Disambiguate overloaded member function pointer being passed as template parameter

Question

I am attempting to recreate the Observer pattern where I can perfectly forward parameters to a given member function of the observers.

If I attempt

Answer 1

The issue is here:

l.call(&foo::func, "hello");
l.call(&foo::func, 0.5);

For both lines, the compiler doesn't know which foo::func you are referring to. Hence, you have to disambiguate yourself by providing the type information that is missing (i.e., the type of foo:func) through casts:

l.call(static_cast<void (foo::*)(const std::string&)>(&foo::func), "hello");
l.call(static_cast<void (foo::*)(const double      )>(&foo::func), 0.5);

Alternatively, you can provide the template arguments that the compiler cannot deduce and that define the type of func:

l.call<void, const std::string&>(&foo::func, "hello");
l.call<void, double            >(&foo::func, 0.5);

Notice that you have to use double and not const double above. The reason is that generally double and const double are two different types. However, there's one situation where double and const double are considered as if they were the same type: as function arguments. For instance,

void bar(const double);
void bar(double);

are not two different overloads but are actually the same function.