No argument constructor and function prototype

[bb]

Hi,

Class Test {
public:
Test() { cout << "Hello from Test" << endl; }
~Test() { cout << "Bye from Test" << endl; }
};

int main(int argc, char** argv) {

Test t1; // Does call the constructor and i can see the output.
Test t2(); // Compiler (mis)takes it for a function prototype? no
output. why? how to avoid it?

}

Thanks.

[Victor Bazarov]

Re: No argument constructor and function prototype

bb wrote:[color=blue]
> Class Test {
> public:
> Test() { cout << "Hello from Test" << endl; }
> ~Test() { cout << "Bye from Test" << endl; }
> };
>
> int main(int argc, char** argv) {
>
> Test t1; // Does call the constructor and i can see the output.
> Test t2(); // Compiler (mis)takes it for a function prototype? no
> output. why? how to avoid it?[/color]

It's not a mistake. It's a function declaration. Why? Because such is
the rule. Of the language. You just stepped in a puddle. No big deal.
Remember where it is and don't step into it again.
[color=blue]
>
> }
>
> Thanks.
>[/color]

V

[Mike Wahler]

Re: No argument constructor and function prototype


"bb" <muralibala68@g mail.com> wrote in message
news:1137535044 .318564.169010@ f14g2000cwb.goo glegroups.com.. .[color=blue]
> Hi,
>
> Class Test {
> public:
> Test() { cout << "Hello from Test" << endl; }
> ~Test() { cout << "Bye from Test" << endl; }
> };
>
> int main(int argc, char** argv) {
>
> Test t1; // Does call the constructor and i can see the output.
> Test t2(); // Compiler (mis)takes it for a function prototype?[/color]

No mistake. It *is* a function prototype. It declares
a function named 't2' which returns a type 'Test' value
and has no parameters. The rule is "if it looks like a
prototype, it is a prototype."
[color=blue]
> no
> output. why?[/color]

Prototypes do not generate output.
[color=blue]
> how to avoid it?[/color]

Don't write a prototype if you mean something else.

-Mike