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

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Mon, 8 Mar 2010 16:06:19 CST
On 8 Mrz., 20:17, Paul Bibbings <> wrote:

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

    #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()

Why does the function not have a const-qualifier?

       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_;

    int main()
       const char *c_str = "c_str";
       outer<const char *> out(c_str);

       std::string s("std::string");
       outer<std::string> out2(s);
       out2.equals(s); // line 31

    16:22:32 Paul Bibbings@JIJOU
    /cygdrive/d/CPPProjects/nano/test_link $g++ -o test test.cpp
    test.cpp: In member function 'bool outer<T>::equals(T)
      [with T = std::basic_string<char, std::char_traits<char>,
       std::allocator<char> >]':
    test.cpp:31: instantiated from here
    test.cpp:9: error: no match for 'operator==' in 't ==
       ((outer<std::basic_string<char, std::char_traits<char>,
       std::allocator<char> > >*)this)->outer<std::basic_string<char,
       std::char_traits<char>, std::allocator<char> > >::h_'

If I uncomment the friend op== then the code compiles successfully
but, as is, whilst the conversion operator holder::operator T&() is
made use of in the selection of the appropriate op== for the call
out.equals(c_str) [c_str = const char *], it is *not* for the later
call out2.equals(s) [s = std::string]. If my understanding is
correct, this failure is because op== for std::string is a template
function parameterized on CharT (e.g.), and conversion operators are
not considered in instantiating template functions (I think).

This is correct.

Having added the op== to account for this, however, I'm wondering now
about the options for declaring/defining it. Here, I have defined it
inline as a friend of holder, but this looks a little odd to me. I
have done it this way, perhaps in ignorance of all the possibilities.

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.

I don't see the need for the symmetry of equality between
holder and T, because holder is only a private member
class of outer. Given this, you could define

       struct holder {
          bool operator==(const T& t) const {
            return t == t_;

as a member operator== overload of holder, but you need to
exchange the arguments in your equals function (I added
a const qualification):

bool equals(T t) const {
  return h_ == t;

If you would really like to have a symmetric function,
I don't see the reason why holder should not simply
provide a static (non-operator) member function:

       bool equals(T t) const {
         return holder::equals(t, h_);

       struct holder {
          static bool equals(const T& t, const holder& h) {
            return t == h.t_;

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.

There are several possible solutions to your problem.

HTH & Greetings from Bremen,

Daniel Kr?gler

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