问题
This question is related to this one except that rather than dealing with typename template parameters, I am trying to use an enum non-type template parameter.
Is it possible to have a templated (class member function) with only specializations, no general (working) definition in the case of non-type template parameter?
I was able to get one version working, by declaration in the class body and providing specializations only, but any misuse calling with a non-defined template parameter doesn't produce an error until linking. What's worse is the missing symbol cryptically refers to the enum's integral value and not its name, so it would be confusing to other developers.
I was able to get the
BOOST_STATIC_ASSERTtechnique from the referenced question to work for typename template parameter only.
This code demonstrates the idea. I don't want the CAT-version call to compile:
#include <iostream>
#include <boost/static_assert.hpp>
// CLASS HEADER FILE:
struct foo_class
{
enum AllowedTypes { DOG, CAT };
template <AllowedTypes type>
void add_one_third( double bar ) const
{
BOOST_STATIC_ASSERT_MSG(sizeof(type)==0, "enum type not supported.");
}
};
// CLASS SOURCE FILE
template<>
void foo_class::add_one_third<foo_class::DOG>( double bar ) const
{
std::cout << "DOG specialization: " << bar + 1./3. << std::endl;
}
// USER SOURCE FILE
int main()
{
std::cout << "Template Specialization!\n\n";
foo_class a;
a.add_one_third<foo_class::DOG>(3.0); // should succeed
// Compilation fails with or without the following line:
a.add_one_third<foo_class::CAT>(3.0); // should fail at compile-time
return 0;
}
Background: I have a class member function that takes an enum "ArgType" and a name.
void declareKernelArgument( ArgType type, std::string name );
The definition has turned into an if..else..if..else list for the half-dozen or so allowed ArgType cases. I also have to have final case that throws an exception for an not-allowed ArgType. I'm thinking it would be cleaner to move ArgType to a template parameter, and provide a specialization for each allowed ArgType. Misuse would be caught at compile-time.
回答1:
With partial specialization of a structure inside the class:
#include <iostream>
class foo_class
{
public:
enum AllowedTypes { T_DOUBLE, T_INT };
private:
template <AllowedTypes type, typename T>
struct AddOneThird;
template <typename T>
struct AddOneThird<T_DOUBLE, T> {
static void apply(T bar) {
std::cout << "T_DOUBLE specialization: " << bar + 1.0/3.0 << std::endl;
}
};
public:
template <AllowedTypes type>
void add_one_third( double bar ) const {
AddOneThird<type, double>::apply(bar);
}
};
int main() {
foo_class a;
a.add_one_third<foo_class::T_DOUBLE>(3.0);
// error: incomplete type ‘foo_class::AddOneThird<(foo_class::AllowedTypes)1u
// a.add_one_third<foo_class::T_INT>(3.0); // should fail at compile-time
return 0;
}
With full specialization of a (friend) class:
#include <iostream>
class foo_class
{
public:
enum AllowedTypes { T_DOUBLE, T_INT };
// if needed
// template<AllowedTypes> friend struct AddOneThird;
public:
template <AllowedTypes type> void add_one_third( double bar ) const;
};
template <foo_class::AllowedTypes>
struct AddOneThird;
template <>
struct AddOneThird<foo_class::T_DOUBLE> {
static void apply(double bar) {
std::cout << "T_DOUBLE specialization: " << bar + 1.0/3.0 << std::endl;
}
};
template <foo_class::AllowedTypes type>
void foo_class::add_one_third( double bar) const {
AddOneThird<type>::apply(bar);
}
int main() {
foo_class a;
a.add_one_third<foo_class::T_DOUBLE>(3.0);
// error: incomplete type ‘AddOneThird<(foo_class::AllowedTypes)1u>’ used
// in nested name specifier
//a.add_one_third<foo_class::T_INT>(3.0); // should fail at compile-time
return 0;
}
Utilizing C++11 or boost::enable_if:
#include <iostream>
#include <type_traits>
class foo_class
{
public:
enum AllowedTypes { T_DOUBLE, T_INT };
template <AllowedTypes type>
typename std::enable_if<type == T_DOUBLE>::type
add_one_third( double bar ) const {
std::cout << "T_DOUBLE specialization: " << bar + 1.0/3.0 << std::endl;
}
};
int main() {
foo_class a;
a.add_one_third<foo_class::T_DOUBLE>(3.0);
// error: no matching function for call to ‘foo_class::add_one_third(double)’
//a.add_one_third<foo_class::T_INT>(3.0); // should fail at compile-time
return 0;
}
回答2:
From Herb Sutter
It's a lot less intuitive to specialize function templates. For one thing, you can't partially specialize them -- pretty much just because the language says you can't.[2] For another thing, function template specializations don't overload. This means that any specializations you write will not affect which template gets used, which runs counter to what most people would intuitively expect. After all, if you had written a nontemplate function with the identical signature instead of a function template specialization, the nontemplate function would always be selected because it's always considered to be a better match than a template.
If you're writing a function template, prefer to write it as a single function template that should never be specialized or overloaded, and implement the function template entirely in terms of a class template. This is the proverbial level of indirection that steers you well clear of the limitations and dark corners of function templates. This way, programmers using your template will be able to partially specialize and explicitly specialize the class template to their heart's content without affecting the expected operation of the function template. This avoids both the limitation that function templates can't be partially specialized, and the sometimes surprising effect that function template specializations don't overload. Problem solved.
Your enum type sizeof is not 0, change that to 4 at least. Otherwise this will not work. A enum element size is not 0.
Without that everything runs
#include <iostream>
struct foo_class
{
enum AllowedTypes { DOG, CAT };
template <AllowedTypes type>
void add_one_third( double bar ) const
{
std::cout << "YES" << std::endl;
}
};
template<>
void foo_class::add_one_third<foo_class::DOG>( double bar ) const
{
std::cout << "DOG specialization: " << bar + 1./3. << std::endl;
}
int main()
{
std::cout << "Template Specialization!\n\n";
foo_class a;
a.add_one_third<foo_class::DOG>(3.0); // should succeed
// Compilation fails with or without the following line:
//a.add_one_third<foo_class::CAT>(3.0); // should fail at compile-time
return 0;
}
回答3:
The main difference between the enum case and the referenced question using a typename parameter is that the default definition will be compiled for any use. So, a working solution is as simple as modifying the BOOST_STATIC_ASSERT condition to check allowed enum values.
#include <iostream>
#include <stdexcept>
#include <boost/static_assert.hpp>
// CLASS HEADER FILE:
struct foo_class
{
enum AllowedTypes { DOG, CAT, MOUSE };
template <AllowedTypes type>
void give_bath() const
{
// compile fails if ever attempting to use this function with CAT parameter.
BOOST_STATIC_ASSERT_MSG( (type==DOG) || (type==MOUSE) , "enum type not supported.");
throw std::runtime_error("Unexpected. Above list inconsistent with specializations.");
}
};
// CLASS SOURCE FILE
template<>
void foo_class::give_bath<foo_class::DOG>() const
{
std::cout << "DOG is bathed." << std::endl;
}
template<>
void foo_class::give_bath<foo_class::MOUSE>() const
{
std::cout << "MOUSE is bathed." << std::endl;
}
// USER SOURCE FILE
int main()
{
std::cout << "Template Specialization!\n\n";
foo_class a;
a.give_bath<foo_class::DOG>(); //success
a.give_bath<foo_class::MOUSE>(); // success
// Compilation fails with the following line:
//a.give_bath<foo_class::CAT>(); // fails at compile-time as intended.
return 0;
}
Of course, the whole design smells bad and could likely be handled more elegantly with AllowedTypes being a struct/class with inherited specializations. But this gets to the question at hand.
来源：https://stackoverflow.com/questions/22721201/specializations-only-for-c-template-function-with-enum-non-type-template-param