How to explicitly instantiate a template for all m

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