How to call function pointer in STL - c++

Question

I am curious about how to call function pointer in a map structure. Here is the details:
#include<iostream>
#include<map>
#include<vector>
#include<string.h>
using namespace std;
class FuncP;
typedef int(FuncP::*func) (int, int);
class FuncP
{
public:
map<int, func> fmap;
map<int, string> fstring;
public:
FuncP(){}
void initial();
int max(int x, int y);
int min(int x, int y);
int call(int op, int x, int y)
{
return (this->*fmap[op])(x, y);
}
};
void FuncP::initial()
{
fmap[0] = &FuncP::max;
fmap[1] = &FuncP::min;
fstring[0] = "fdsfaf";
}
int FuncP::min(int x, int y)
{
return (x<y)?x:y;
}
int FuncP::max(int x, int y)
{
return (x<y)?y:x;
}
int main()
{
func h = &FuncP::max;
FuncP *handle = new FuncP();
handle->initial();
cout<< handle->call(0, 1, 4); //1
cout<< (handle->FuncP::*fmap)[0](1,5); //2
return 0;
}
For the number 2 (handle->FuncP::*fmap)0; The compiler gives a error:
‘fmap’ was not declared in this scope
I am not sure why it happened. What the difference of the number 1 and 2 call methods?

Answer 1

As commented by Piotr, a correct way would be
(handle->*(handle->fmap[0]))(1, 5);
Explanation:
handle->fmap[0] gives you the function pointer. To call it, you need to dereference it, giving *(handle->fmap[0]) (parentheses optional)
and call it on the respecting object (handle), leaving us with the expression above.
This is essentially the same as your above statement (this->*fmap[op])(x, y) except of handle->fmap[0]instead offmap[op].