Error in function overloading

[CitricAcid]

I just started learning C++. Here is a code on the overloading of functions. My problem is I am unable to find the error. Can some one please help me on this?

 # include <iostream> 
  # include <conio.h> 

  void test (int n = 0, float x = 2.5) 
  {
       cout << "Function No. 1:"; 
       cout << "n =" <<n << "x =" <<x << "\ n"; 
  }

  void test (float x = 4.1, int n = 2) 
  { 
    cout << "Function No. 2:"; 
    cout<< "n= " <<n<< " x = " <<x<< " \n " ;
  } 
  void main () 
  { 
       int i = 5; float r = 3.2; 
       test (i, r) // function N1 
       test (r, i) // function N2 
       test (i) // function N1 
       test (r) // function N2 
     
       }

I get the error:

18 C:\Users\Nas\Desktop\Projects\Exo 1.5\main.cpp `main' must return `int'
C:\Users\Nas\Desktop\Projects\Exo 1.5\Makefile.win [Build Error] [main.o] Error 1

[fellah]

The compiler gives you the answer. The main () must return an integer, so ...

1 - you rename your main() into "int main ()"
2 - you put a "return 0" at the end, if you do not want to return anything

[Lemog]

"int f ()" indicates that the function "f ()" returns an int value. That is why we need to add a return statement at the end (eg return 0; means that the function returns the integer zero)

"void f ()" means that the function f () does not returns any value.

However the main function is a little different because it is the entry point of the program. Some compilers requires that a C++ program should return a value, so that's why it is necessary that the function main () returns an int.

Therefore the function "void main ()" is not legal in C++.

[Modifier]

One thing that should be understandable initially is that overloading function only works if you have a different number or types of arguments.

Thus, a surcharge may be in a form close to

  void foo () 
  void foo (int) 
  void foo (int, int)

but you can not consider overloading

  void foo (int i = 3, int j = 4) 
 {
      / * What should be done * / 
 } 
  void foo (int i = 5, int j = 6)

because the number and type of argument is identical.

Remember that if you give default values for both arguments (as example), you find yourself in a reality with the option of using prototypes

  void foo () / * the default values are used * / 
  void foo (int i) / * the default value of j is used * / 
  void foo (int i, int j) / * no default is used * /