enable_if + disable_if combination provokes an ambiguous call

问题

While trying to answer this question I wanted to suggest the use of enable_if + disable_if to allow the overload of a method based on the fact that a type was (or not) polymorphic.

So I created a small test file:

template <class T>
void* address_of(T* p,
                 boost::enable_if< boost::is_polymorphic<T> >* dummy = 0)
{ return dynamic_cast<void*>(p); }

template <class T>
void* address_of(T* p,
                 boost::disable_if< boost::is_polymorphic<T> >* dummy = 0)
{ return static_cast<void*>(p); }

struct N { int x; };


int main(int argc, char* argv[])
{
  N n;
  std::cout << address_of(&n) << std::endl;
  return 0;
}

which seems quite tame.

However gcc (3.4 ...) choke on this:

test.cpp: In function int main(int, char**):
test.cpp:29: error: call of overloaded address_of(N*) is ambiguous
test.cpp:17: note: candidates are: void* address_of(T*, boost::enable_if<boost::is_polymorphic<T>, void>*) [with T = N]
test.cpp:20: note: void* address_of(T*, boost::disable_if<boost::is_polymorphic<T>, void>*) [with T = N]

It seems rather clear to my human mind which overload should be used here. I mean it seems clear that I have defined an alternative and only one function can be used at a time... and I would have thought that SFINAE would take care of invalidating the unnecessary overload.

I patched it up by using ... (ellipsis) instead of disable_if and requiring a dummy second argument... but I am still interested in why the compiler choke on this.

回答1:

The compiler choked because you forgot the trailing ::type on enable_if and disable_if. The templates are always defined; it is just that the member type is present if and only if the expression is true (for enable_if) or false (for disable_if).

template <class T>
void* address_of(T* p,
                 typename boost::enable_if< boost::is_polymorphic<T> >::type* dummy = 0)
{ return dynamic_cast<void*>(p); }

template <class T>
void* address_of(T* p,
                 typename boost::disable_if< boost::is_polymorphic<T> >::type* dummy = 0)
{ return static_cast<void*>(p); }

Without the trailing ::type, your function templates just create overloads that take pointers to instances of enable_if or disable_if as the second parameter. With the trailing ::type, the templates either create an overload with a second parameter of type void*, or the overload is removed (i.e. the desired behaviour).

回答2:

Using the "return type" version of enable_if works in 3.4.4: gcc version 3.4.4 (cygming special, gdc 0.12, using dmd 0.125)

#include <boost/utility/enable_if.hpp>
#include <boost/type_traits/is_polymorphic.hpp>
#include <iostream>

template <class T>
typename boost::enable_if< boost::is_polymorphic<T>, void* >::type
address_of(T* p)
{ return dynamic_cast<void*>(p); }

template <class T>
typename boost::disable_if< boost::is_polymorphic<T>, void* >::type
address_of(T* p)
{ return static_cast<void*>(p); }

struct N { int x; };


int main(int argc, char* argv[])
{
  N n;
  std::cout << address_of(&n) << std::endl;
  return 0;
}

来源：https://stackoverflow.com/questions/3491848/enable-if-disable-if-combination-provokes-an-ambiguous-call

标签

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boost

enable-if