Using type of function pointer as template argument

Question

I'm trying to write a template which gets the type of a functionpointer as templateargument and the corresponding functionpointer as a function argument, so as a simplyfied example I'm doing this right now:

int myfunc(int a)
{ return a; }

template<typename T, typename Func> struct Test
{
    typedef typeof(myfunc) Fun;
    static T MyFunc(T val, Func f)
    {
        return f(val);
    }
};

int main(void)
{
    std::cout<<Test<int, typeof(myfunc)>::MyFunc(5, myfunc)<<std::endl;
}

However this code instantly crashes. If I change the type of f to Fun it works perfectly. So what am I doing wrong and how can I get this to work?

I'm using mingw under windows vista if that behaviour in case that behaviour is compiler dependent.

typeof is not part of C++, so that doesn't even compile on my compiler — Andreas Bonini
– Andreas Bonini, Commented Dec 19, 2009 at 20:42
typeof is a GCC extension. I assume it's not available on MSVC. — ZoogieZork
– ZoogieZork, Commented Dec 19, 2009 at 20:54
decltype isn't in the current version of C++, but is due in the next version. gcc generats the expected errors so long as it is in C++ mode (e.g. -std=c++98). — Lara Bailey
– Lara Bailey, Commented Dec 19, 2009 at 21:45

xtofl · Accepted Answer · 2009-12-19 20:49:06Z

2

You don't really need the Test class for this scenario, and Why use the typeof function here?

template< typename T, typename fT > 
T TestMyFunc( T t, fT f ) {
   return f(t);
};

will do, and no fiddling with function pointer types:

std::cout << TestMyFunc(5,myfunc) << std::endl;

The template arguments are deduced automatically for functions!

answered Dec 19, 2009 at 20:49

xtofl

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Comments

Lara Bailey · Accepted Answer · 2009-12-19 20:45:28Z

2

typeof isn't valid C++, but if I replace your two lines involving them with these, then your code works fine with either Fun or Func in the definition of MyFunc.

typedef int (*Fun)(int);

and

std::cout<<Test<int, int(*)(int)>::MyFunc(5, myfunc)<<std::endl;

answered Dec 19, 2009 at 20:45

Lara Bailey

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