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Functions that cannot be overloaded in C++

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  • Difficulty Level : Easy
  • Last Updated : 23 Jun, 2022

In C++, following function declarations cannot be overloaded. 1) Function declarations that differ only in the return type. For example, the following program fails in compilation. 

CPP




#include<iostream>
int foo() {
  return 10;
}
 
char foo() {
  return 'a';
}
 
int main()
{
   char x = foo();
   getchar();
   return 0;
}


2) Member function declarations with the same name and the name parameter-type-list cannot be overloaded if any of them is a static member function declaration. For example, following program fails in compilation. 

CPP




#include<iostream>
class Test {
   static void fun(int i) {}
   void fun(int i) {}  
};
 
int main()
{
   Test t;
   getchar();
   return 0;
}


3) Parameter declarations that differ only in a pointer * versus an array [] are equivalent. That is, the array declaration is adjusted to become a pointer declaration. Only the second and subsequent array dimensions are significant in parameter types. For example, following two function declarations are equivalent. 

C




int fun(int *ptr);
int fun(int ptr[]); // redeclaration of fun(int *ptr)


4) Parameter declarations that differ only in that one is a function type and the other is a pointer to the same function type are equivalent. 

CPP




void h(int ());
void h(int (*)()); // redeclaration of h(int())


5) Parameter declarations that differ only in the presence or absence of const and/or volatile are equivalent. That is, the const and volatile type-specifiers for each parameter type are ignored when determining which function is being declared, defined, or called. For example, following program fails in compilation with error “redefinition of `int f(int)’ “ Example: 

CPP




#include<iostream>
#include<stdio.h>
  
using namespace std;
  
int f ( int x) {
    return x+10;
}
 
int f ( const int x) {
    return x+10;
}
 
int main() {    
  getchar();
  return 0;
}


Only the const and volatile type-specifiers at the outermost level of the parameter type specification are ignored in this fashion; const and volatile type-specifiers buried within a parameter type specification are significant and can be used to distinguish overloaded function declarations. In particular, for any type T, “pointer to T,” “pointer to const T,” and “pointer to volatile T” are considered distinct parameter types, as are “reference to T,” “reference to const T,” and “reference to volatile T.” For example, see the example in this comment posted by Venki. 6) Two parameter declarations that differ only in their default arguments are equivalent. For example, following program fails in compilation with error “redefinition of `int f(int, int)’ “ 

CPP




#include<iostream>
#include<stdio.h>
  
using namespace std;
  
int f ( int x, int y) {
    return x+10;
}
 
int f ( int x, int y = 10) {
    return x+y;
}
 
int main() {    
  getchar();
  return 0;
}


Function overloading and Namespaces

You can overload functions across namespaces. For example:

C++




#include <iostream>
using namespace std;
 
// Original X.h:
int f(int);
// Original Y.h:
int f(char);
 
// Original program.c:
#include "X.h"
#include "Y.h"
 
int main(){
  f('a');   // calls f(char) from Y.h
}


Namespaces can be introduced to the previous example without drastically changing the source code:

C++




#include <iostream>
using namespace std;
 
i// New X.h:
namespace X {
  f(int);
}
 
// New Y.h:
namespace Y {
  f(char);
}
 
// New program.c:
#include "X.h"
#include "Y.h"
 
using namespace X;
using namespace Y;
 
int main(){
  f('a');   // calls f() from Y.h
}


In program.c, the main function calls function f(), which is a member of namespace Y. If you place the using directives in the header files, the source code for program.c remains unchanged.

References: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1905.pdf Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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