How To Get the Type of a Variable in Cpp

Sometimes we need to determine the variable type at runtime in C++ code. This tutorial will discuss how to get the type of a variable in C++.

The typeid operator allows us to determine the type of variable/object at runtime.

The syntax of typeid operator

typeid ( type )

typeid ( expression )

Parameters:

Typeid operator accepts one parameter :

Type – Variable or object.
Expression – Any valid expression.

Returns:

The typeid operator returns an lvalue of type const std::type_info that represents the type of expression.

To use the typeid operator we must include the STL header<typeinfo>.

Examples

Let’s check a number variable.

#include <iostream>
#include <typeinfo>

int main() {

    int a = 20;
    auto b = a * 2 ;
    std::cout << "b has type -> "<<typeid(b).name()<<"\n";
    
    return 0;
}

Output

b has type -> i

Let’s check a string variable.

#include <iostream>
#include <string>
#include <typeinfo>

int main() {

    std::string message = "Hello from NoloWiz";
    std::cout << message <<"\n";
    std::cout << "message has type -> "<<typeid(message).name()<<"\n";
    
    return 0;
}

Output

Hello from NoloWiz
message has type -> NSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEE

Let’s check a float pointer.

#include <iostream>
#include <typeinfo>

int main() {

    float* percentage = nullptr;
    std::cout << "percentage has type -> "<<typeid(percentage).name()<<"\n";
    
    return 0;
}

Output

percentage has type -> Pf

Non-polymorphic object

#include <iostream>
#include <typeinfo>

struct Base {}; 
struct Derived : Base {};

int main() {

    Derived drv;
    Base& base = drv;
    std::cout << "Reference to non-polymorphic base: " << typeid(base).name() << '\n';
    
    return 0;
}

Output

Reference to non-polymorphic base: 4Base

Polymorphic object

So what about the polymorphic object? let’s check that.

#include <iostream>
#include <typeinfo>

struct Base { virtual void foo() {} }; //polymorphic
struct Derived : Base {};

int main() {

    Derived drv;
    Base& base = drv;
    std::cout << "Reference to polymorphic base: " << typeid(base).name() << '\n';
    
    return 0;
}

Output

Reference to polymorphic base: 7Derived

Expression as parameter

Let’s pass an expression as a parameter to typeid.

#include <iostream>
#include <typeinfo>

int main() {

    int abc = 42;
    
    const std::type_info& typeInfo = typeid((10 + abc) / 2.0);
    std::cout <<"Type of expression (10 + abc) / 2.0 : "<<typeInfo.name();
    
    return 0;
}

Output

Type of expression (10 + abc) / 2.0 : d

The syntax of typeid operator

Examples

Expression as parameter

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