对于那些(像我一样)

使用c++ 17 想要添加最少的样板文件/重复和 不介意使用宏(在等待元类时…)，

下面是另一种说法:

#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)...));              \
    }

它基本上是@Pait， @DavidStone和@sh1的答案的混合(编辑:和@cdhowie的改进)。它向表中添加的是，你只需要额外的一行代码，它只是简单地命名函数(但没有参数或返回类型重复):

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

注意:gcc在8.1之前编译失败，clang-5及以上版本以及MSVC-19都很高兴(根据编译器资源管理器)。

是的，可以避免代码重复。你需要使用const成员函数来拥有逻辑，并让非const成员函数调用const成员函数，并将返回值重新转换为非const引用(或指针，如果函数返回指针):

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
};

注意:重要的是，不要将逻辑放在非const函数中，并让const函数调用非const函数——这可能会导致未定义的行为。原因是常量类实例被转换为非常量实例。非const成员函数可能会意外地修改类，c++标准状态将导致未定义的行为。

我建议使用私有helper静态函数模板，如下所示:

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);
    }
};

很好的问题和答案。我有另一个解决方案，不使用类型转换:

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);
    }

};

但是，它需要一个静态成员，并且需要使用其中的实例变量。

我没有考虑到这个解决方案的所有可能(负面)影响。如果有，请告诉我。

c++ 17更新了这个问题的最佳答案:

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

这样做的好处是:

很明显发生了什么 有最小的代码开销——它适合单行 很难出错(只能抛弃不稳定的偶然，但不稳定是一个罕见的限定词)

如果你想要走完整的演绎路线，那么可以通过一个辅助函数来完成

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;

现在你甚至不能搞混volatile，它的用法看起来就像

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

比Meyers说得啰嗦一点，但我可能会这样做:

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); }
};

私有方法有一个不受欢迎的属性，它为const实例返回一个非const的Z&，这就是为什么它是私有的。私有方法可能会破坏外部接口的不变量(在这种情况下，所需的不变量是“一个const对象不能通过引用它所拥有的对象来修改”)。

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.

顺便说一下，我很欣赏梅耶斯的解决方案。我对此没有哲学上的异议。不过，就我个人而言，我更喜欢一点点受控的重复，以及只能在某些严格控制的情况下调用的私有方法，而不是看起来像线噪声的方法。选择你的毒药并坚持下去。

[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.]