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binary operator overloading; implicit type conversion

class my_bool {
  private:
    bool value;
  public:
    my_bool(bool value) : value(value) {}
    operator bool();

    friend my_bool operator==(const my_bool & instance_1, const my_bool & instance_2);
};

void main(){
  my_bool a = true;
  bool b = false;

  if(a == b){
    // do something
  }
}

The compiler says that comparison operator is ambiguous. Compiler cannot decide whether a should be converted to bool or b should be converted to my_bool . Is there a way I can solve this problem without writing down 3 overloads ( my_bool , my_bool ), ( bool , my_bool ), ( my_bool , bool ) of the same comparison operator?

Remove the const qualifer on the second second parameter to get rid of the ambiguity:

friend my_bool operator==(const my_bool & instance_1, my_bool & instance_2);

http://ideone.com/30VfO1

Or use explicit

 explicit operator bool();

Or use a different == overload that make more sense like this:

class my_bool
{
private:
    bool value;
public:
    my_bool(bool value) : value(value) {}
    operator bool() { return value; }

    bool operator == (bool val)
    {
        return this->value == val;
    }
};

http://ideone.com/fBaiKp

You could explicitly convert a to a bool

  if(((bool) a) == b){ // yes, this is a cast.
      // do something
  } 

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