从 class C++ 外部访问私有 static 方法

Question

How exactly can you access a private static method from outside the class. Say, I have a class

    Class ABC {
    private:
        static void print(string str) {
            cout << "It works!!!" << endl;
        }
    };

Now, I just to call print() function say from another function like:

    void doSomething() {
         string str = "1776a0";
         // Call to print() here
    }

I have searched the inte.net and stackoverflow for such a problem but I couldn't find much. So, please point me in the right direction as to if this is possible or not and if so how to do it.

I am currently using GCC.

Thank you all in advance.

Answer 1

You can't. That is exactly what private means. If it is intended to be callable from outside the class, make it public .

Instead of making it public you could call it from another function that is publicly accessible. This other function could either be a public member of ABC , or a friend function.

Both cases require changing the class definition of ABC .

If the other function just does nothing besides call print() then you have achieved the same effect as making print() public. But presumably print is private for a reason, eg it relies on some preconditions. You could make the other function allow for that. For example:

void abc_printer(string printer_name, string str_to_print)
{
    open_printer(printer_name);          
    ABC::print(str);                         
    close_printer();
}

and inside the class definition of ABC :

friend void abc_printer(string, string);

Answer 2

You can if you know the address of function:

typedef void (*funcPtr)();

class beforeABC
{
public:
    static int getRange() 
    {
        char* funcAddr = (char*)&(beforeABC::func);
        char* endAddr = (char*)&(beforeABC::end);
        return endAddr - funcAddr;
    }
    static void func() { };
    static void end() { };
};

class ABC
{
private:
    static void print()
    {
        cout << "It works!!!" << endl;
    }
public:
    static void func() 
    { 
        // does not works without this line. Can someone explain this?
        if(0 == (unsigned long int)&print);
    };
};

int main()
{
    ABC::func();

    funcPtr f = (funcPtr)((char*)&(beforeABC::end) + beforeABC::getRange());
    f();

    return 0;
}    

It is just a hack. Very unsafe and compiler\\system dependent. Do not use it in real projects.

Answer 3

谢谢，问一个朋友，我发现了另一种方法：通过使doSomething（）函数成为ABC类的朋友。

Answer 4

A great hack (just for proof of concept, DO NOT CODE LIKE THIS.:!) is for example put a break point in the context you need for example main.cpp:21

let us assume our function is static void for simplicty

so I will declare in main.cpp:22

void (*func_p) () = nullptr;
func_p();

I will advance until line 22 with the break point and will print the address of my static function

in GDB you can use print giladStruct::myfunc or b giladStruct::myfunc this will return the address something like

{void (void)} 0x7fffcf5bf112 <giladStruct::myfunc()>

then you need to set func_p = 0x7fffcf5bf112 and continue to run.

The address will change but for a POC/simple hack this will work.

Answer 5

I'm a newcomer to C++, but you can access private static methods via pointers passed as callbacks, which still allows one to assure the methods cannot be called directly. This can be done without requiring a friend function, since the callback pointer is registered using the constructor of the class. For example, the following would make the "print" method available as a call back, without allowing the direct invocation of the print method:

#include <iostream>
using namespace std;

void RegisterCallback(void (* const fn)(void));

class ABC 
{
public:
    ABC() {RegisterCallback((void (*)(void))&ABC::print);}
private:
    static void print(void) 
    {
        cout << "It works!!!" << endl;
    }
};

int main(void)
{
    ABC abc;
    return 0;
}

void RegisterCallback(void (* const fn)(void))
{
    fn();
}