What namespaces are you using.
You have declared the friend operator as:
std::ostream& operator<<(...)
and implemented it as:
std::ostream& jme::operator<<(...)
Without a maching output operator, the compiler will emit the pointer to
the object instead.
/Per W
On Thu, 30 Jun 2005 jalkadir@xxxxxxxxxx wrote:
> In my class, when I use the "getter" to display the value I get the right
> output, but when I try using the overloaded extractor operator (<<) I get
> some hex value displayed. Here is some of the code:
> --- snip
> Main.cpp
> int main() {
> Money money("12.21");
>
> std::cout << "Value is: " << money.getAmount() << std::endl; // 12.21
> std::cout << "Value is: " << money << std::endl; // hex value????
> return 0;
> }
>
>
>
> The class goes like this:
> ---snip
> Money.hpp
> class Money {
> protected:
> float amount; //!< This variable holds the numerical value
> public:
> //! Constor
> Money(const std::string&);
>
> //!Copy constructor
> Money( const Money& );
>
> //!Destructor
> ~Money() { std::cout << amount << std::endl;} //<<=== 12.21
>
> // Setters
> ...
> // Getters
> const float getAmount() const;
>
> // Overloaded operators
> .....
> friend std::ostream& operator<<( std::ostream&, const jme::Money& );
> friend std::istream& operator>>( std::istream&, jme::Money& );
> ....
> } ; //Money
>
>
>
> and the code looks like this
> --- snip
> jme::Money::Money( const std::string& x){
> this->setAmount(x); // this metho converts the std::string to a float
> }
> const float jme::Money::getAmount() const {
> return this->amount;
> }
> std::ostream&
> jme::operator<<( std::ostream& os, const jme::Money& obj ) {
> os << obj.getAmount(); //<<==== 0x4b1120
> return os;
> }
> ----- end of snip
> As you can see when using the member method getAmount() the value
> displayed is in the right format, however the extractor operator is not
> doing
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