Is it possible to explicitly specialize template to match lambda?

Question

Suppose I have a header wrapper.h:

template <typename Func> void wrapper(const Func func);

and a file wrapper.cpp containing:

#include "wrapper.h"
template <typename Func>
void wrapper(const Func  func)
{
  func();
}

And a file main.cpp containing:

#include "wrapper.h"
#include <iostream>

int main()
{
  wrapper( [](){std::cout<<"hello."<<std::endl;} );
}

If I compile these together (e.g., cat wrapper.cpp main.cpp | g++ -std=c++11 -o main -x c++ -), I get no linker errors.

But if I compile them separately (e.g., g++ -std=c++11 -o wrapper.o -c wrapper.cpp && g++ -std=c++11 -o main main.cpp wrapper.o), I --- of course --- get a linker error:

Undefined symbols for architecture x86_64:
  "void wrapper<main::$_0>(main::$_0)", referenced from:
      _main in main-5f3a90.o

Normally, I could explicitly specialize wrapper and add something like this to wrapper.cpp:

template void wrapper<void(*)()>(void(*)())

But this particular template specialization doesn't work.

Is it possible to specialize a template on a lambda?

Answer 1

First, I assume you know about Why can templates only be implemented in the header file?

To your question:

Is it possible to specialize a template on a lambda?

Unfortunately No, template specializations work with exact match, and a lambda is a unique unnamed type. The problem is specializing for that type which you do not know.

Your best bet is to use std::function; or as you have done, then additionally force the lambda to be converted into a function pointer by adding +

int main()
{
  wrapper(+[](){std::cout<<"hello."<<std::endl;} );
}

Full example:

#include <iostream>

template <typename Func>
void wrapper(const Func  func)
{
    std::cout << "PRIMARY\n";
    func();
}

template <>
void wrapper<void(*)()>(void(*func)())
{
    std::cout << "SPECIALIZATION\n";
    func();
}

int main()
{
     wrapper([](){std::cout<<"hello\n"<<std::endl;} );
     wrapper(+[](){std::cout<<"world."<<std::endl;} );
}

This will print

PRIMARY
hello

SPECIALIZATION
world

---

Also, decltype facility wouldn't help, if it does, it will take away the flexibility of your need for lambda