Re: [Dev-C++] ostringstream and insertion operator problem

[Jonathan W. <jon...@gm...>]

Hi,

A reference to an object is actually a pointer, which is a number (to which
we have attached a special meaning)
Testing this reference tests the pointer with 0, or NULL. I think you will
find that << returns NULL if it fails.

HTH,
Jonathan

On 11/1/07, A D <a_...@ho...> wrote:
>
>
> Hi! I'm trying to learn c++ and stumbled upon part of a code which I'm
> having
> difficulties to understand:
>
>
> class BadConversion : public std::runtime_error {
> public:
>    BadConversion(const std::string& s)
>      : std::runtime_error(s)
>      { }
> };
>
> inline std::string stringify(double x)
> {
>    std::ostringstream o;
>    if (!(o << x))
>      throw BadConversion("stringify(double)");
>    return o.str();
> }
>
>
> the function stringify() converts a double number to string And returns
> the string.
> I'm having problem with the following line:
>
>
> if (!(o << x)) //<-----------------------------------problem
>      throw BadConversion("stringify(double)");
>
> according to the api ostringstream is a typedef of basic_ostringstream
> class. This insertion (<
>      basic_ostream&
>     basic_ostream::operator<put(*this, *this, this->fill(), __n).failed())
>            this->setstate(ios_base::badbit);
>          }
>        catch(...)
>          {
>            // 27.6.1.2.1 Common requirements.
>            // Turn this on without causing an ios::failure to be thrown.
>            this->_M_setstate(ios_base::badbit);
>            if ((this->exceptions() & ios_base::badbit) != 0)
>          __throw_exception_again;
>          }
>      }
>        return *this;
>      }
>
>
> You can  clearly see from the definition that binary operator "<<" outputs
> an
> object of basic_ostream reference type. My question is can the operator <<
> return a zero in this case? Because by definition this operator clearly
> returns a
> reference object of type basic_ostream&.  If it returns zero(which is
> tested in
> order for a exception to be thrown in the former code shown above),
> shouldn't
> it contradict the fact that this operator can only return an object not a
> number?
> Thanks.
>
>
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