How to print function pointers with cout?

Question

I want to print out a function pointer using cout, and found it did not work. But it worked after I converting the function pointer to (void *), so does printf with %p, such

Answer 1

Regarding your edit, you can print out contents of anything by accessing it via unsigned char pointer. An example for pointers to member functions:

#include 
#include 

struct foo { virtual void bar(){} };
struct foo2 { };
struct foo3 : foo2, foo { virtual void bar(){} };

int main()
{
    void (foo3::*p)() = &foo::bar;

    unsigned char const * first = reinterpret_cast(&p);
    unsigned char const * last = reinterpret_cast(&p + 1);

    for (; first != last; ++first)
    {
        std::cout << std::hex << std::setw(2) << std::setfill('0')
            << (int)*first << ' ';
    }
    std::cout << std::endl;
}