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Why doesn't this example work?


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#1 noatom   Members   -  Reputation: 769

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Posted 21 June 2013 - 01:40 AM

#include <iostream>
#include <typeinfo>
using namespace std;

class Shape { public: virtual ~Shape() {}; };
class Circle : public Shape {};
class Square : public Shape {};
class Other {};

int main() {
  Circle c;
  Shape* s = &c; // Upcast: normal and OK
  // More explicit but unnecessary:
  s = static_cast<Shape*>(&c);
  // (Since upcasting is such a safe and common
  // operation, the cast becomes cluttering)
  Circle* cp = 0;
  Square* sp = 0;
  // Static Navigation of class hierarchies
  // requires extra type information:
  if(typeid(s) == typeid(cp)) // C++ RTTI
    cp = static_cast<Circle*>(s);
  if(typeid(s) == typeid(sp))
    sp = static_cast<Square*>(s);
  if(cp != 0)
    cout << "It's a circle!" << endl;
  if(sp != 0)
    cout << "It's a square!" << endl;
  // Static navigation is ONLY an efficiency hack;
  // dynamic_cast is always safer. However:
  // Other* op = static_cast<Other*>(s);
  // Conveniently gives an error message, while
  Other* op2 = (Other*)s;
  // does not

I copy pasted the code from a book's source,yet I see nothing writen in the console! Why is that?



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#2 Paradigm Shifter   Crossbones+   -  Reputation: 5153

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Posted 21 June 2013 - 01:50 AM

You may have to turn on RTTI in the project settings if it is Visual Studio? It didn't used to be on by default but I don't know if it produces an error if it is not turned on.

 

Other possibility is the console window disappeared before any text was printed... use cout.flush()? EDIT: Probably not that since endl does a flush.

 

Why don't you step through looking at the variables in the debugger?


Edited by Paradigm Shifter, 21 June 2013 - 01:52 AM.

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#3 Wooh   Members   -  Reputation: 574

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Posted 21 June 2013 - 03:19 AM

Looks like typeid(s) == typeid(cp) and typeid(s) == typeid(sp) doesn't return true so cp and sp remains null.

#4 plainoldcj   Members   -  Reputation: 702

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Posted 21 June 2013 - 03:55 AM

s is of type Shape*. cp is of type Circle*. sp is of type Square*. Hence the type ids are different and you get no output.

Verify this by trying

Circle* cp = 0; ... if(typeid(Circle*) == typeid(cp)) cout << "okay!";


#5 BloodyEpi   Members   -  Reputation: 622

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Posted 21 June 2013 - 04:09 AM

You can use typeid(*s) if you want to know the type it points to.

 

Also, you can use your debugger to step into the typeinfo::operator==(...) and see for yourself what it is doing.



#6 Matt-D   Crossbones+   -  Reputation: 1410

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Posted 23 June 2013 - 12:14 PM

Note:
 

This will only work if you dereference the pointers to the objects. Only then will typeid work with the dynamic type of objects, instead of just giving you Circle* and Shape* back. – Seth Carnegie

 

 

http://stackoverflow.com/questions/12588264/static-cast-and-rtti-vs-dynamic-cast

 

Try this instead: http://ideone.com/dDRNFY

 

See how this works in the example here: http://en.cppreference.com/w/cpp/types/type_info/operator_cmp

 

Also used C++11's nullptr pointer literal:

http://en.cppreference.com/w/cpp/language/nullptr

#include <iostream>
#include <typeinfo>

class Shape { public: virtual ~Shape() {}; };
class Circle : public Shape {};
class Square : public Shape {};
class Other {};

int main() {
  Circle c;
  Shape* s = &c; // Upcast: normal and OK
  // More explicit but unnecessary:
  s = static_cast<Shape*>(&c);
  // (Since upcasting is such a safe and common
  // operation, the cast becomes cluttering)
  Circle* cp = nullptr;
  Square* sp = nullptr;
  // Static Navigation of class hierarchies
  // requires extra type information:
  // C++ RTTI
  if (typeid(*s) == typeid(Circle))
    cp = static_cast<Circle*>(s);
  if (typeid(*s) == typeid(Square))
    sp = static_cast<Square*>(s);
  std::cout << "cp = " << cp << std::endl;
  std::cout << "sp = " << sp << std::endl;
  if (cp) std::cout << "It's a circle!" << std::endl;
  if (sp) std::cout << "It's a square!" << std::endl;
  // Static navigation is ONLY an efficiency hack;
  // dynamic_cast is always safer. However:
  // Other* op = static_cast<Other*>(s);
  // Conveniently gives an error message, while
  // that would be undefined behavior (UB)!
  // Other* op2 = (Other*)s;
  // does not
  // AND *nothing* is guaranteed if we uncomment the UB
  // INCLUDING *anything* about the lines *before* the UB (yes, really!)
}

Edited by Matt-D, 23 June 2013 - 12:16 PM.


#7 Matt-D   Crossbones+   -  Reputation: 1410

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Posted 23 June 2013 - 12:25 PM

BTW, I think the example may be somewhat simpler without all this pointer cruft (dynamic polymorphism works perfectly fine with references) and unnecessary access specifiers (let's use structs, this will give us public-by-default for free): http://ideone.com/sxNT7o

#include <iostream>
#include <typeinfo>

struct Shape { virtual ~Shape() {}; };
struct Circle : Shape {};
struct Square : Shape {};
struct Other {};

int main() {
  Circle c;
  Shape & s = c; // Upcast: normal and OK
  // More explicit but unnecessary:
  s = static_cast<Shape&>(c);
  std::cout << "Address of `s` = " << &s << std::endl;
  
  // (Since upcasting is such a safe and common
  // operation, the cast becomes cluttering)

  // Static Navigation of class hierarchies
  // requires extra type information:
  // C++ RTTI
  if (typeid(s) == typeid(Circle))
  {
    Circle & cp = static_cast<Circle&>(s);
    std::cout << "It's a circle!" << std::endl;
    std::cout << "Address = " << &cp << std::endl;
  }
  if (typeid(s) == typeid(Square))
  {
    Square & sp = static_cast<Square&>(s);
    std::cout << "It's a square!" << std::endl;
    std::cout << "Address = " << &sp << std::endl;
  }
  // Static navigation is ONLY an efficiency hack;
  // dynamic_cast is always safer. However:
  // Other & op = static_cast<Other&>(s);
  // Conveniently gives an error message, while
  // that would be undefined behavior (UB)!
  // Other& op2 = (Other&)s;
  // does not
  // AND *nothing* is guaranteed if we uncomment the UB
  // INCLUDING *anything* about the lines *before* the UB (yes, really!)
}





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