How to call a templated function if it exists, and

问题:

I want to do something like

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

I thought that something using enable_if would do the job here, splitting up foo into two pieces, but I can't seem to work out the details. What's the simplest way of achieving this?

回答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);
}

回答2:

litb has given you a very good answer. However, I wonder whether, given more context, we couldn't come up with something that's less generic, but also less, um, elaborate?

For example, what types can be T? Anything? A few types? A very restricted set which you have control over? Some classes you design in conjunction with the function foo? Given the latter, you could simple put something like

typedef boolean<true> has_bar_func;

into the types and then switch to different foo overloads based on that:

template <typename T>
void foo_impl(const T& t, boolean<true> /*has_bar_func*/);
template <typename T>
void foo_impl(const T& t, boolean<false> /*has_bar_func*/);

template <typename T>
void foo(const T& t) {
  foo_impl( t, typename T::has_bar_func() );
}

Also, can the bar/baz function have just about any signature, is there a somewhat restricted set, or is there just one valid signature? If the latter, litb's (excellent) fallback idea, in conjunction with a meta-function employing sizeof might be a bit simpler. But this I haven't explored, so it's just a thought.

回答3:

I think litb's solution works, but is overly complex. The reason is that he's introducing a function fallback::bar(...) which acts as a "function of last resort", and then goes to great lengths NOT to call it. Why? It seems we have a perfect behavior for it:

namespace fallback {
    template<typename T>
    inline void bar(T const& t, ...)
    {
        baz(t);
    }
}
template<typename T>
void foo(T const& t)
{
    using namespace fallback;
    bar(t);
}

But as I indicated in a comment to litb's original post, there are many reasons why bar(t) could fail to compile, and I'm not certain this solution handles the same cases. It certainly will fail on a private bar::bar(T t)

回答4:

If you're willing to limit yourself to Visual C++, you can use the __if_exists and __if_not_exists statements.

Handy in a pinch, but platform specific.

回答5:

EDIT: I spoke too soon! litb's answer shows how this can actually be done (at the possible cost of your sanity... :-P)

Unfortunately I think the general case of checking "would this compile" is out of reach of function template argument deduction + SFINAE, which is the usual trick for this stuff. I think the best you can do is to create a "backup" function template:

template <typename T>
void bar(T t) {   // "Backup" bar() template
    baz(t);
}

And then change foo() to simply:

template <typename T>
void foo(const T& t) {
    bar(t);
}

This will work for most cases. Because the bar() template's parameter type is T, it will be deemed "less specialised" when compared with any other function or function template named bar() and will therefore cede priority to that pre-existing function or function template during overload resolution. Except that:

• If the pre-existing bar() is itself a function template taking a template parameter of type T, an ambiguity will arise because neither template is more specialised than the other, and the compiler will complain.
• Implicit conversions also won't work, and will lead to hard-to-diagnose problems: Suppose there is a pre-existing bar(long) but foo(123) is called. In this case, the compiler will quietly choose to instantiate the "backup" bar() template with T = int instead of performing the int->long promotion, even though the latter would have compiled and worked fine!

In short: there's no easy, complete solution, and I'm pretty sure there's not even a tricky-as-hell, complete solution. :(

回答6:

//default

////////////////////////////////////////// 
    template <class T>
    void foo(const T& t){
        baz(t);
    }

//specializations
//////////////////////////////////////////  

    template <>
    void foo(const specialization_1& t){
        bar(t);
    }
    ....
    template <>
    void foo(const specialization_n& t){
        bar(t);
    }

回答7:

Are you not able to use full specialisation here (or overloading) on foo. By say having the function template call bar but for certain types fully specialise it to call baz?