getting the name of an object

[Thomas Baier]

Hi there,

I've written a simple tree class and want to make a function that shows a
full tree with all its nodes.
My problem is that I do not know how to get the name of an object. I'd like
to have something like
class Tree
{

char *name;

Tree() {
name = this->variable_that_contains_name_of_object;
}
}
Tree one;
cout << one.name;

that results in the output:
one

Thanks for help


Thomas

 

[Mike Wahler]

Hi there,

I've written a simple tree class and want to make a function that shows a
full tree with all its nodes.
My problem is that I do not know how to get the name of an object. I'd like
to have something like
class Tree
{

char *name;

Tree() {
name = this->variable_that_contains_name_of_object;
}
}
Tree one;
cout << one.name;

that results in the output:
one

This cannot be done 'directly', since once the code is
compiled, there are no more 'names' of objects. Everything
is an address or an offset.

You could, however, create a data member which you store
your 'name' in as a string.

#include <string>

class X
{
std::string m_name;
public:
X(const std::string& n) : m_name(n) {}
const std::string& name() const { return m_name; }
};

int main()
{
X obj("one"); /* create object with 'name' of "one" */

std::string s(obj.name()); /* retrieve object's name and store
in another string */

return 0;
}

-Mike

 

[Thomas Baier]

Isn't there another way, without giving the name to the constructor as an
argument?

 

[Ron Natalie]

Isn't there another way, without giving the name to the constructor as an
argument?

No. The language provides no way to get the identifier names. In general, they
do not exist at runtime.

 

[Mike Wahler]

Isn't there another way, without giving the name to the constructor as an
argument?

No there is not. I already told you that.
Unless you're perhaps using an interpreter instead
of a compiler, there *are* no identifiers in the
executable, so of course they cannot be 'looked up'.
An interpreter might have means to do this, but that's
outside the purview of the language.

-Mike

 

[tom_usenet]

Hi there,

I've written a simple tree class and want to make a function that shows a
full tree with all its nodes.
My problem is that I do not know how to get the name of an object. I'd like
to have something like
class Tree
{

char *name;

Tree() {
name = this->variable_that_contains_name_of_object;
}
}
Tree one;
cout << one.name;

that results in the output:
one

How is this useful? You're passing something about the source code
organization into the output of the executable. This sounds like it
might be related to debugging, in which case the solution is to use a
debugger.

It would be helpful if you described exactly what you want to use this
facility for, and why you need it.

Tom

 

[Thomas Matthews]

Isn't there another way, without giving the name to the constructor as an
argument?

Yes, supply the name as a static constant:

class MyClass
{
public:
const char * get_class_name(void)
{ return class_name;} // option #2
static const char class_name[];
};

const char MyClass::class_name[] = "MyClass";


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Thomas Matthews

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Other sites:
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[Mike Wahler]

Isn't there another way, without giving the name to the constructor as an
argument?

Yes, supply the name as a static constant:

class MyClass
{
public:
const char * get_class_name(void)
{ return class_name;} // option #2
static const char class_name[];
};

const char MyClass::class_name[] = "MyClass";

I think he's looking for the identifier of each
specific instance, not for the name of the class.

-Mike

 

[Thomas Baier]

Isn't there another way, without giving the name to the constructor as an
argument?

Yes, supply the name as a static constant:

class MyClass
{
public:
const char * get_class_name(void)
{ return class_name;} // option #2
static const char class_name[];
};

const char MyClass::class_name[] = "MyClass";

I think he's looking for the identifier of each
specific instance, not for the name of the class.

-Mike

Right. I want to have a Tree class were you can get the name of every single
tree in order to show the hole tree graphically.

 

[Demanet Michael]

Isn't there another way, without giving the name to the constructor as an
argument?

Yes, supply the name as a static constant:

class MyClass
{
public:
const char * get_class_name(void)
{ return class_name;} // option #2
static const char class_name[];
};

const char MyClass::class_name[] = "MyClass";

In that cas, isn't it better using RTTI mechanism instead of your method ?

Michael

 

[Ron Natalie]

In that cas, isn't it better using RTTI mechanism instead of your method ?

Not if you want the names to be meaningful. The typeinfo::name() field isn't
guaranteed to be overly useful.