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if(typeid(int) == typeid(const int))
cout << "Same types"<< endl;
PROGRAM OUTPUT:
Same types
am I missing something? these are not same types lol.
627
const int * And int const * are the same. const int * const And int const * const are the same. If you ever face confusion in reading such symbols, remember the Spiral rule: Start from the name of the variable and move clockwise to the next pointer or type.
The typeid operator returns an lvalue of type const std::type_info that represents the type of expression expr. You must include the standard template library header <typeinfo> to use the typeid operator. Classes A and B are polymorphic; classes C and D are not.
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.
int const* is pointer to constant integer This means that the variable being declared is a pointer, pointing to a constant integer. Effectively, this implies that the pointer is pointing to a value that shouldn't be changed.
They aren't the same type, but the typeid operator strips const and volatile.
From section 5.2.8 [expr.typeid]:
The top-level cv-qualifiers of the glvalue expression or the type-id that is the operand of
typeidare always ignored.
162
You probably want this instead:
#include <type_traits>
if (std::is_same<int, const int>::value)
std::cout << "same types\n";
else
std::cout << "different types\n";
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