How do I create and array of pointers to functions?

Question

gcc 6.4.0 win 7 Netbeans 8.2

I can't figure what I'm doing wrong.

#ifndef TYPEDEF_H
#define TYPEDEF_H

class Typedef {
   class Alphabet { };
   class Tuple { };
   class Operations { };
public:
   void test();
private:
   Tuple constant(long one, long two, Alphabet& bet, Operations& op) { return Tuple(); } ;
   Tuple expression(long one, long two, Alphabet& bet, Operations& op) {return Tuple(); } ;
};

#endif /* TYPEDEF_H */

===================== .cpp file ========================
#include "Typedef.h"

void Typedef::test() {
   typedef Tuple (*fn)(long, long, Alphabet&, Operations&);
   fn func[]      = { constant, expression };
   Tuple (*fnc[2])= { constant, expression };
}

Typedef.cpp: In member function 'void Typedef::test()':

Typedef.cpp:6:44: error: cannot convert 'Typedef::constant' from type 'Typedef::Tuple (Typedef::)(long int, long int, Typedef::Alphabet&, Typedef::Operations&)' to type 'fn {aka Typedef::Tuple (*)(long int, long int, Typedef::Alphabet&, Typedef::Operations&)}' fn func[] = { constant, expression };

The error message is repeated for all four instances. I've tried &constant and as expected it didn't work.

Member functions aren't functions...

Kerrek SB
– Kerrek SB

2017-09-29 14:44:05 +00:00
Commented Sep 29, 2017 at 14:44 — Kerrek SB
– Kerrek SB, Commented Sep 29, 2017 at 14:44
You may be interested in std::function.

François Andrieux
– François Andrieux

2017-09-29 14:44:06 +00:00
Commented Sep 29, 2017 at 14:44 — François Andrieux
– François Andrieux, Commented Sep 29, 2017 at 14:44

rustyx · Accepted Answer · 2017-09-29 15:37:54Z

2

A class member function is special in that it accepts an additional invisible argument, the this* pointer. Also, a member function and a free function are completely different types, even if the latter had the extra argument compatible with the this* type.

To capture a pointer to the member function, use the member pointer syntax (Typedef::*), like this:

void Typedef::test() {
  typedef Tuple (Typedef::*fn)(long, long, Alphabet&, Operations&);
  fn func[] = { &Typedef::constant, &Typedef::expression };
}

In addition, an address of a member function can be taken only on fully-qualified names (reference: [expr.unary.op]/4), so you have to use &Typedef::constant, &Typedef::expression.

edited Sep 29, 2017 at 15:37

answered Sep 29, 2017 at 14:50

rustyx

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2 Comments

lostbits Over a year ago

Thanks. I don't understand why the Typdef:: is required since the functions are visible.

rustyx Over a year ago

@ArthurSchwarez see my updates. Typedef::* here is not for visibility but a member pointer syntax.

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