C++: Calling Member Function via Pointer

How to call through a member function pointer?

More parentheses are required:

(bigCat.*pcat)();
^            ^

The function call (()) has higher precedence than the pointer-to-member binding operator (.*). The unary operators have higher precedence than the binary operators.

C++: Calling member function via pointer

You need to tell the compiler which class the foos are coming from (otherwise it thinks they're functions from global scope):

void A::setPtr(int v){
    if(v == 1){
        _currentPtr = &A::foo1;
                  //  ^^^^
    } else {
        _currentPtr = &A::foo2;
                  //  ^^^^
    }
}

and you need a set of parentheses here:

std::cout << (this->*_currentPtr)(4,5);
          // ^                  ^

Use pointer to member function to determine which function to call

Syntax to call a member function via member function pointer is

(this->*memf)();

You cannot magically turn the string into a member function pointer. Sloppy speaking, names of functions do not exist at runtime. If you want such mapping you need to provide it yourself. No way around that. What you can avoid is the "forest of if-else" by using a std::unordered_map:

#include <unordered_map>
#include <string>
#include <iostream>

class Karen
{
    public:
        void complain(std::string level) {
            static const std::unordered_map<std::string, void(Karen::*)() const>  m{
                {"debug",&Karen::debug},
                {"info",&Karen::info},
                {"warning",&Karen::warning},
                {"error",&Karen::error}
            };
            auto it = m.find(level);
            if (it == m.end()) return;
            (this->*(it->second))();
        }

    private:
        void debug(void) const { std::cout << "debug\n"; }
        void info(void) const { std::cout << "info\n"; }
        void warning(void) const { std::cout << "warning\n"; }
        void error(void) const { std::cout << "error\n"; }
};

int main() {
    Karen k;
    k.complain("info");
}

Live Demo

As mentioned in comments, you could use an enum in place of the string. When possible you should use the help of the compiler, which can diagnose a typo in an enum but not in a string. Alternatively you could directly pass a member function pointer to complain. Then implementation of complain would be trivial, no branching needed. Though this would require the methods to be public and the caller would have to deal with member function pointers.

If you are not allowed to use C++11 or newer you should have a serious talk with your teacher. Soon C++20 will be the de facto standard and things have changed quite a lot. I am not fluent in C++98 anymore, so here is just a quick fix of the above to get it working somehow. You cannot use std::unordered_map but there is std::map and initialization of the map is rather cumbersome:

#include <map>
#include <string>
#include <iostream>

class Karen
{
    typedef void(Karen::*memf_t)() const; 
    typedef std::map<std::string,void(Karen::*)() const> map_t;

    public:
        void complain(std::string level) {
            map_t::const_iterator it = get_map().find(level);
            if (it == get_map().end()) return;
            (this->*(it->second))();
        }

    private:
        const map_t& get_map(){
            static const map_t m = construct_map();
            return m;
        }
        const map_t construct_map() {
            map_t m;
            m["debug"] = &Karen::debug;
            m["info"] = &Karen::info;
            m["warning"] = &Karen::warning;
            m["error"] = &Karen::error;
            return m;
        }
        void debug(void) const { std::cout << "debug\n"; }
        void info(void) const { std::cout << "info\n"; }
        void warning(void) const { std::cout << "warning\n"; }
        void error(void) const { std::cout << "error\n"; }
};

int main() {
    Karen k;
    k.complain("info");
}

Live Demo

Calling a function pointer on a member function

myMethod is a member function, so you need to call it on an instance of the class it is a member of. Since you are already in a member function, you can use this:

(this->*myMethod)()

C++ Calling Member Function by Function Pointer Error

I think the syntax to invoke that member function by pointer should be e.g.

( this->*FunctionPointer )( Parameters, ReturnValue );

C++ has two operators, .* and ->*, specifically created to access class members by pointer. Here's an article on the topic. The ubiquitous * operator just doesn't work with class member pointers.

Calling C++ member functions via a function pointer

Read this for detail :

// 1 define a function pointer and initialize to NULL

int (TMyClass::*pt2ConstMember)(float, char, char) const = NULL;

// C++

class TMyClass
{
public:
   int DoIt(float a, char b, char c){ cout << "TMyClass::DoIt"<< endl; return a+b+c;};
   int DoMore(float a, char b, char c) const
         { cout << "TMyClass::DoMore" << endl; return a-b+c; };

   /* more of TMyClass */
};
pt2ConstMember = &TMyClass::DoIt; // note: <pt2Member> may also legally point to &DoMore

// Calling Function using Function Pointer

(*this.*pt2ConstMember)(12, 'a', 'b');

How to call private member function by using a pointer

Add a public method to your class that returns a pointer to the private function:

class A
{
private:
    void error(void);
public:
    void callError(void);

    auto get_error_ptr() {
        return &A::error;
    }
};

int main(void)
{
    void (A::*abc)(void) = &A::callError;
    A test;

    (test.*abc)();

    void (A::*error_ptr)(void) = test.get_error_ptr();
    (test.*error_ptr)();

    return (0);
}

But I wouldn't suggest actually using this kind of code in a real application, it is extremely confusing and error-prone.

How do I call pointer-to-member functions from another class?

You need an instance to call it:

void Game::tick(){
    (this->*(button1->click))();
}

Demo

C++ Call Pointer To Member Function

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);