Question

Possible Duplicate:
Error on calling default constructor with empty set of brackets

When I run this I get the compiler warning:34 [Warning] the address of`Rational test4(), will always evaluate as true. but I am trying to make it so that the default constructor is the rational number 0/1. Line 34 is is int main() the line: cout << test4;.

#include <iostream>

using namespace std;

class Rational
{
public:
       Rational();

       friend ostream& operator <<(ostream& out,Rational rational1);


private:
        int numerator;
        int denominator;

};

int main()
{
    //Rational test1(24,6), test2(24);
    Rational test4();
    //cout << test1<< endl;
    //cout << test2<< endl;
    cout << test4;
    system("pause");
}

Rational::Rational() : numerator(0), denominator(1)
{ 
    //empty body
}

ostream& operator <<(ostream& out,Rational rational1)
{
       out << rational1.numerator <<"/"<<rational1.denominator;
       return out;
}

Answer 1

The reason your program prints “1” is because Rational test4(); declares a function pointer. So how does std::cout print a function pointer? It involves automatic conversions. First look to plain old data pointers. The I/O machinery does not have a mechanism for printing a double* pointer, or a MyClass* pointer, or any of the myriad number of pointers that can arise. What the I/O machinery can do have is a mechanism to print a void* pointer. Thanks to implicit conversions, that same mechanism works for double* and MyClass* pointers because all pointers convert to void* pointers.

Function pointers don’t convert to void*. Function pointers are not pointers! The only available conversion is to a boolean. That conversion to boolean is what let’s you do stuff like if (function_pointer) do_something(); Your function pointer isn’t null, so on conversion to bool it becomes true, which prints as 1.

The solution is simple: Change that Rational test4(); to Rationaltest4;`,