function-pointers

A lot of C++ books and tutorials explain how to do this, but I haven't seen one that gives a convincing reason to choose to do this. I understand very well why function pointers were necessary in C (e.g., when using some POSIX facilities). However, AFAIK you can't send them a member function because of the "this" parameter. But if you're already using classes and objects, why not just use an object oriented solution like functors? Real world examples of where you had to use such function pointers would be appreciated. Update: I appreciate everyone's answers. I have to say, though, that none of

What is NSComparisonResult and NSComparator ? I've seen one of the type definitions, something like that: typedef NSComparisonResult (^NSComparator)(id obj1, id obj2); Is it any different from a function pointer? Also, I can't even guess what the ^ symbol means. ^ signifies a block type , similar in concept to a function pointer. typedef NSComparisonResult (^NSComparator)(id obj1, id obj2); // ^ ^ ^ // return type of block type name arguments This means that the type NSComparator is a block that takes in two objects of type id called obj1 and obj2 , and returns an NSComparisonResult .

I'm trying to recursively dereference a pointer in C++. If an object is passed that is not a pointer (this includes smart pointers), I just want to return the object itself, by reference if possible. I have this code: template<typename T> static T &dereference(T &v) { return v; } template<typename T> static const T &dereference(const T &v) { return v; } template<typename T> static T &dereference(T *v) { return dereference(*v); } My code seems to work fine in most cases, but it breaks when given function pointers , because dereferencing a function pointer results in the same exact type of

Is it possible to take the address of a function that would be found through ADL? For example: template<class T> void (*get_swap())(T &, T &) { return & _________; // how do I take the address of T's swap() function? } int main() { typedef some_type T; get_swap<T>(); } Cassio Neri Honestly, I don't know but I tend towards saying that this is not possible. Depending on what you want to achieve I can suggest a workaround. More precisely, if you just need the address of a function that has the same semantics as swap called through ADL then you can use this: template <typename T> void (*get_swap()

#include <functional> int func(int x, int y) { return x+y; } int main() { typedef std::function<int(int, int)> Funcp; Funcp funcp = func; return 0; } But is it possible to point to a template function? #include <functional> template<class T> T func(T x, T y) { return x+y; } int main() { typedef std::function<?(?, ?)> Funcp; Funcp funcp = func; return 0; } No. A template function is exactly that, a template. It's not a real function. You can point a std::function to a specific instantiation of the template function, e.g. func<int,int> There is no such thing as a template function in C++ — what

I know this is very basic but it is little bit confusing to me. I've read: a pointer is nothing more than an address , and a pointer variable is just a variable that can store an address . When we store the address of a variable i in the pointer variable p , we say that p points to i . int i, *p = &i; p points to i . To gain access to the object that a pointer points to, we use the * (indirection) operator. If p is a pointer then *p represents the object to which p currently points. Now I am confused that what should I call p -- pointer or pointer variable ? Additional: Is a pointer always the