You’re not actually calling pr
in your code, you’re passing the function pointer to cout
. pr
is then being converted to a bool
when being passed to cout
. If you put cout << boolalpha
beforehand you will output true
instead of 1
.
EDIT:
With C++11 you can write the following overload:
template <class RType, class ... ArgTypes>
std::ostream & operator<<(std::ostream & s, RType(*func)(ArgTypes...))
{
return s << "(func_ptr=" << (void*)func << ")(num_args="
<< sizeof...(ArgTypes) << ")";
}
which means the call cout << pr
will print (func_ptr=<address of pr>)(num_args=0)
. The function itself can do whatever you want obviously, this is just to demonstrate that with C++11’s variadic templates, you can match function pointers of arbitrary arity. This still won’t work for overloaded functions and function templates without specifying which overload you want (usually via a cast).