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



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 gosonic ca 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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