Re: [Dev-C++] Getting the wrong output

std::xx is for symbols declared inside of the std namespace. Your own
operator is definitely not defined in the std namespace - and you should
never try to define anyting in that namespace. It is only used for
standardized variables, types, methods etc.

You only add the xx:: prefix to a symbol when referencing symbols outside
of the current namespace or any namespace you have specifically said you
are using.

If you don't use any namespaces yourself, just write:

friend std::ostream& operator<<(std::ostream&,const jme::Money&);
std::ostream& operator<<(std::ostream& os,const jme::Money& obj) {
    ...
}

If the operator was defined within the jme namespace, then you have to
also do the friend declaration specifying:
friend std::ostream& jme::operator<<(std::ostream&,const jme::Money&);

and somewhere else specify the implementation within the jme namespace:
:
namespace jme {
    std::ostream& operator<<(std::ostream& os,const Money&) {
        ...
    }
}

/Per W

On Thu, 30 Jun 2005 jal...@go... wrote:

> Yes, I just notice that, so this are the changes I have made:
>
> --- Money.hpp
> friend std::ostream& std::operator<<( std::ostream&, const jme::Money& );
> friend std::istream& std::operator>>( std::istream&, jme::Money& );
>
>
> --- Money.cpp
> std::operator<<( std::ostream& os, const jme::Money& obj ) {
>     os << obj.getAmount();
>     return os;
> }
> std::istream&
> std::operator>>( std::istream& is, jme::Money& obj ) {
>     is >> obj.amount;
>     return is;
> }
> --------------
> But this is giving me linker errors sayng:
>
>
> =========  ERROR ===========
> In file included from money.cpp:2:
> money.hpp:67: error: `std::ostream& std::operator<<(std::ostream&, const
> jme::Money&)' should have been declared inside `std'
> money.hpp:68: error: `std::istream& std::operator>>(std::istream&,
> jme::Money&)' should have been declared inside `std'
>
>
> What am I doing wrong!!?
>
> Again thanks for you help.
>
>
>
>
> > 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 jal...@go... 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