How to call a templated function if it exists, and something else otherwise?

Question

I want to do something like

template 
void foo(const T& t) {
   IF bar(t) would compile
      bar(t);
   ELSE
      baz(t);
}

Answer 1

There are two lookups that are done for the name bar. One is the unqualified lookup at the definition context of foo. The other is argument dependent lookup at each instantiation context (but the result of the lookup at each instantiation context is not allowed to change behavior between two different instantiation contexts).

To get the desired behavior, you could go and define a fallback function in a fallback namespace that returns some unique type

namespace fallback {
  // sizeof > 1
  struct flag { char c[2]; };
  flag bar(...);
}

The bar function will be called if nothing else matches because the ellipsis has worst conversion cost. Now, include that candidates into your function by a using directive of fallback, so that fallback::bar is included as candidate into the call to bar.

Now, to see whether a call to bar resolves to your function, you will call it, and check whether the return type is flag. The return type of an otherwise chosen function could be void, so you have to do some comma operator tricks to get around that.

namespace fallback {
  int operator,(flag, flag);

  // map everything else to void
  template<typename T> 
  void operator,(flag, T const&);

  // sizeof 1
  char operator,(int, flag);
}

If our function was selected then the comma operator invocation will return a reference to int. If not or if the selected function returned void, then the invocation returns void in turn. Then the next invocation with flag as second argument will return a type that has sizeof 1 if our fallback was selected, and a sizeof greater 1 (the built-in comma operator will be used because void is in the mix) if something else was selected.

We compare the sizeof and delegate to a struct.

template<bool>
struct foo_impl;

/* bar available */
template<>
struct foo_impl<true> {
  template<typename T>
  static void foo(T const &t) {
    bar(t);
  }
};

/* bar not available */
template<>
struct foo_impl<false> {
  template<typename T>
  static void foo(T const&) {
    std::cout << "not available, calling baz...";
  }
};

template <typename T>
void foo(const T& t) {
   using namespace fallback;

   foo_impl<sizeof (fallback::flag(), bar(t), fallback::flag()) != 1>
     ::foo(t);
}

This solution is ambiguous if the existing function has an ellipsis too. But that seems to be rather unlikely. Test using the fallback:

struct C { };
int main() {
  // => "not available, calling baz..."
  foo(C());
}

And if a candidate is found using argument dependent lookup

struct C { };
void bar(C) {
  std::cout << "called!";
}
int main() {
  // => "called!"
  foo(C());
}

To test unqualified lookup at definition context, let's define the following function above foo_impl and foo (put the foo_impl template above foo, so they have both the same definition context)

void bar(double d) {
  std::cout << "bar(double) called!";
}

// ... foo template ...

int main() {
  // => "bar(double) called!"
  foo(12);
}