Re: declaring/defining op== for private nested class

"Johannes Schaub (litb)" <>
Mon, 8 Mar 2010 16:01:39 CST
Paul Bibbings wrote:

Consider the following sample code and output from an attempt to
compile it:

    16:22:28 Paul Bibbings@JIJOU
    /cygdrive/d/CPPProjects/nano/test_link $cat test.cpp
    // file: test.cpp

    #include <string>

    template<typename T>
    class outer {
       explicit outer(T t) : h_(t) { }
       bool equals(T t) { return t == h_; } // uses holder::op T&()/T()
       struct holder {
          explicit holder(T t) : t_(t) { }
          operator T&() { return t_; }
          operator T() const { return t_; }
          T t_;
          // friend bool operator==(const T& t, const holder& h) {
          // return t == h.t_;
          // }
       holder h_;


It seems odd because, though declared as a friend of holder, op== does
not 'require' this friendship in the sense that holder's data is
public. Yet, without defining it as an inline friend I can't see what
other options there are. I'm thinking that it cannot be defined in
the nearest surrounding namespace scope as a non-friend because holder
is a *private* nested class of outer. Furthermore, in real code, I
would want to balance this with its counterparts op==(const holder&,
const T&) and op==(const holder&, const holder&), so it seems that it
must be a free function rather than a member.

So, my question is: is it the language, or is it my understanding,
that is seemingly limiting me in having to provide op== here as an
inline-defined friend of my nested private class.

You could create a base that knows the Holder type.

    template<typename T, typename Holder> struct holder_base {
      T const& get_t_() const
      { return static_cast<Holder const&>(*this).t_; }

    template<typename T>
    class outer {
       explicit outer(T t) : h_(t) { }
       bool equals(T t) { return t == h_; }
       struct holder : holder_base<T, holder> {
          explicit holder(T t) : t_(t) { }
          T t_;
       holder h_;

    template<typename T, typename Holder>
    bool operator==(const T& t, holder_base<T, Holder> const& h)
       return t == h.get_t_();

I don't know tho - this looks worse than the friend function approach you
took above. It's got a higher conversion cost for the second parameter
(derived->base) and some noise around :)

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