问题
Consider the following header file:
// Foo.h
class Foo {
public:
template <typename T>
void read(T& value);
};
I want to explicitly instantiate the Foo::read
member function template in a source file for all types included in a boost::mpl::vector
:
// Foo.cc
#include <boost/mpl/vector.hpp>
#include <boost/mpl/begin_end.hpp>
#include "Foo.h"
template <typename T>
void Foo::read(T& value) { /* do something */ }
typedef boost::mpl::vector<int, long, float> types;
// template Foo::read<int >(int&);
// template Foo::read<long >(long&);
// template Foo::read<float>(float&);
// instantiate automatically ???
Is it possible? Thanks in advance, Daniel.
EDIT
I found some solution - it seems that assigning a pointer to Foo::read<T>
in the constructor of a struct, of which variable is then declared, cause instantiation:
// intermezzo
template <typename T> struct Bar {
Bar<T>() {
void (Foo::*funPtr)(T&) = &Foo::read<T>;
}
};
static Bar<int > bar1;
static Bar<long > bar2;
static Bar<float> bar3;
So then the process can be automatized as follows:
// Foo.cc continued
template <typename B, typename E>
struct my_for_each {
my_for_each<B, E>() {
typedef typename B::type T; // vector member
typedef void (Foo::*FunPtr)(T&); // pointer to Foo member function
FunPtr funPtr = &Foo::read<T>; // cause instantiation?
}
my_for_each<typename boost::mpl::next<B>::type, E> next;
};
template<typename E>
struct my_for_each<E, E> {};
static my_for_each< boost::mpl::begin<types>::type,
boost::mpl::end<types>::type > first;
But I don't know if this solution is portable and standard-conformant? (Works with Intel and GNU compilers.)
回答1:
I've had the same requirement not long ago, and had good results with a simple function template instantiation:
template <class... T>
void forceInstantiation(typedef boost::mpl::vector<T...>*) {
using ex = int[];
(void)ex{(void(&Foo::read<T>), 0)..., 0};
// C++17
// (void)((void(&Foo::read<T>), ...));
}
template void forceInstantiation(types*);
回答2:
I am not sure if this is the solution to your problem, but maybe you can do with a template specialization.
New header:
// Foo.h
template < typename T >
struct RealRead;
class Foo {
public:
template <typename T>
void read(T& value);
};
template <typename T>
void Foo::read(T& value)
{
RealRead< T >::read( value );
}
New source :
template < >
struct RealRead< int >
{
static void read( int & v )
{
// do read
}
};
template < >
struct RealRead< float >
{
static void read( float & v )
{
// do read
}
};
//etc
// explicitly instantiate templates
template struct RealRead< int >;
template struct RealRead< float >;
回答3:
You can explicitly instantiate Foo for a given T template parameter with template class Foo<T>;
As for batch instantiation, I don't think it is possible. Maybe with variadic templates it is possible to create an Instantiate class so something like Instantiate<Foo, int, short, long, float, etc>
would instantiate the appropriate templates, but other than that, you have to resort to manual instantiation.
回答4:
explicit instantiation has special grammar and special meaning to complier, so cannot be done with meta programming.
your solution cause a instantiation, but not a explicit instantiation.
回答5:
I don't think it is necessary, nor is it possible.
You can directly use (call) the function Foo:read(bar), for variable bar of any type, as long as the type is well-defined in your template function implementation. The compiler will automatically morph your argument into type "T".
For example:
template <class T>
Foo::read(T & var)
{
std::cin >> var;
}
T is well-defined when T is a streaming type supported by cin.
The example will be self-contained, if "Foo::" is removed. I mean, for "Foo::", you should have somewhere defined a class Foo, or a namespace Foo, to make it work.
Yet please note that template should always go inside a .h file, not a .cpp file (just search the web with keyword "c++ template can not be implemented in cpp file"
回答6:
If you intend to use your class only in a single module (i.e. you won't export it) you can use boost/mpl/for_each. The template function defined this way (using mpl/for_each) won't be exported (even if you declare __declspec(export) before class name or function signature):
// Foo.cpp
#include <boost/mpl/vector.hpp>
#include <boost/mpl/for_each.hpp>
template<class T>
void read(T& value)
{
...
}
using types = boost::mpl::vector<long, int>;
//template instantiation
struct call_read {
template <class T>
void operator()(T)
{
T t; //You should make sure that T can be created this way
((Foo*)nullptr)->read<T>(t); //this line tells to compiler with templates it should instantiate
}
};
void instantiate()
{
boost::mpl::for_each<types>(call_read());
}
If you need export/import you structure and template methods there is the solution using boost/preprocessor
// Foo.h
#ifdef <preprocessor definition specific to DLL>
# define API __declspec(dllexport)
#else
# define API __declspec(dllimport)
#endif
class API Foo {
public:
template<class T> void read(T& value);
};
// Foo.cpp
#include <boost/preprocessor/seq/for_each.hpp>
#include <boost/preprocessor/seq/enum.hpp>
#include <boost/mpl/vector.hpp>
template<class T>
void read(T& value)
{
...
}
//using this macro you can define both boost::mpl structure AND instantiate explicitly your template function
#define VARIANT_LIST (std::wstring)(long)(int)
using types = boost::mpl::vector<BOOST_PP_SEQ_ENUM(VARIANT_LIST)>;
//Here we should use our API macro
#define EXPLICIT_INSTANTIATION(r, d, __type__) \
template API void Foo::read<__type__>(__type__&);
BOOST_PP_SEQ_FOR_EACH(EXPLICIT_INSTANTIATION, _, VARIANT_LIST)
If you don't need this extra functionality the first solution is much cleaner I guess
来源:https://stackoverflow.com/questions/5715586/how-to-explicitly-instantiate-a-template-for-all-members-of-mpl-vector-in-c