Operators that cannot be overloaded in C++
C++ provides a rich set of operators that can be overloaded to give specific meaning to user-defined classes. However, some operators cannot be overloaded in C++. This article will discuss the operators that cannot be overloaded in C++ and the reasons behind their limitations.
The operators that cannot be overloaded in C++ are
- the conditional operator (?:)
- the scope resolution operator (::)
- the sizeof operator
- the member selection operator (.)
- the member selection operator (->)
- the scope resolution operator with pointer to member (.)
- the scope resolution operator with pointer to member function (->)
The conditional operator (? : )
The conditional operator in C++ is a ternary operator that takes three operands. The first operand is a condition, the second operand is the expression to be evaluated if the condition is true, and the third operand is the expression to be evaluated if the condition is false. The conditional operator cannot be overloaded in C++ because it has a fixed semantic meaning.
int x = 5, y = 10;
int result = (x < y) ? x : y;
Scope resolution operator (::)
The scope resolution operator in C++ is used to access a member of a namespace or a class. It cannot be overloaded because it is a fundamental operator that is used to resolve scope.
Example:
#include<iostream>
using namespace std;
namespace my_namespace {
int my_variable = 10;
}
int main() {
int my_variable = 20;
std::cout << my_namespace::my_variable << std::endl;
return 0;
}
Explanation:
- In this C++ program, a variable called my_variable is defined both inside a namespace and inside the main() function.
- The program uses the scope resolution operator ‘::’ to distinguish between the two variables and access the one defined inside the namespace.
- The my_namespace namespace is defined at the global scope, and it contains a single variable called my_variable, which is initialized to 10.
- Inside the main() function, another variable called my_variable is defined and initialized to 20.
- This variable is local to the main() function and has no relationship to the my_variable defined inside the namespace.
- To access the variable defined in the my_namespace namespace, the program uses the scope resolution operator (::).
- The ‘std::cout’ statement then prints the value of ‘my_namespace::my_variable’ to the console. Since this is the variable defined inside the my_namespace namespace, its value is 10, not 20.
Program Output:
Sizeof operator:
The sizeof operator in C++ is used to determine the size of a type or an object. It cannot be overloaded because it is a compile-time operator that is evaluated by the compiler.
Example:
int arr[10];
std::cout << sizeof(arr) << std::endl;
Member selection operator:
In C++, the member selection operator is used to access a member of a class or structure. There are two types of member selection operators: the dot operator (.) and the arrow operator (->).
The dot operator is used to access a member of an object directly. For example, if we have a class Person with a member variable name, we can access it using the dot operator like this:
Person p;
p.name = "John";
The arrow operator is used to access a member of an object through a pointer to that object. For example, if we have a class Person with a member variable name, and a pointer p to an object of that class, we can access it using the arrow operator like this:
Person* p = new Person;
p->name = "John";
The dot operator cannot be overloaded in C++ because it has a fixed semantic meaning. It is used to access a member of an object directly, and there is no way to change that meaning.
Example:
#include<iostream>
using namespace std;
struct my_struct {
int my_variable;
};
int main() {
my_struct s;
s.my_variable = 10;
std::cout << s.my_variable << std::endl;
return 0;
}
Explanation:
- This C++ program defines a struct named my_struct, which has a single integer member variable named my_variable.
- In the main function, an instance of my_struct is declared and named ‘s’. The value of its ‘my_variable’ member is set to 10 using the member selection operator (.), which allows access to the individual members of a struct.
- Then, the value of ‘s.my_variable’ is printed to the console using cout and endl.
- The output of this program will be 10, which is the value assigned to ‘s.my_variable’.
- Overall, this program demonstrates the basic syntax and use of the member selection operator to access and modify struct member variables.
Program Output:
The arrow operator cannot be overloaded in C++ because it is used to access a member of an object through a pointer, and there is no way to change that meaning.
Example:
#include<iostream>
using namespace std;
struct my_struct {
int my_variable;
};
int main() {
my_struct* s = new my_struct;
s->my_variable = 10;
std::cout << s->my_variable << std::endl;
delete s;
return 0;
}
Explanation:
- This program defines a struct called my_struct which has a single member variable of type int called my_variable.
- In the main() function, it declares a pointer variable s of type my_struct* and dynamically allocates memory for a my_struct object using the new operator.
- The address of the allocated object is then assigned to the pointer variable s.
- The program then uses the member selection operator (->) to set the value of the my_variable member of the my_struct object that s points to, to 10.
- The member selection operator is used to access the members of a struct or class through a pointer to that struct or class.
- Finally, the program prints the value of my_variable using std::cout, and deallocates the dynamically allocated memory using the delete operator.
Program Output:
Scope resolution operator with pointer to member (.*)
The scope resolution operator with a pointer to a member in C++ is used to access a member of an object through a pointer to a member. It cannot be overloaded because it has a fixed semantic meaning.
#include<iostream>
using namespace std;
struct my_struct {
int my_variable;
};
int main() {
my_struct s;
my_struct* p = &s;
int my_struct::*ptr = &my_struct::my_variable;
p->*ptr = 10;
std::cout << s.my_variable << std::endl;
return 0;
}
Explanation:
- This program demonstrates the use of pointer to the member in C++.
- First, it defines a struct my_struct with a single member variable my_variable.
- In the main() function, it declares a my_struct object called ‘s’. It then declares a pointer variable p of type my_struct* and initializes it to the address of ‘s’. This creates a pointer to the my_struct object.
- Next, it declares a pointer to member ptr of type int my_struct::* and initializes it to the address of my_struct::my_variable. This creates a pointer to the my_variable member of the my_struct class.
- The program then uses the pointer to member ptr with the scope resolution operator ‘.* ’ to set the value of my_variable of the my_struct object pointed to by ‘p’ to 10. The syntax p->*ptr is equivalent to (*p).*ptr.
- Finally, the program prints the value of my_variable of the my_struct object ‘s’ using std::cout. Since ‘p’ is a pointer to ‘s’, the modification made by p->*ptr = 10 affects the my_variable member of ‘s’.
Program Output:
Scope resolution operator with pointer to member function (->*)
The scope resolution operator with a pointer to a member function in C++ is used to access a member function of an object through a pointer to a member function. It cannot be overloaded because it has a fixed semantic meaning.
Example:
#include<iostream>
using namespace std;
struct my_struct {
void my_function() {
std::cout << "my_function called" << std::endl;
}
};
int main() {
my_struct s;
my_struct* p = &s;
void (my_struct::*ptr)() = &my_struct::my_function;
(p->*ptr)();
return 0;
}
Explanation:
- This program demonstrates the use of a pointer to a member function in C++.
- It defines a struct my_struct with a member function my_function() that simply prints a message to the console.
- In the main() function, it declares a my_struct object called ‘s’. It then declares a pointer variable ‘p’ of type ‘my_struct*’ and initializes it to the address of ‘s’. This creates a pointer to the my_struct object.
- Next, it declares a pointer to a member function ptr of type void ‘(my_struct::*)()’ and initializes it to the address of ‘my_struct::my_function’. This creates a pointer to the my_function member function of the my_struct class.
- The program then uses the pointer to member function ptr with the pointer-to-member operator ‘->*’ to call the my_function member function on the my_struct object pointed to by ‘p’.
Program Output:
Conclusion:
In conclusion, while C++ provides the ability to overload many operators to give specific meaning to user-defined classes, there are several operators that cannot be overloaded in C++.
These operators are essential to the language and have a fixed semantic meaning.
Understanding the limitations of these operators is essential to effectively use C++ and create high-quality programs.