How to define typedef of function pointer which has template arguments

Question

I would like to make typedef for function pointer which has stl container as argument and this container has unknown type. Something like this:

typedef void (* TouchCallBack)(GLRenderer*, const MotionEvent&, std::vector<T >);

it's possible? (especially in c++ 03)

Not in C++03, no, unless Boost or something has a trick up its sleeve. — chris, Feb 13 '13 at 08:02
There's nothing wrong with what you've written but you have to be in a context (function or class template) where `T` is declared. — CB Bailey, Feb 13 '13 at 08:08

score 40 · Accepted Answer · answered Feb 13 '13 at 08:03

I don't know of any C++03 solution exactly like that, and it's not built into the language, but in C++11, this is possible with using aliases:

template<typename T>
using TouchCallBack = void (*)(GLRenderer*, const MotionEvent&, std::vector<T >);

One workaround for C++03 is using a struct:

template<typename T>
struct TouchCallBack {
    typedef void (*type)(GLRenderer*, const MotionEvent&, std::vector<T >);
};

//use like TouchCallBack<int>::type

How to define typedef of function pointer which has template arguments

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