typeid() returns extra characters in g++

Question

class foo
{
public:
  void say_type_name()
  {
    std::cout << typeid(this).name() << std::endl;
  }
};

int main()
{
  foo f;;
  f.say_type_name();
}

Above code prints P3foo on my ubuntu machine with g++. I am not getting why it is printing P3foo instead of just foo. If I change the code like

    std::cout << typeid(*this).name() << std::endl;

it prints 3foo.

Any thoughts?

Answer 1

Because it is a pointer to foo. And foo has 3 characters. So it becomes P3foo. The other one has type foo, so it becomes 3foo. Note that the text is implementation dependent, and in this case GCC just gives you the internal, mangled name.

Enter that mangled name into the program c++filt to get the unmangled name:

$ c++filt -t P3foo
foo*

Answer 2

Is there a portable solution to demangle it runtime (e.g. demangle(typeid(anything).name())?)

Answer 3

std::type_info::name() returns an implementation specific name. AFAIK, there is no portable way to get a "nice" name, although GCC has one. Look at abi::__cxa_demangle().

int status;
char *realname = abi::__cxa_demangle(typeid(obj).name(), 0, 0, &status);
std::cout << realname;
free(realname);

Answer 4

Is there a portable solution

workaround would be to make a template hack to return all harcoded type names as char*

which platform dont have #include ?

-- http://rzr.online.fr/q/cxx