (T::*mf)(Args…)> … Read more
So, here the address of function and the address of first variable in function is not same. Why so? Why it would be so? A function pointer is a pointer that points to the function. It does not point to the first variable inside the function, anyway. To elaborate, a function (or subroutine) is a … Read more
Note that there is a difference between equality and identity. The operators == and != in Go1 are comparing the values for equivalence (except when comparing channels), not for identity. Because these operators are trying not to mix equality and identity, Go1 is more consistent than pre-Go1 in this respect. Function equality is different from … Read more
Pointers to non-static member functions are a unique beast with unique calling syntax. Calling those functions require you to supply not just named parameters, but also a this pointer, so you must have the Box pointer handy that will be used as this. (box->*h)(xPos, yPos, width, height);
Basically a typedef has exactly the same syntax as an object declaration except that it is prefixed with typedef. Doing that changes the meaning of the declaration so that the new identifier declares an alias for the type that the object that would have been declared, had it been a normal declaration, would have had. … Read more
Start with your declaration for f1: int (*f1)(float); You want f2 to be a pointer to a function returning f1, so substitute f1 in the declaration above with the declaration for f2: int (* f1 )(float); | +—–+—–+ | | v v int (*(*f2)(double))(float); The declaration reads as f2 — f2 *f2 — is a … Read more
class A{ public: typedef int (A::*method)(); method p; A(){ p = &A::foo; (this->*p)(); // <- trick 1, inner call } int foo(){ printf(“foo\n”); return 0; } }; void main() { A a; (a.*a.p)(); // <- trick 2, outer call }
From C99 spec, section 6.2.5, paragraph 27: A pointer to void shall have the same representation and alignment requirements as a pointer to a character type. Similarly, pointers to qualified or unqualified versions of compatible types shall have the same representation and alignment requirements. All pointers to structure types shall have the same representation and … Read more
You can make Base be a class template that takes its function pointers from its template argument: extern “C” { struct CStruct { void (*funA)(int, char const*); int (*funB)(void); }; } template <typename T> class Base { public: CStruct myStruct; void FillPointers() { myStruct.funA = &T::myFunA; myStruct.funB = &T::myFunB; } Base() { FillPointers(); } }; … Read more
Your question isn’t clear, but I think you might want something like this: int foo(int i){ return i + 1;} typedef int (*g)(int); // Declare typedef g func = &foo; // Define function-pointer variable, and initialise int hvar = func(3); // Call function through pointer