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

Question

Let's say I have the following class X where I want to return access to an internal member:

class Z
{
    // details
};

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

public:
    Z& Z(size_t index)
    {
        // massive amounts of code for validating index

        Z& ret = vecZ[index];

        // even more code for determining that the Z instance
        // at index is *exactly* the right sort of Z (a process
        // which involves calculating leap years in which
        // religious holidays fall on Tuesdays for
        // the next thousand years or so)

        return ret;
    }
    const Z& Z(size_t index) const
    {
        // identical to non-const X::Z(), except printed in
        // a lighter shade of gray since
        // we're running low on toner by this point
    }
};

The two member functions X::Z() and X::Z() const have identical code inside the braces. This is duplicate code and can cause maintenance problems for long functions with complex logic.

Is there a way to avoid this code duplication?

Answer 1

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.)

Answer 2

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());
}

Answer 3

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.

Answer 4

C++23 has updated the best answer for this question thanks to explicit object parameters.

struct s {
    auto && f(this auto && self) {
        // all the common code goes here
    }
};

A single function template is callable as a normal member function and deduces the correct reference type for you. No casting to get wrong, no writing multiple functions for something that is conceptually one thing.

---

Note: this feature was added by P0847: Deducing this.

Answer 5

I think Scott Meyers' solution can be improved in C++11 by using a tempate helper function. This makes the intent much more obvious and can be reused for many other getters.

template <typename T>
struct NonConst {typedef T type;};
template <typename T>
struct NonConst<T const> {typedef T type;}; //by value
template <typename T>
struct NonConst<T const&> {typedef T& type;}; //by reference
template <typename T>
struct NonConst<T const*> {typedef T* type;}; //by pointer
template <typename T>
struct NonConst<T const&&> {typedef T&& type;}; //by rvalue-reference

template<typename TConstReturn, class TObj, typename... TArgs>
typename NonConst<TConstReturn>::type likeConstVersion(
   TObj const* obj,
   TConstReturn (TObj::* memFun)(TArgs...) const,
   TArgs&&... args) {
      return const_cast<typename NonConst<TConstReturn>::type>(
         (obj->*memFun)(std::forward<TArgs>(args)...));
}

This helper function can be used the following way.

struct T {
   int arr[100];

   int const& getElement(size_t i) const{
      return arr[i];
   }

   int& getElement(size_t i) {
      return likeConstVersion(this, &T::getElement, i);
   }
};

The first argument is always the this-pointer. The second is the pointer to the member function to call. After that an arbitrary amount of additional arguments can be passed so that they can be forwarded to the function. This needs C++11 because of the variadic templates.

Answer 6

Nice question and nice answers. I have another solution, that uses no casts:

class X {

private:

    std::vector<Z> v;

    template<typename InstanceType>
    static auto get(InstanceType& instance, std::size_t i) -> decltype(instance.get(i)) {
        // massive amounts of code for validating index
        // the instance variable has to be used to access class members
        return instance.v[i];
    }

public:

    const Z& get(std::size_t i) const {
        return get(*this, i);
    }

    Z& get(std::size_t i) {
        return get(*this, i);
    }

};

However, it has the ugliness of requiring a static member and the need of using the instance variable inside it.

I did not consider all the possible (negative) implications of this solution. Please let me know if any.

Answer 7

A bit more verbose than Meyers, but I might do this:

class X {

    private:

    // This method MUST NOT be called except from boilerplate accessors.
    Z &_getZ(size_t index) const {
        return something;
    }

    // boilerplate accessors
    public:
    Z &getZ(size_t index)             { return _getZ(index); }
    const Z &getZ(size_t index) const { return _getZ(index); }
};

The private method has the undesirable property that it returns a non-const Z& for a const instance, which is why it's private. Private methods may break invariants of the external interface (in this case the desired invariant is "a const object cannot be modified via references obtained through it to objects it has-a").

Note that the comments are part of the pattern - _getZ's interface specifies that it is never valid to call it (aside from the accessors, obviously): there's no conceivable benefit to doing so anyway, because it's 1 more character to type and won't result in smaller or faster code. Calling the method is equivalent to calling one of the accessors with a const_cast, and you wouldn't want to do that either. If you're worried about making errors obvious (and that's a fair goal), then call it const_cast_getZ instead of _getZ.

By the way, I appreciate Meyers's solution. I have no philosophical objection to it. Personally, though, I prefer a tiny bit of controlled repetition, and a private method that must only be called in certain tightly-controlled circumstances, over a method that looks like line noise. Pick your poison and stick with it.

[Edit: Kevin has rightly pointed out that _getZ might want to call a further method (say generateZ) which is const-specialised in the same way getZ is. In this case, _getZ would see a const Z& and have to const_cast it before return. That's still safe, since the boilerplate accessor polices everything, but it's not outstandingly obvious that it's safe. Furthermore, if you do that and then later change generateZ to always return const, then you also need to change getZ to always return const, but the compiler won't tell you that you do.

That latter point about the compiler is also true of Meyers's recommended pattern, but the first point about a non-obvious const_cast isn't. So on balance I think that if _getZ turns out to need a const_cast for its return value, then this pattern loses a lot of its value over Meyers's. Since it also suffers disadvantages compared to Meyers's, I think I would switch to his in that situation. Refactoring from one to the other is easy -- it doesn't affect any other valid code in the class, since only invalid code and the boilerplate calls _getZ.]

Answer 8

For those (like me) who

• use c++17
• want to add the least amount of boilerplate/repetition and
• don't mind using macros (while waiting for meta-classes...),

here is another take:

#include <utility>
#include <type_traits>

template <typename T> struct NonConst;
template <typename T> struct NonConst<T const&> {using type = T&;};
template <typename T> struct NonConst<T const*> {using type = T*;};

#define NON_CONST(func)                                                     \
    template <typename... T> auto func(T&&... a)                            \
        -> typename NonConst<decltype(func(std::forward<T>(a)...))>::type   \
    {                                                                       \
        return const_cast<decltype(func(std::forward<T>(a)...))>(           \
            std::as_const(*this).func(std::forward<T>(a)...));              \
    }

It is basically a mix of the answers from @Pait, @DavidStone and @sh1 (EDIT: and an improvement from @cdhowie). What it adds to the table is that you get away with only one extra line of code which simply names the function (but no argument or return type duplication):

class X
{
    const Z& get(size_t index) const { ... }
    NON_CONST(get)
};

Note: gcc fails to compile this prior to 8.1, clang-5 and upwards as well as MSVC-19 are happy (according to the compiler explorer).

Answer 9

You could also solve this with templates. This solution is slightly ugly (but the ugliness is hidden in the .cpp file) but it does provide compiler checking of constness, and no code duplication.

.h file:

#include <vector>

class Z
{
    // details
};

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

public:
    const std::vector<Z>& GetVector() const { return vecZ; }
    std::vector<Z>& GetVector() { return vecZ; }

    Z& GetZ( size_t index );
    const Z& GetZ( size_t index ) const;
};

.cpp file:

#include "constnonconst.h"

template< class ParentPtr, class Child >
Child& GetZImpl( ParentPtr parent, size_t index )
{
    // ... massive amounts of code ...

    // Note you may only use methods of X here that are
    // available in both const and non-const varieties.

    Child& ret = parent->GetVector()[index];

    // ... even more code ...

    return ret;
}

Z& X::GetZ( size_t index )
{
    return GetZImpl< X*, Z >( this, index );
}

const Z& X::GetZ( size_t index ) const
{
    return GetZImpl< const X*, const Z >( this, index );
}

The main disadvantage I can see is that because all the complex implementation of the method is in a global function, you either need to get hold of the members of X using public methods like GetVector() above (of which there always need to be a const and non-const version) or you could make this function a friend. But I don't like friends.

[Edit: removed unneeded include of cstdio added during testing.]

Answer 10

If you don't like const casting, I use this C++17 version of the template static helper function suggested by another answer, with and optional SFINAE test.

#include <type_traits>

#define REQUIRES(...)         class = std::enable_if_t<(__VA_ARGS__)>
#define REQUIRES_CV_OF(A,B)   REQUIRES( std::is_same_v< std::remove_cv_t< A >, B > )

class Foobar {
private:
    int something;

    template<class FOOBAR, REQUIRES_CV_OF(FOOBAR, Foobar)>
    static auto& _getSomething(FOOBAR& self, int index) {
        // big, non-trivial chunk of code...
        return self.something;
    }

public:
    auto& getSomething(int index)       { return _getSomething(*this, index); }
    auto& getSomething(int index) const { return _getSomething(*this, index); }
};

Full version: https://godbolt.org/z/mMK4r3

Answer 11

While most of answers here suggest to use a const_cast, CppCoreGuidelines have a section about that:

Instead, prefer to share implementations. Normally, you can just have the non-const function call the const function. However, when there is complex logic this can lead to the following pattern that still resorts to a const_cast:

class Foo {
public:
    // not great, non-const calls const version but resorts to const_cast
    Bar& get_bar()
    {
        return const_cast<Bar&>(static_cast<const Foo&>(*this).get_bar());
    }
    const Bar& get_bar() const
    {
        /* the complex logic around getting a const reference to my_bar */
    }
private:
    Bar my_bar;
};

Although this pattern is safe when applied correctly, because the caller must have had a non-const object to begin with, it's not ideal because the safety is hard to enforce automatically as a checker rule.

Instead, prefer to put the common code in a common helper function -- and make it a template so that it deduces const. This doesn't use any const_cast at all:

class Foo {
public:                         // good
          Bar& get_bar()       { return get_bar_impl(*this); }
    const Bar& get_bar() const { return get_bar_impl(*this); }
private:
    Bar my_bar;

    template<class T>           // good, deduces whether T is const or non-const
    static auto& get_bar_impl(T& t)
        { /* the complex logic around getting a possibly-const reference to my_bar */ }
};

Note: Don't do large non-dependent work inside a template, which leads to code bloat. For example, a further improvement would be if all or part of get_bar_impl can be non-dependent and factored out into a common non-template function, for a potentially big reduction in code size.

Answer 12

How about moving the logic into a private method, and only doing the "get the reference and return" stuff inside the getters? Actually, I would be fairly confused about the static and const casts inside a simple getter function, and I'd consider that ugly except for extremely rare circumstances!

Answer 13

I'd suggest a private helper static function template, like this:

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

    // ReturnType is explicitly 'Z&' or 'const Z&'
    // ThisType is deduced to be 'X' or 'const X'
    template <typename ReturnType, typename ThisType>
    static ReturnType Z_impl(ThisType& self, size_t index)
    {
        // massive amounts of code for validating index
        ReturnType ret = self.vecZ[index];
        // even more code for determining, blah, blah...
        return ret;
    }

public:
    Z& Z(size_t index)
    {
        return Z_impl<Z&>(*this, index);
    }
    const Z& Z(size_t index) const
    {
        return Z_impl<const Z&>(*this, index);
    }
};

Answer 14

Is it cheating to use the preprocessor?

struct A {

    #define GETTER_CORE_CODE       \
    /* line 1 of getter code */    \
    /* line 2 of getter code */    \
    /* .....etc............. */    \
    /* line n of getter code */       

    // ^ NOTE: line continuation char '\' on all lines but the last

   B& get() {
        GETTER_CORE_CODE
   }

   const B& get() const {
        GETTER_CORE_CODE
   }

   #undef GETTER_CORE_CODE

};

It's not as fancy as templates or casts, but it does make your intent ("these two functions are to be identical") pretty explicit.

Answer 15

It's surprising to me that there are so many different answers, yet almost all rely on heavy template magic. Templates are powerful, but sometimes macros beat them in conciseness. Maximum versatility is often achieved by combining both.

I wrote a macro FROM_CONST_OVERLOAD() which can be placed in the non-const function to invoke the const function.

Example usage:

class MyClass
{
private:
    std::vector<std::string> data = {"str", "x"};

public:
    // Works for references
    const std::string& GetRef(std::size_t index) const
    {
        return data[index];
    }

    std::string& GetRef(std::size_t index)
    {
        return FROM_CONST_OVERLOAD( GetRef(index) );
    }


    // Works for pointers
    const std::string* GetPtr(std::size_t index) const
    {
        return &data[index];
    }

    std::string* GetPtr(std::size_t index)
    {
        return FROM_CONST_OVERLOAD( GetPtr(index) );
    }
};

Simple and reusable implementation:

template <typename T>
T& WithoutConst(const T& ref)
{
    return const_cast<T&>(ref);
}

template <typename T>
T* WithoutConst(const T* ptr)
{
    return const_cast<T*>(ptr);
}

template <typename T>
const T* WithConst(T* ptr)
{
    return ptr;
}

#define FROM_CONST_OVERLOAD(FunctionCall) \
  WithoutConst(WithConst(this)->FunctionCall)

---

Explanation:

As posted in many answers, the typical pattern to avoid code duplication in a non-const member function is this:

return const_cast<Result&>( static_cast<const MyClass*>(this)->Method(args) );

A lot of this boilerplate can be avoided using type inference. First, const_cast can be encapsulated in WithoutConst(), which infers the type of its argument and removes the const-qualifier. Second, a similar approach can be used in WithConst() to const-qualify the this pointer, which enables calling the const-overloaded method.

The rest is a simple macro that prefixes the call with the correctly qualified this-> and removes const from the result. Since the expression used in the macro is almost always a simple function call with 1:1 forwarded arguments, drawbacks of macros such as multiple evaluation do not kick in. The ellipsis and __VA_ARGS__ could also be used, but should not be needed because commas (as argument separators) occur within parentheses.

This approach has several benefits:

• Minimal and natural syntax -- just wrap the call in FROM_CONST_OVERLOAD( )
• No extra member function required
• Compatible with C++98
• Simple implementation, no template metaprogramming and zero dependencies
• Extensible: other const relations can be added (like const_iterator, std::shared_ptr<const T>, etc.). For this, simply overload WithoutConst() for the corresponding types.

Limitations: this solution is optimized for scenarios where the non-const overload is doing exactly the same as the const overload, so that arguments can be forwarded 1:1. If your logic differs and you are not calling the const version via this->Method(args), you may consider other approaches.

Answer 16

I came up with a macro that generates pairs of const/non-const functions automatically.

class A
{
    int x;    
  public:
    MAYBE_CONST(
        CV int &GetX() CV {return x;}
        CV int &GetY() CV {return y;}
    )

    //   Equivalent to:
    // int &GetX() {return x;}
    // int &GetY() {return y;}
    // const int &GetX() const {return x;}
    // const int &GetY() const {return y;}
};

See the end of the answer for the implementation.

The argument of MAYBE_CONST is duplicated. In the first copy, CV is replaced with nothing; and in the second copy it's replaced with const.

There's no limit on how many times CV can appear in the macro argument.

There's a slight inconvenience though. If CV appears inside of parentheses, this pair of parentheses must be prefixed with CV_IN:

// Doesn't work
MAYBE_CONST( CV int &foo(CV int &); )

// Works, expands to
//         int &foo(      int &);
//   const int &foo(const int &);
MAYBE_CONST( CV int &foo CV_IN(CV int &); )

---

Implementation:

#define MAYBE_CONST(...) IMPL_CV_maybe_const( (IMPL_CV_null,__VA_ARGS__)() )
#define CV )(IMPL_CV_identity,
#define CV_IN(...) )(IMPL_CV_p_open,)(IMPL_CV_null,__VA_ARGS__)(IMPL_CV_p_close,)(IMPL_CV_null,

#define IMPL_CV_null(...)
#define IMPL_CV_identity(...) __VA_ARGS__
#define IMPL_CV_p_open(...) (
#define IMPL_CV_p_close(...) )

#define IMPL_CV_maybe_const(seq) IMPL_CV_a seq IMPL_CV_const_a seq

#define IMPL_CV_body(cv, m, ...) m(cv) __VA_ARGS__

#define IMPL_CV_a(...) __VA_OPT__(IMPL_CV_body(,__VA_ARGS__) IMPL_CV_b)
#define IMPL_CV_b(...) __VA_OPT__(IMPL_CV_body(,__VA_ARGS__) IMPL_CV_a)

#define IMPL_CV_const_a(...) __VA_OPT__(IMPL_CV_body(const,__VA_ARGS__) IMPL_CV_const_b)
#define IMPL_CV_const_b(...) __VA_OPT__(IMPL_CV_body(const,__VA_ARGS__) IMPL_CV_const_a)

---

Pre-C++20 implementation that doesn't support CV_IN:

#define MAYBE_CONST(...) IMPL_MC( ((__VA_ARGS__)) )
#define CV ))((

#define IMPL_MC(seq) \
    IMPL_MC_end(IMPL_MC_a seq) \
    IMPL_MC_end(IMPL_MC_const_0 seq)

#define IMPL_MC_identity(...) __VA_ARGS__
#define IMPL_MC_end(...) IMPL_MC_end_(__VA_ARGS__)
#define IMPL_MC_end_(...) __VA_ARGS__##_end

#define IMPL_MC_a(elem) IMPL_MC_identity elem IMPL_MC_b
#define IMPL_MC_b(elem) IMPL_MC_identity elem IMPL_MC_a
#define IMPL_MC_a_end
#define IMPL_MC_b_end

#define IMPL_MC_const_0(elem)       IMPL_MC_identity elem IMPL_MC_const_a
#define IMPL_MC_const_a(elem) const IMPL_MC_identity elem IMPL_MC_const_b
#define IMPL_MC_const_b(elem) const IMPL_MC_identity elem IMPL_MC_const_a
#define IMPL_MC_const_a_end
#define IMPL_MC_const_b_end

Answer 17

I did this for a friend who rightfully justified the use of const_cast... not knowing about it I probably would have done something like this (not really elegant) :

#include <iostream>

class MyClass
{

public:

    int getI()
    {
        std::cout << "non-const getter" << std::endl;
        return privateGetI<MyClass, int>(*this);
    }

    const int getI() const
    {
        std::cout << "const getter" << std::endl;
        return privateGetI<const MyClass, const int>(*this);
    }

private:

    template <class C, typename T>
    static T privateGetI(C c)
    {
        //do my stuff
        return c._i;
    }

    int _i;
};

int main()
{
    const MyClass myConstClass = MyClass();
    myConstClass.getI();

    MyClass myNonConstClass;
    myNonConstClass.getI();

    return 0;
}

Answer 18

Typically, the member functions for which you need const and non-const versions are getters and setters. Most of the time they are one-liners so code duplication is not an issue.

Answer 19

To add to the solution jwfearn and kevin provided, here's the corresponding solution when the function returns shared_ptr:

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

Answer 20

Didn't find what I was looking for, so I rolled a couple of my own...

This one is a little wordy, but has the advantage of handling many overloaded methods of the same name (and return type) all at once:

struct C {
  int x[10];

  int const* getp() const { return x; }
  int const* getp(int i) const { return &x[i]; }
  int const* getp(int* p) const { return &x[*p]; }

  int const& getr() const { return x[0]; }
  int const& getr(int i) const { return x[i]; }
  int const& getr(int* p) const { return x[*p]; }

  template<typename... Ts>
  auto* getp(Ts... args) {
    auto const* p = this;
    return const_cast<int*>(p->getp(args...));
  }

  template<typename... Ts>
  auto& getr(Ts... args) {
    auto const* p = this;
    return const_cast<int&>(p->getr(args...));
  }
};

If you have only one const method per name, but still plenty of methods to duplicate, then you might prefer this:

  template<typename T, typename... Ts>
  auto* pwrap(T const* (C::*f)(Ts...) const, Ts... args) {
    return const_cast<T*>((this->*f)(args...));
  }

  int* getp_i(int i) { return pwrap(&C::getp_i, i); }
  int* getp_p(int* p) { return pwrap(&C::getp_p, p); }

Unfortunately this breaks down as soon as you start overloading the name (the function pointer argument's argument list seems to be unresolved at that point, so it can't find a match for the function argument). Although you can template your way out of that, too:

  template<typename... Ts>
  auto* getp(Ts... args) { return pwrap<int, Ts...>(&C::getp, args...); }

But reference arguments to the const method fail to match against the apparently by-value arguments to the template and it breaks. Not sure why.Here's why.

Answer 21

I made a macro https://gist.github.com/esnosy/947565a4e651991dc6a25f3e9f467285

#include <utility>

// Based on https://en.cppreference.com/mwiki/index.php?title=cpp/utility/as_const&oldid=151976#Possible_implementation
template <typename T> constexpr T &as_non_const(const T &t) noexcept { return const_cast<T &>(t); }

#define NON_CONST_GETTER(const_getter)                                                                                 \
  auto &const_getter() { return as_non_const(std::as_const(*this).const_getter()); }

// Example
class C {
private:
  int secret = 42;
public:
  const int &get() const { return secret; }
  NON_CONST_GETTER(get);
};

Answer 22

This DDJ article shows a way using template specialization that doesn't require you to use const_cast. For such a simple function it really isn't needed though.

boost::any_cast (at one point, it doesn't any more) uses a const_cast from the const version calling the non-const version to avoid duplication. You can't impose const semantics on the non-const version though so you have to be very careful with that.

In the end some code duplication is okay as long as the two snippets are directly on top of each other.