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.)
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*);
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