Adapting checking for member functions existence to function arguments

Question

I've been trying to adapt this solution for enabling the existence of ordinary (non-member) functions. In my case, I have a lot of global string-utility-type functions that take any string type T such that T has has, say, a char const* c_str() const member function.

The goal is to eliminate weird compiler error messages if the user attempt to pass some type T that has no c_str() member function, i.e., rather than the compiler saying, "c_str(): no such member function", I'd rather the compiler say, "foo(T&): no such function" where foo is the global function.

Here is the adapted code:

template<bool B,typename T = void>
struct enable_if {
  typedef T type;
};

template<typename T>
struct enable_if<false,T> {
};

//
// This macro is adapted from the linked-to question -- this part works fine.
//
#define DECL_HAS_MEM_FN(FN_NAME)                                            \
  template<class ClassType,typename MemFnSig>                               \
  struct has_##FN_NAME##_type {                                             \
    typedef char yes[1];                                                    \
    typedef char no[2];                                                     \
    template<typename T,T> struct type_check;                               \
    template<class T> static yes& has( type_check<MemFnSig,&T::FN_NAME>* ); \
    template<class T> static no&  has( ... );                               \
    enum { value = sizeof( has<ClassType>(0) ) == sizeof( yes ) };          \
  }

// Declare an instance of the above type that checks for "c_str()".
DECL_HAS_MEM_FN(c_str);

// Define another macro just to make life easier.
#define has_c_str(STRINGTYPE) \
  has_c_str_type<STRINGTYPE,char const* (STRINGTYPE::*)() const>::value

//
// A "ValidatedStringType" is a StringType that uses the above machinery to ensure that
// StringType has a c_str() member function
//
template<class StringType>
struct ValidatedStringType {
  typedef typename enable_if<has_c_str(StringType),StringType>::type type;
};

// Here's the global function where I want to accept only validated StringTypes.
template<class StringType>
void f( typename ValidatedStringType<StringType>::type const& s ) {
}

struct S { // Class for testing that has c_str().
  char const* c_str() const {
    return 0;
  }
};

struct N { // Class for testing that does not have c_str().
};

using namespace std;

int main() {
  S s;
  N n;
  cout << has_c_str(S) << endl; // correctly prints '1'
  cout << has_c_str(N) << endl; // correctly prints '0'
  f( s ); // error: no matching function for call to 'f(S&)'
}

However, as shown above, the compiler doesn't "see" the f(S&) -- why not?

FYI: The fact that f() takes an S const& rather than just an S makes no difference. Also making 's' const makes no difference. Lastly, eliminating the '&' makes no difference. — Paul J. Lucas, Jan 23 '11 at 04:35

score 1 · Accepted Answer · answered Jan 23 '11 at 12:51

1

If I understand the question correctly, applying enable_if to f itself like the following will solve the problem:

template<class StringType>
typename enable_if<has_c_str(StringType),StringType>::type
f( StringType const& s ) {....}

Hope this helps.

answered Jan 23 '11 at 12:51

Ise Wisteria

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But f's return type is supposed to be void. If the ",StringType" is omitted, it's correct and works. :) But I'd still like to know why the original doesn't work. – Paul J. Lucas Jan 23 '11 at 15:44
@PaulJ.Lucas: The parameter `ValidatedStringType::type` is a kind of nested name specifier, and its type isn't deduced while overload resolution. This is called non-deduced context (14.8.2.4 -4- in the standard). – Ise Wisteria Jan 23 '11 at 19:12

Daniel Gallagher · Answer 2 · 2011-01-23T04:29:27.260

I can't answer your question ("why not?") but I can provide a work-around. But man, it's ugly. You end up defining a structure and a function for every function:

template<class StringType, class T = typename ValidatedStringType<StringType>::type>
struct f_helper {
  void operator()( T const& s ) {
      // Put the contents of f() here
  }
};

template<class StringType>
void f(StringType const &s) {
  f_helper<StringType>()( s );
}

I'm sure there's some preprocessing magic you could write to eliminate some of the boilerplate, but it would also be quite ugly:

#define DECL_VALIDATED_FUNC( RetType, name, Validator, contents ) \
  template<class StringType, class T = typename Validator<StringType>::type> \
  struct name ## _helper { \
    RetType operator()( T const& s ) contents \
  }; \
  \
  template<class StringType> \
  RetType name(StringType const& s) { \
     name ## _helper<StringType>()( s ); \
}

DECL_VALIDATED_FUNC( void, f, ValidatedStringType, {
   // put contents of f() here
}) // note the closing parenthesis

Unfortunately, you can't specify default template arguments to free functions, or else you could have done:

template<class StringType, class T = typename ValidatedStringType<StringType>::type>
void f( T const& s ) {}

Yeah, it's a bit too ugly and too much work just to make error messages better. — Paul J. Lucas, Jan 23 '11 at 04:32

Adapting checking for member functions existence to function arguments

2 Answers2