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Home/ Questions/Q 208703
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T17:50:58+00:00 2026-05-11T17:50:58+00:00

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

  • 0
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?

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  1. Editorial Team
    Editorial Team
    2026-05-11T17:50:59+00:00Added an answer on May 11, 2026 at 5:50 pm

    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*
    
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