Why this weired c++ template syntax is compiled successfully?

Question

I Recently came across this c++ weired syntax used with function and compiled successfully. I could not able to make sense out of it , what this really does.

How "*member" does not gave me undefined error or something else, because its not declared anywhere else.

Can anyone let me how to call this function?

template<class POS, class META>
size_t test11( META POS:: *member)
{
    /********/
} 

Thanks everyone.

Answer 1

When you wrote

META POS:: *member

this means that member is a pointer to a member of class POS that has type META.

Now coming to your second question,

Can anyone let me how to call this function?

One possible example is given below:

#include <iostream>
#include <string>
struct Name
{
    std::string name;
    
    Name() = default;
    Name(std::string pname): name(pname)
    {
        std::cout<<"Constructor called"<<std::endl;
    }
};

template<class POS, class META>
size_t test11( META POS:: *member)
{
    //create object of type Name
    Name n("Anoop");
    
    //create a pointer to object n
    Name *ptrN = &n;
    
    //Now let's use the variable member 
    std::cout<<ptrN->*member<<std::endl; 
    
    return 5;//don't forget to return something since the function has non-void return type
}
int main()
{
    //create a pointer to non-static member
    std::string Name::*ptrData = &Name::name; 
    
    //pass the pointer ptrData to the template function test11<>
    test11(ptrData);
    
    
}

The output of the program is:

Constructor called
Anoop

which can be verified here.

Explanation

When i wrote

test11(ptrData);

in the above code sample, then using template argument deduction, POS is deduced to be the type Name and META is deduced to be the type std::string.