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Home/ Questions/Q 317477
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T08:28:46+00:00 2026-05-12T08:28:46+00:00

I’m trying to create a lookup table of member functions in my code, but

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I’m trying to create a lookup table of member functions in my code, but it seems to be trying to call my copy constructor, which I’ve blocked by extending an “uncopyable” class. What I have is something like the following.

enum {FUN1_IDX, FUN2_IDX, ..., NUM_FUNS };

class Foo {
  fun1(Bar b){ ... }
  fun2(Bar b){ ... }
  ...
  void (Foo::*lookup_table[NUM_FUNS])(Bar b);
  Foo(){ 
    lookup_table[FUN1_IDX] = &Foo::fun1;
    lookup_table[FUN2_IDX] = &Foo::fun2;
  }

  void doLookup(int fun_num, Bar b) {
    (this->*lookup_table[fun_num])(b);
  }
};

The error is that the ‘(this->…’ line tries to call the copy constructor, which is not visible. Why is it trying to do this, and what do I have to change so it won’t?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T08:28:46+00:00Added an answer on May 12, 2026 at 8:28 am

    Make them reference parameters.

    enum {FUN1_IDX, FUN2_IDX, ..., NUM_FUNS };
    
    class Foo {
      fun1(Bar &b){ ... }
      fun2(Bar &b){ ... }
      ...
      void (Foo::*lookup_table[NUM_FUNS])(Bar &b);
      Foo(){ 
        lookup_table[FUN1_IDX] = &Foo::fun1;
        lookup_table[FUN2_IDX] = &Foo::fun2;
      }
    
      void doLookup(int fun_num, Bar &b) {
        (this->*lookup_table[fun_num])(b);
      }
    };
    

    In C++, otherwise such plain parameters don’t just reference objects, but they are those objects themselves. Making them reference parameters will merely reference what is passed. In this matter, C++ has the same semantics as C (in which you would use pointers for that).

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