RTTI typeid question

[Mike]

I want to use typeid() in a base class function to determine the name of the
derived class. typeid(this) returns the name of the base class (which is an
abstract class) rather than the derived class. I'd rather not require the
typeid call in every derived class implementation. How do I get around
this?

I'm depending on the statement in Microsoft Visual Studio documentation
which states :

"
typeid( expression )
....
....
If the expression points to a base class type, yet the object is actually of
a type derived from that base class, a type_info reference for the derived
class is the result. The expression must point to a polymorphic type, that
is, a class with virtual functions. "

example

class foo
{
void create();
void virtfunction()= 0;
const char *m_type;
};

class x : public foo
{
create();
void virtfunction;
}

void foo::create()
{
const type_info *T = typeid(this); // this produces type
information for foo, not the derived class!!
m_type = T->name();
}

void x::Create()
{
// I'd rather not put the typeid call here although things work if I do
foo::create();
}

void x::virtfunction ()
{
// does something
}

[Victor Bazarov]

Re: RTTI typeid question

Mike wrote:[color=blue]
> I want to use typeid() in a base class function to determine the name of the
> derived class. typeid(this) returns the name of the base class (which is an
> abstract class) rather than the derived class. I'd rather not require the
> typeid call in every derived class implementation. How do I get around
> this?
>
> I'm depending on the statement in Microsoft Visual Studio documentation
> which states :
>
> "
> typeid( expression )
> ...
> ...
> If the expression points to a base class type, yet the object is actually of
> a type derived from that base class, a type_info reference for the derived
> class is the result. The expression must point to a polymorphic type, that
> is, a class with virtual functions. "[/color]

Your class 'foot' does not have virtual functions.
[color=blue]
>
> example
>
> class foo
> {
> void create();
> void virtfunction()= 0;[/color]

Did you mean to say

virtual void virtfunction()= 0;

?
[color=blue]
> const char *m_type;
> };
>
> class x : public foo
> {
> create();
> void virtfunction;[/color]

Did you mean to say

void virtfunction();

??
[color=blue]
> }
>
> void foo::create()
> {
> const type_info *T = typeid(this); // this produces type
> information for foo, not the derived class!!
> m_type = T->name();
> }
>
> void x::Create()
> {
> // I'd rather not put the typeid call here although things work if I do
> foo::create();
> }
>
> void x::virtfunction ()
> {
> // does something
> }[/color]

Try to always post the _actual_code_ that doesn't work or doesn't
compile, instead of typing some nonsense while in your newsreader.

Victor

[Carl Ribbegaardh]

Re: RTTI typeid question

Try: typeid(*this)

:)

"Mike" <mike@somewhere .com> wrote in message
news:boMtc.1488 8$Ly.14192@attb i_s01...[color=blue]
> I want to use typeid() in a base class function to determine the name of[/color]
the[color=blue]
> derived class. typeid(this) returns the name of the base class (which is[/color]
an[color=blue]
> abstract class) rather than the derived class. I'd rather not require the
> typeid call in every derived class implementation. How do I get around
> this?
>
> I'm depending on the statement in Microsoft Visual Studio documentation
> which states :
>
> "
> typeid( expression )
> ...
> ...
> If the expression points to a base class type, yet the object is actually[/color]
of[color=blue]
> a type derived from that base class, a type_info reference for the derived
> class is the result. The expression must point to a polymorphic type, that
> is, a class with virtual functions. "
>
> example
>
> class foo
> {
> void create();
> void virtfunction()= 0;
> const char *m_type;
> };
>
> class x : public foo
> {
> create();
> void virtfunction;
> }
>
> void foo::create()
> {
> const type_info *T = typeid(this); // this produces type
> information for foo, not the derived class!!
> m_type = T->name();
> }
>
> void x::Create()
> {
> // I'd rather not put the typeid call here although things work if I[/color]
do[color=blue]
> foo::create();
> }
>
> void x::virtfunction ()
> {
> // does something
> }
>
>[/color]

[Mike]

Re: RTTI typeid question

Thanks, Carl. That did the trick.

"Carl Ribbegaardh" <carl_ribbegaar dh.nospam@hotma il.com> wrote in message
news:2hpkd0Feeg 56U1@uni-berlin.de...[color=blue]
> Try: typeid(*this)
>
> :)
>
> "Mike" <mike@somewhere .com> wrote in message
> news:boMtc.1488 8$Ly.14192@attb i_s01...[color=green]
> > I want to use typeid() in a base class function to determine the name of[/color]
> the[color=green]
> > derived class. typeid(this) returns the name of the base class (which[/color][/color]
is[color=blue]
> an[color=green]
> > abstract class) rather than the derived class. I'd rather not require[/color][/color]
the[color=blue][color=green]
> > typeid call in every derived class implementation. How do I get around
> > this?
> >
> > I'm depending on the statement in Microsoft Visual Studio documentation
> > which states :
> >
> > "
> > typeid( expression )
> > ...
> > ...
> > If the expression points to a base class type, yet the object is[/color][/color]
actually[color=blue]
> of[color=green]
> > a type derived from that base class, a type_info reference for the[/color][/color]
derived[color=blue][color=green]
> > class is the result. The expression must point to a polymorphic type,[/color][/color]
that[color=blue][color=green]
> > is, a class with virtual functions. "
> >
> > example
> >
> > class foo
> > {
> > void create();
> > void virtfunction()= 0;
> > const char *m_type;
> > };
> >
> > class x : public foo
> > {
> > create();
> > void virtfunction;
> > }
> >
> > void foo::create()
> > {
> > const type_info *T = typeid(this); // this produces type
> > information for foo, not the derived class!!
> > m_type = T->name();
> > }
> >
> > void x::Create()
> > {
> > // I'd rather not put the typeid call here although things work if I[/color]
> do[color=green]
> > foo::create();
> > }
> >
> > void x::virtfunction ()
> > {
> > // does something
> > }
> >
> >[/color]
>
>[/color]