typeid Operator
The latest version of this topic can be found at typeid Operator.
Syntax
typeid(
type-id
)
typeid( expression )
Remarks
The typeid operator allows the type of an object to be determined at run time.
The result of typeid is a const type_info&. The value is a reference to a type_info object that represents either the type-id or the type of the expression, depending on which form of typeid is used. See type_info Class for more information.
The typeid operator does not work with managed types (abstract declarators or instances), see typeid for information on getting the Type of a specified type.
The typeid operator does a run-time check when applied to an l-value of a polymorphic class type, where the true type of the object cannot be determined by the static information provided. Such cases are:
A reference to a class
A pointer, dereferenced with *
A subscripted pointer (i.e. [ ]). (Note that it is generally not safe to use a subscript with a pointer to a polymorphic type.)
If the expression points to a base class type, yet the object is actually of a type derived from that base class, a type_info reference for the derived class is the result. The expression must point to a polymorphic type (a class with virtual functions). Otherwise, the result is the type_info for the static class referred to in the expression. Further, the pointer must be dereferenced so that the object it points to is used. Without dereferencing the pointer, the result will be the type_info for the pointer, not what it points to. For example:
// expre_typeid_Operator.cpp
// compile with: /GR /EHsc
#include <iostream>
#include <typeinfo.h>
class Base {
public:
virtual void vvfunc() {}
};
class Derived : public Base {};
using namespace std;
int main() {
Derived* pd = new Derived;
Base* pb = pd;
cout << typeid( pb ).name() << endl; //prints "class Base *"
cout << typeid( *pb ).name() << endl; //prints "class Derived"
cout << typeid( pd ).name() << endl; //prints "class Derived *"
cout << typeid( *pd ).name() << endl; //prints "class Derived"
delete pd;
}
If the expression is dereferencing a pointer, and that pointer's value is zero, typeid throws a bad_typeid exception. If the pointer does not point to a valid object, a __non_rtti_object exception is thrown, indicating an attempt to analyze the RTTI that triggered a fault (like access violation), because the object is somehow invalid (bad pointer or the code wasn't compiled with /GR).
If the expression is neither a pointer nor a reference to a base class of the object, the result is a type_info reference representing the static type of the expression. The static type of an expression refers to the type of an expression as it is known at compile time. Execution semantics are ignored when evaluating the static type of an expression. Furthermore, references are ignored when possible when determining the static type of an expression:
// expre_typeid_Operator_2.cpp
#include <typeinfo>
int main()
{
typeid(int) == typeid(int&); // evaluates to true
}
typeid can also be used in templates to determine the type of a template parameter:
// expre_typeid_Operator_3.cpp
// compile with: /c
#include <typeinfo>
template < typename T >
T max( T arg1, T arg2 ) {
cout << typeid( T ).name() << "s compared." << endl;
return ( arg1 > arg2 ? arg1 : arg2 );
}