How to print function pointers with cout?

There actually is an overload of the << operator that looks something like:

ostream & operator <<( ostream &, const void * );

which does what you expect - outputs in hex. There can be no such standard library overload for function pointers, because there are infinite number of types of them. So the pointer gets converted to another type, which in this case seems to be a bool - I can't offhand remember the rules for this.

Edit: The C++ Standard specifies:

4.12 Boolean conversions

1 An rvalue of arithmetic, enumeration, pointer, or pointer to member type can be converted to an rvalue of type bool.

This is the only conversion specified for function pointers.

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 <iostream>
#include <iomanip>

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

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

You can think of a function pointer as being the address of the first instruction in that function's machine code. Any pointer can be treated as a bool: 0 is false and everything else is true. As you observed, when cast to void * and given as an argument to the stream insertion operator (<<), the address is printed. (Viewed strictly, casting a pointer-to-function to void * is undefined.)

Without the cast, the story is a little complex. For matching overloaded functions ("overload resolution"), a C++ compiler gathers a set of candidate functions and from these candidates selects the "best viable" one, using implicit conversions if necessary. The wrinkle is the matching rules form a partial order, so multiple best-viable matches cause an ambiguity error.

In order of preference, the standard conversions (and of course there also user-defined and ellipsis conversions, not detailed) are

exact match (i.e., no conversion necessary)
promotion (e.g., int to float)
other conversions

The last category includes boolean conversions, and any pointer type may be converted to bool: 0 (or NULL) is false and everything else is true. The latter shows up as 1 when passed to the stream insertion operator.

To get 0 instead, change your initialization to

pf = 0;

Remember that initializing a pointer with a zero-valued constant expression yields the null pointer.

In C++11 one could modify this behavior by defining a variadic template overload of operator<< (whether that is recommendable or not is another topic):

#include<iostream>
namespace function_display{
template<class Ret, class... Args>
std::ostream& operator <<(std::ostream& os, Ret(*p)(Args...) ){ // star * is optional
    return os << "funptr " << (void*)p;
}
}

// example code:
void fun_void_void(){};
void fun_void_double(double d){};
double fun_double_double(double d){return d;}

int main(){
    using namespace function_display;
    // ampersands & are optional
    std::cout << "1. " << &fun_void_void << std::endl; // prints "1. funptr 0x40cb58"
    std::cout << "2. " << &fun_void_double << std::endl; // prints "2. funptr 0x40cb5e"
    std::cout << "3. " << &fun_double_double << std::endl; // prints "3. funptr 0x40cb69"
}

Casting pointers to (void*) to print them to cout is the right thing (TM) to do in C++ if you want to see their values.

How to print function pointers with cout?

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