pointer to const member function typedef

Question

I know it's possible to separate to create a pointer to member function like this

struct K { void func() {} }; typedef void FuncType(); typedef FuncType K::* MemFuncType; MemFuncType pF = &K::func; 

Is there similar way to construct a pointer to a const function? I've tried adding const in various places with no success. I've played around with gcc some and if you do template deduction on something like

template <typename Sig, typename Klass> void deduce(Sig Klass::*); 

It will show Sig with as a function signature with const just tacked on the end. If to do this in code it will complain that you can't have qualifiers on a function type. Seems like it should be possible somehow because the deduction works.

Answer 1

You want this:

typedef void (K::*MemFuncType)() const; 

If you want to still base MemFuncType on FuncType, you need to change FuncType:

typedef void FuncType() const; typedef FuncType K::* MemFuncType; 

Answer 2

A slight refinement showing how to do it without a typedef. In a deduced context like the following, you can't use a typedef.

template <typename Class, typename Field> Field extract_field(const Class& obj, Field (Class::*getter)() const) {    return (obj.*getter)(); } 

applied to some class with a const getter:

class Foo {  public:   int get_int() const; };  Foo obj; int sz = extract_field(obj, &Foo::get_int);