Prototype or instantiation?

[John Doe]

Please have a look at this thread:

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&safe=off&threadm=q15zU3EADHA.2280%40cpmsftngxa06.phx.gbl&rnum=2&prev=/groups%3Fq%3DC4930%2Bprototyped%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26safe%3Doff%26selm%3Dq15zU3EADHA.2280%2540cpmsftngxa06.phx.gbl%26rnum%3D2

When I read it I couldn't believe that! I mean.... for years I have done
things like this for instantiating objects:

--------------------------------------
class CL
{
public:
CL (std::string s) {i_=s.length(); }
std::string get_a_string() {return std::string("wo w!");}
protected:
int i_;
};



.....
int main(....)
{
CL cl(std::string( "whatever") );
std::cout << cl.get_a_string ();
.....

}
--------------------------------------

with VC++6 and it would work! Today I have tried again with VC7.1 and it
still works!

So what is the above googlegroups thread speaking about? I can't see the
difference between the code posted there and my code, which DOES
instantiate the object.

Thanks in advance

[John Harrison]

Re: Prototype or instantiation?

>[color=blue]
> ....
> int main(....)
> {
> CL cl(std::string( "whatever") );
> std::cout << cl.get_a_string ();
> ....
>
> }[/color]

That works because "whatever" cannot be mistaken for a parameter name. But
try this

CL cl(std::string( whatever));

That declares a function called cl, returning a CL type object and taking a
single parameter of type std::string.

The parameter name in this prototype is whatever and it has a superfluous
but legal pair of brackets around it.

john

[John Harrison]

Re: Prototype or instantiation?

>[color=blue]
> The parameter name in this prototype is whatever and it has a superfluous
> but legal pair of brackets around it.
>[/color]

Maybe you'll find it easier to grasp if you realise that

int (a);

is a perfectly legal way to declare the variable a.

john