Error in function overloading

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?

Code:

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# 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
   
      }

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I get the error:

Quote:

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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

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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

"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++.

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

Code:

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

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but you can not consider overloading

Code:

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

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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

Code:

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  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 * /

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