How do I remove code duplication between similar const and non-const member functions?

For a detailed explanation, please see the heading "Avoid Duplication in const and Non-const Member Function," on p. 23, in Item 3 "Use const whenever possible," in Effective C++, 3d ed by Scott Meyers, ISBN-13: 9780321334879.

Here's Meyers' solution (simplified):

struct C {
  const char & get() const {
    return c;
  }
  char & get() {
    return const_cast<char &>(static_cast<const C &>(*this).get());
  }
  char c;
};

The two casts and function call may be ugly, but it's correct in a non-const method as that implies the object was not const to begin with. (Meyers has a thorough discussion of this.)

Yes, it is possible to avoid the code duplication. You need to use the const member function to have the logic and have the non-const member function call the const member function and re-cast the return value to a non-const reference (or pointer if the functions returns a pointer):

class X
{
   std::vector<Z> vecZ;

public:
   const Z& z(size_t index) const
   {
      // same really-really-really long access 
      // and checking code as in OP
      // ...
      return vecZ[index];
   }

   Z& z(size_t index)
   {
      // One line. One ugly, ugly line - but just one line!
      return const_cast<Z&>( static_cast<const X&>(*this).z(index) );
   }

 #if 0 // A slightly less-ugly version
   Z& Z(size_t index)
   {
      // Two lines -- one cast. This is slightly less ugly but takes an extra line.
      const X& constMe = *this;
      return const_cast<Z&>( constMe.z(index) );
   }
 #endif
};

NOTE: It is important that you do NOT put the logic in the non-const function and have the const-function call the non-const function -- it may result in undefined behavior. The reason is that a constant class instance gets cast as a non-constant instance. The non-const member function may accidentally modify the class, which the C++ standard states will result in undefined behavior.

C++17 has updated the best answer for this question:

T const & f() const {
    return something_complicated();
}
T & f() {
    return const_cast<T &>(std::as_const(*this).f());
}

This has the advantages that it:

• Is obvious what is going on
• Has minimal code overhead -- it fits in a single line
• Is hard to get wrong (can only cast away volatile by accident, but volatile is a rare qualifier)

If you want to go the full deduction route then that can be accomplished by having a helper function

template<typename T>
constexpr T & as_mutable(T const & value) noexcept {
    return const_cast<T &>(value);
}
template<typename T>
constexpr T * as_mutable(T const * value) noexcept {
    return const_cast<T *>(value);
}
template<typename T>
constexpr T * as_mutable(T * value) noexcept {
    return value;
}
template<typename T>
void as_mutable(T const &&) = delete;

Now you can't even mess up volatile, and the usage looks like

decltype(auto) f() const {
    return something_complicated();
}
decltype(auto) f() {
    return as_mutable(std::as_const(*this).f());
}