How can I create a variable with the same type as a given function?

Question

I have a C++ function like

int f( const std::string &s, double d );

Now I'd like to create a variable which holds a pointer to f. This variable should have the correct type (int (*)( const std::string &, double ) - but I don't want to write out that type explicitely. I'd like to deduce it from f so that I don't repeat the type signature. Eventually, I'd like to be able to write something along the lines of:

TypeOf<f>::Result x = f;

To achieve this, I tried to do something like this:

// Never implemented, only used to deduce the return type into something which can be typedef'ed
template <typename T> T deduceType( T fn ); 

template <typename T>
struct TypeOf {
    typedef T Result;
};

// ...
TypeOf<deduceType(f)>::Result x = f;

My hope was that maybe the return type of a function (deduceType, in this case) could be used as a template argument but alas - it seems you can't do that.

Does anybody know how to do this? I'm looking for a C++03 solution.

Answer 1

C++0x added decltype which does what you want (if I understood correctly).

Another option might be Boost::Typeof which is intended to provide the same functionality until decltype is supported in all compilers.