about overloading function

[seacret]

#include<iostream>
using namespace std;

void f1()                  {cout << "overload def#1 of f1()" << endl;} // def#1
void f1(int     x)         {cout << "overload def#2 of f1()" << endl;} // def#2
void f1(double  x)         {cout << "overload def#3 of f1()" << endl;} // def#3
void f1(double  x, char y) {cout << "overload def#4 of f1()" << endl;} // def#4
void f1(int     x, char y) {cout << "overload def#5 of f1()" << endl;} // def#5
void f1(double &x)         {cout << "overload def#6 of f1()" << endl;} // def#6

int main()
{
  /*1.*/ f1(2);
  /*2.*/ f1();
  /*3.*/ f1(2.5, 'A');
  /*4.*/ f1(2,   'A');

  double a = 2.5;

  /*5.*/ f1(2.5);
  /*6.*/ f1(a);
  return 0;
} // end main()

I can't understand why function call 6 can't call def#3.isn't their type the same?

and if comment the function call at 6 i give default value to parameter x of def#2 ,it will be error function call 2.

can someone explain to me how is it ambiguous?. or give me the chapter name so i can go re-study.
at least give me a hint before i got more misunderstanding.

sorry for my english.

[GCDEF]

It can't tell if you want to call 3 or 6.

[seacret]

isn't it reference passing?

#include<iostream>
using namespace std;

void f1()                  {cout << "overload def#1 of f1()" << endl;} // def#1
void f1(int     x=3)         {cout << "overload def#2 of f1()" << endl;} // def#2
void f1(double  x)         {cout << "overload def#3 of f1()" << endl;} // def#3
void f1(double  x, char y) {cout << "overload def#4 of f1()" << endl;} // def#4
void f1(int     x, char y) {cout << "overload def#5 of f1()" << endl;} // def#5
void f1(double &x)         {cout << "overload def#6 of f1()" << endl;} // def#6

int main()
{
  /*1.*/ f1(2);    
  /*2.*/ f1();  /*<<<<this got error*/
  /*3.*/ f1(2.5, 'A');
  /*4.*/ f1(2,   'A');

  double a = 2.5;

  /*5.*/ f1(2.5);

  return 0;
} // end main()

this got error.it say ambiguous. why f1() call def#1.

thanks for answer

[Igor Vartanov]

I can't understand why function call 6 can't call def#3.isn't their type the same?

When overloaded function is met, compiler deduces the closest matching variant from the function call. Calling like f1(a) matches to both f1(double) and f1(double&), so there's an ambiguity which cannot be solved by compiler.

To solve this, you need to signify which exact f1 version you need.

  double a = 2.5;

  /*5.*/ f1(2.5);
  /*6.*/ ((void (*)(double &))f1)(a);

Or better you never introduce such ambiguous overloaded function prototypes.

[Paul McKenzie]

this got error.it say ambiguous. why f1() call def#1.

void f1();
void f1(int x=3);
//...
f1();  // which one will be called?

Both of the f1() functions can be called with no arguments. That's why it is ambiguous.

Regards,

Paul McKenzie

[seacret]

thank you all of you .