Thread: Is typeid is still in use ?

Code:

#include <iostream>
#include <typeinfo>

using namespace std;

class point 
   {
   int  x, y ;
public:
   point(int Ix, int Iy) : x(Ix), y(Iy)
      {
      }
   void show(void)
      {
      cout << "x = " << x << " y = " << y << endl;
      }
};
   

int main(void)
{
point  p1(3,3);
point  p2(99, 3922);

cout << "point type = " << typeid(p1).name() << endl;
cout << "point type = " << (typeid(p2)).name() << endl;
cout << "int type = " << (typeid(int)).name() << endl;
cout << "int type = " << typeid(int).name() << endl;
cout << "unsigned int type = " << typeid(unsigned int).name() << endl;


cout << "Press to continue ...." ;
cin.get();
return EXIT_SUCCESS;
}

The output :
point type = 5point
point type = 5point
int type = i
int type = i
unsigned int type = j
Press to continue ....


why is it 5point and not point ?
why the "unsigned int type = j" ? what is j ?

I am using bloodshed 4.9.9.2

AFAIK, the output of type_info::name() is not covered in the standard. It is down to the implmentation. So each compiler might have a different idea about what to return from that function

Originally Posted by dude543

You mean to say that every compiler will return somthing out of the blue ?

Is only type_info::name() is not covered in the standard.
The rest of class type_info is covered by the standard ?

It is in the standard, but the standard says that how thew function works is implementation specific.

A quick google suugests this

18.5.1 Class type_info
const char* name() const;
Returns: an implementation-defined NTBS."
(NTBS means null-terminated byte string)

If that's correct, then technically, every call could return "Eat at Joe's" and still not fall foul of the standard. Not very helpful huh?