Re: How to build a string on the fly, using std::ostringstream? [Reposting]

On Mon, 28 Apr 2008 01:56:19 CST, Vidar Hasfjord wrote:

> On Apr 27, 2:29 pm, Brendan Miller <catph...@[EMAIL PROTECTED]
> wrote:
>>>>       using ::operator <<; // Note: Ugly disambiguation.
>> Is this using statement really necessary? What ambiguity exists?
> 
> See my reply to Niels. I was unclear on the reason for the using
> statement. As you point out its only use is for disambiguation. The
> global namespace is of course always searched, but with the using
> declaration any matching global operator becomes the preferred.
> 
> I have since removed the using statement because I found better
> solutions in the client code were lookup problems occurred.
> 
>>>> Usage:
>>>>   string s = S () << "i = " << i;
>>
>>> Cool!!!  It looks much better than my approach!
>>
>> Yes and much better than mine, which I was proud of up until now. This
>> strikes me as the One True Way of inline stringbuilding.
> 
> In the search for the ultimate terseness:
> 
>    const struct StringBuilderFactory
>    {
>      struct StringBuilder
>      {
>        std::auto_ptr <std::ostringstream> s;

Well, if you are confident that RVO is performed, you can get rid of the
auto_ptr and write the copy constructor as
StringBuilder(const StringBuilder& sb);

with no body. However, this will fail if your compiler isn't very
agressive
about RVO. In particular, in debug mode of VC9, you get a linker error,
but
release mode it links fine. I've seen this trick pulled in GCC to some
effect.

By default I think that the auto_ptr is fine, and definitely more
****table,
but if you need to do some optimizing and your compiler sup****ts RVO
(recent GCC and VC++ do if optimization is on), you can use this little
hack to avoid dynamic memory allocation. However, I doubt that the
auto_ptr
here will ever become so expensive that it is worth the hack to avoid it.

> 
>        StringBuilder ()
>          : s (new std::ostringstream ())
>        {}
> 
>        template <typename T>
>        inline StringBuilder& operator << (const T& v)
>        {
>          *s << v;
>          return *this;
>        }
> 
>        inline operator std::string () const
>        {return s->str ();}
>      };
> 
>      template <typename T>
>      StringBuilder operator << (const T& obj) const
>      {return StringBuilder () << obj;}
>    }
>    S;
> 
> Usage:
> 
>    string s = S << "i = " << i;

If I try:
S << "asdf" << endl;

For some reason it can't find an overload for << endl.

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