Templates, Function Pointers and C++0x

问题

One of my personal experiments to understand some of the C++0x features: I'm trying to pass a function pointer to a template function to execute. Eventually the execution is supposed to happen in a different thread. But with all the different types of functions, I can't get the templates to work.

#include <functional>

int foo(void) {return 2;}

class bar {
public:
    int operator() (void) {return 4;};
    int something(int a) {return a;};
};

template <class C>
int func(C&& c)
{
    //typedef typename std::result_of< C() >::type result_type;
    typedef typename std::conditional< 
        std::is_pointer< C >::value,
        std::result_of< C() >::type,
        std::conditional<
            std::is_object< C >::value,
            std::result_of< typename C::operator() >::type,
            void>
        >::type result_type;
    result_type result = c();
    return result;
}

int main(int argc, char* argv[])
{
    // call with a function pointer
    func(foo);

    // call with a member function
    bar b;
    func(b);

    // call with a bind expression
    func(std::bind(&bar::something, b, 42));

    // call with a lambda expression
    func( [](void)->int {return 12;} );

    return 0;
}

The result_of template alone doesn't seem to be able to find the operator() in class bar and the clunky conditional I created doesn't compile. Any ideas? Will I have additional problems with const functions?

回答1:

How about using decltype?

template <class C>
auto func(C&& c) -> decltype(c()) {
    auto result = c();
    return result;
}

回答2:

If i understand the C++0x draft right the following should actually be sufficient:

typedef typename std::result_of<C()>::type result_type;

Using that instead of your conditional expression, it compiles fine on gcc4.5 - maybe you have found a bug in whatever compiler you are using?

回答3:

I got your template to instantiate, but GCC complains about possibly every use of result_of.

template <class C>
int func(C&& c)
{
    //typedef typename std::result_of< C() >::type result_type;
    typedef typename std::conditional<
        std::is_pointer< C >::value,
         // C++0x still requires "typename" sprinkles:
        typename std::result_of< C() >::type,
        typename std::conditional<
            std::is_object< C >::value,
             // result_of takes a *type* as an argument, not an object:
            //typename std::result_of< decltype( &C::operator() ) >::type,
             // Or better:
            typename std::result_of< C >::type,
            void>
        >::type result_type;
    result_type result = c();
    return result;
}

int main(int argc, char* argv[])
{
    // according to GCC, func(foo) passes a function reference.
    func(foo);

First error message:

rof.cpp:23:17: error: invalid use of incomplete type 'struct std::result_of<int (&)()>'

result_of is implemented as specified in the standard, so it appears that GCC cannot match the pseudo-prototype syntax in the partial specialization declaration.

来源：https://stackoverflow.com/questions/2735294/templates-function-pointers-and-c0x

标签

c++

templates

c++11

function-pointers