C++ – Why can’t templates be within extern “C” blocks

cexternlinkagetemplates

This is a follow-up question to an answer to Is it possible to typedef a pointer-to-extern-“C”-function type within a template?

This code fails to compile with g++, Visual C/C++, and Comeau C/C++ with basically the same error message:

#include <cstdlib>

extern "C" {
    static int do_stuff(int) {
        return 3;
    }

    template <typename return_t_, typename arg1_t_>
    struct test {
        static void foo(return_t_ (*)(arg1_t_)) { }
    };
}

int main()
{
    test<int, int>::foo(&do_stuff);
    return EXIT_SUCCESS;
}

g++ says "error: template with C linkage", Visual C/C++ emits compiler error C2894, and Comeau C/C++ says "error: this declaration may not have extern "C" linkage".

The thing is, all are happy with:

#include <cstdlib>

extern "C" {
    static int do_stuff(int) {
        return 3;
    }

    struct test {
        static void foo(int (*)(int)) { }
    };
}

int main()
{
    test::foo(&do_stuff);
    return EXIT_SUCCESS;
}

Section 7.5, Linkage specifications, of the C++ Standard states:

A C language linkage is ignored for the names of class members and the member function
type of class member functions.

And it even gives the example:

extern "C" {
    class X {
        void mf(); // the name of the function mf and the member
                // function's type have C++ language linkage
        void mf2(void(*)()); // the name of the function mf2 has C++ language
                // linkage; the parameter has type pointer to C function
    };
}

If templates were allowed in extern "C" blocks, then the member functions of the instantiations would have C++ linkage.

Why, then, does chapter 14, Templates, of the C++98 Standard state:

A template name may have linkage (3.5). A template, a template explicit specialization (14.7.3), and a class template partial specialization shall not have C linkage.

What does it mean that a template "may" have linkage? What is template linkage?

Why is it explicitly forbidden to have a template with C linkage, when a class is okay, and all member functions of instantiations of the template (the default constructor, destructor, and assignment operator overload) would have C++ linkage?

Best Answer

Templates aren't actual code, they're just guidelines to the compiler for how to generate the code once the template parameters are known. As such they don't actually exist until you try to use them. You can't provide linkage to something that doesn't exist.

Related Solutions

C++ – Why can’t variables be declared in a switch statement

Case statements are only labels. This means the compiler will interpret this as a jump directly to the label. In C++, the problem here is one of scope. Your curly brackets define the scope as everything inside the switch statement. This means that you are left with a scope where a jump will be performed further into the code skipping the initialization.

The correct way to handle this is to define a scope specific to that case statement and define your variable within it:

switch (val)
{   
case VAL:  
{
  // This will work
  int newVal = 42;  
  break;
}
case ANOTHER_VAL:  
...
break;
}

C++ – Why can templates only be implemented in the header file

Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer.

Anyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. For example:

template<typename T>
struct Foo
{
    T bar;
    void doSomething(T param) {/* do stuff using T */}
};

// somewhere in a .cpp
Foo<int> f;

When reading this line, the compiler will create a new class (let's call it FooInt), which is equivalent to the following:

struct FooInt
{
    int bar;
    void doSomething(int param) {/* do stuff using int */}
}

Consequently, the compiler needs to have access to the implementation of the methods, to instantiate them with the template argument (in this case int). If these implementations were not in the header, they wouldn't be accessible, and therefore the compiler wouldn't be able to instantiate the template.

A common solution to this is to write the template declaration in a header file, then implement the class in an implementation file (for example .tpp), and include this implementation file at the end of the header.

Foo.h

template <typename T>
struct Foo
{
    void doSomething(T param);
};

#include "Foo.tpp"

Foo.tpp

template <typename T>
void Foo<T>::doSomething(T param)
{
    //implementation
}

This way, implementation is still separated from declaration, but is accessible to the compiler.

Alternative solution

Another solution is to keep the implementation separated, and explicitly instantiate all the template instances you'll need:

Foo.h

// no implementation
template <typename T> struct Foo { ... };

Foo.cpp

// implementation of Foo's methods

// explicit instantiations
template class Foo<int>;
template class Foo<float>;
// You will only be able to use Foo with int or float

If my explanation isn't clear enough, you can have a look at the C++ Super-FAQ on this subject.

Best Answer

Related Solutions

C++ – Why can’t variables be declared in a switch statement

C++ – Why can templates only be implemented in the header file

Alternative solution

Related Topic