Function signature-like expressions as C++ template arguments

Question

I was looking at Don Clugston's FastDelegate mini-library and noticed a weird syntactical trick with the following structure:

TemplateClass< void( int, int ) > Object; 

It almost appears as if a function signature is being used as an argument to a template instance declaration.

This technique (whose presence in FastDelegate is apparently due to one Jody Hagins) was used to simplify the declaration of template instances with a semi-arbitrary number of template parameters.

To wit, it allowed this something like the following:

// A template with one parameter template<typename _T1> struct Object1 {     _T1 m_member1; };  // A template with two parameters template<typename _T1, typename _T2> struct Object2 {     _T1 m_member1;     _T2 m_member2; };  // A forward declaration template<typename _Signature> struct Object;  // Some derived types using "function signature"-style template parameters template<typename _Dummy, typename _T1> struct Object<_Dummy(_T1)> : public Object1<_T1> {};  template<typename _Dummy, typename _T1, typename _T2> struct Object<_Dummy(_T1, _T2)> : public Object2<_T1, _T2> {};  // A. "Vanilla" object declarations Object1<int> IntObjectA; Object2<int, char> IntCharObjectA;  // B. Nifty, but equivalent, object declarations typedef void UnusedType; Object< UnusedType(int) > IntObjectB; Object< UnusedType(int, char) > IntCharObjectB;  // C. Even niftier, and still equivalent, object declarations #define DeclareObject( ... ) Object< UnusedType( __VA_ARGS__ ) > DeclareObject( int ) IntObjectC; DeclareObject( int, char ) IntCharObjectC; 

Despite the real whiff of hackiness, I find this kind of spoofy emulation of variadic template arguments to be pretty mind-blowing.

The real meat of this trick seems to be the fact that I can pass textual constructs like "Type1(Type2, Type3)" as arguments to templates. So here are my questions: How exactly does the compiler interpret this construct? Is it a function signature? Or, is it just a text pattern with parentheses in it? If the former, then does this imply that any arbitrary function signature is a valid type as far as the template processor is concerned?

A follow-up question would be that since the above code sample is valid code, why doesn't the C++ standard just allow you to do something like the following, which does not compile?

template<typename _T1> struct Object {     _T1 m_member1; };  // Note the class identifier is also "Object" template<typename _T1, typename _T2> struct Object {     _T1 m_member1;     _T2 m_member2; };  Object<int> IntObject; Object<int, char> IntCharObject; 

Answer 1

With regards to your first question - about the type int(char, float) - this is a valid C++ type and is the type of a function that takes in a char and a float and returns an int. Note that this is the type of the actual function, not a function pointer, which would be an int (*) (char, float). The actual type of any function is this unusual type. For example, the type of

void DoSomething() {     /* ... */ } 

is void ().

The reason that this doesn't come up much during routine programming is that in most circumstances you can't declare variables of this type. For example, this code is illegal:

void MyFunction() {      void function() = DoSomething; // Error! } 

However, one case where you do actually see function types used is for passing function pointers around:

void MyFunction(void FunctionArgument()) {      /* ... */ } 

It's more common to see this sort of function written to take in a function pointer, but it's perfectly fine to take in the function itself. It gets casted behind-the-scenes.

As for your second question, why it's illegal to have the same template written with different numbers of arguments, I don't know the exactly wording in the spec that prohibits it, but it has something to do with the fact that once you've declared a class template, you can't change the number of arguments to it. However, you can provide a partial specialization over that template that has a different number of arguments, provided of course that the partial specialization only specializes over the original number of arguments. For example:

template <typename T> class Function; template <typename Arg, typename Ret> class Function<Ret (Arg)> {      /* ... */ }; 

Here, Function always takes one parameter. The template specialization takes in two arguments, but the specialization is still only over one type (specifically, Ret (Arg)).

Answer 2

int* int_pointer;    // int_pointer   has type "int*" int& int_reference;  // int_reference has type "int&" int  int_value;      // int_value     has type "int"  void (*function_pointer)(int, int);    // function_pointer has type                                        // "void (*)(int, int)" void (&function_reference)(int, int);  // function_reference has type                                        // "void (&)(int ,int)" void function(int, int);               // function has type                                        // "void(int, int)"  template<> struct Object1<void(int, int)> {     void m_member1(int, int);  // wait, what?? not a value you can initialize. };