函数签名式表达式作为C ++模板参数

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.
};