typeid operator

< cpp‎ | language

Queries information of a type.

Used where the dynamic type of a polymorphic object must be known and for static type identification.


[edit] Syntax

typeid( type ) (1)
typeid( expression ) (2)

The header <typeinfo> must be included before using typeid (if the header is not included, every use of the keyword typeid makes the program ill-formed.)

The typeid expression is lvalue expression which refers to an object with static storage duration, of the polymorphic type const std::type_info or of some type derived from it.

[edit] Explanation

1) Refers to a std::type_info object representing the type type. If type is a reference type, the result refers to a std::type_info object representing the referenced type.
2) Examines the expression expression
a) If expression is a glvalue expression that identifies an object of a polymorphic type (that is, a class that declares or inherits at least one virtual function), the typeid expression evaluates the expression and then refers to the std::type_info object that represents the dynamic type of the expression. If the glvalue expression is obtained by applying the unary * operator to a pointer and the pointer is a null pointer value, an exception of type std::bad_typeid or a type derived from std::bad_typeid is thrown.
b) If expression is not a glvalue expression of polymorphic type, typeid does not evaluate the expression, and the std::type_info object it identifies represents the static type of the expression. Lvalue-to-rvalue, array-to-pointer, or function-to-pointer conversions are not performed. Temporary materialization, however, is (formally) performed for prvalue arguments: typeid determines the type of the result object. (since C++17)

In all cases, cv-qualifiers are ignored by typeid (that is, typeid(T) == typeid(const T))

If the operand to typeid is a class type or a reference to a class type, then that class type must not be an incomplete type.

If typeid is used on an object under construction or destruction (in a destructor or in a constructor, including constructor's initializer list or default member initializers), then the std::type_info object referred to by this typeid represents the class that is being constructed or destroyed even if it is not the most-derived class.

[edit] Keywords


[edit] Notes

When applied to an expression of polymorphic type, evaluation of a typeid expression may involve runtime overhead (a virtual table lookup), otherwise typeid expression is resolved at compile time.

It is unspecified whether the destructor for the object referred to by typeid is executed at the end of the program.

There is no guarantee that the same std::type_info instance will be referred to by all evaluations of the typeid expression on the same type, although std::type_info::hash_code of those type_info objects would be identical, as would be their std::type_index.

const std::type_info& ti1 = typeid(A);
const std::type_info& ti2 = typeid(A);
assert(&ti1 == &ti2); // not guaranteed
assert(ti1.hash_code() == ti2.hash_code()); // guaranteed
assert(std::type_index(ti1) == std::type_index(ti2)); // guaranteed

[edit] Example

The example showing output using one of the implementations where type_info::name prints full type names; filter through c++filt -t if using gcc or similar.

#include <iostream>
#include <string>
#include <typeinfo>
struct Base {}; // non-polymorphic
struct Derived : Base {};
struct Base2 { virtual void foo() {} }; // polymorphic
struct Derived2 : Base2 {};
int main() {
    int myint = 50;
    std::string mystr = "string";
    double *mydoubleptr = nullptr;
    std::cout << "myint has type: " << typeid(myint).name() << '\n'
              << "mystr has type: " << typeid(mystr).name() << '\n'
              << "mydoubleptr has type: " << typeid(mydoubleptr).name() << '\n';
    // std::cout << myint is a glvalue expression of polymorphic type; it is evaluated
    const std::type_info& r1 = typeid(std::cout << myint);
    std::cout << "std::cout<<myint has type : " << << '\n';
    // std::printf() is not a glvalue expression of polymorphic type; NOT evaluated
    const std::type_info& r2 = typeid(std::printf("%d\n", myint));
    std::cout << "printf(\"%d\\n\",myint) has type : " << << '\n';
    // Non-polymorphic lvalue is a static type
    Derived d1;
    Base& b1 = d1;
    std::cout << "reference to non-polymorphic base: " << typeid(b1).name() << '\n';
    Derived2 d2;
    Base2& b2 = d2;
    std::cout << "reference to polymorphic base: " << typeid(b2).name() << '\n';
    try {
        // dereferencing a null pointer: okay for a non-polymoprhic expression
        std::cout << "mydoubleptr points to " << typeid(*mydoubleptr).name() << '\n'; 
        // dereferencing a null pointer: not okay for a polymorphic lvalue
        Derived2* bad_ptr = nullptr;
        std::cout << "bad_ptr points to... ";
        std::cout << typeid(*bad_ptr).name() << '\n';
    } catch (const std::bad_typeid& e) {
         std::cout << " caught " << e.what() << '\n';

Possible output:

myint has type: int
mystr has type: std::basic_string<char, std::char_traits<char>, std::allocator<char> >
mydoubleptr has type: double*
50std::cout<<myint has type : std::basic_ostream<char, std::char_traits<char> >
printf("%d\n",myint) has type : int
reference to non-polymorphic base: Base
reference to polymorphic base: Derived2
mydoubleptr points to double
bad_ptr points to...  caught std::bad_typeid