dot (.) operator in C/C++
The dot (.) operator is used for direct member selection via object name. In other words, it is used to access the child object.
Syntax:
object.member;
For example:
#include <stdio.h> struct Point { int x, y;}; int main(){ struct Point p1 = { 0, 1 }; // Accessing members of point p1 // using the dot operator p1.x = 20; printf("x = %d, y = %d", p1.x, p1.y); return 0;} |
x = 20, y = 1
Is dot (.) actually an Operator?
Yes, dot (.) is actually an operator in C/C++ which is used for direct member selection via object name. It has the highest precedence in Operator Precedence and Associativity Chart after the Brackets.
Is there any other Operator like dot(.) operator?
Yes. There is another such operator (->). It is called as “Indirect member selection” operator and it has precedence just lower to dot (.) operator. It is used to access the members indirectly with the help of pointers.
Example:
void addXtoList(struct Node* node, int x){ while (node != NULL) { node->data = node->data + x; node = node->next; }} |
Can dot (.) operator be overloaded?
No, Dot (.) operator can’t be overloaded. Doing so will cause an error.
Example:
// C++ program to illustrate// Overloading this .(dot) operator #include <iostream>using namespace std; class cantover {public: void fun();}; // assume that you can overload . operator// Class X below overloads the . operatorclass X { cantover* p; // Overloading the . operator cantover& operator.() { return *p; } void fun();}; void g(X& x){ // Now trying to access the fun() method // using the . operator // But this will throw an error // as we have overloaded the . operator above // Hence compiler won't allow doing so x.fun();} |
Output:
prog.cpp:11:20: error: expected type-specifier before '.' token
cantover& operator.()
^
prog.cpp:11:12: error: expected ';' at end of member declaration
cantover& operator.()
^
prog.cpp:11:20: error: expected unqualified-id before '.' token
cantover& operator.()
^
prog.cpp: In function 'void g(X&)':
prog.cpp:15:7: error: 'void X::fun()' is private
void fun();
^
prog.cpp:19:8: error: within this context
x.fun(); // X::fun or cantover::fun or error?
^
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