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RTTI : Base class type name v/s Derived


Hello there,
    Take a look at the code below

#include <stdio.h>
#include <typeinfo>

class Base {
};

class Drv : public Base {
};

void
printType (char* realName, Base* pbase) {
    printf ("\n%s ==> %s\n\n", realName, typeid(pbase).name());
}

int
main(){
    printType("base", new Base());
    printType("drv", new Drv());
    return 0;
}


If you run it (w/msdev) you'll get the following output

base ==> class Base *
drv ==> class Base *

Apparently, the typeinfo contains the info about the local variable of printType() function.
Is it possible to get the type info of the actual object  instead ?

Something like this:

base ==> class Base *
drv ==> class Drv *

thanks
~J
0
mxjijo
Asked:
mxjijo
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1 Solution
 
rstaveleyCommented:
You need to dereference the pointer because you want the type that's pointed to. You will need a virtual function in the base class too to enable polymorphism. Also compile with /GR to enable RTTI.

--------8<--------
#include <stdio.h>
#include <typeinfo>

class Base {
public:
virtual ~Base() {}
};

class Drv : public Base {
};

void
printType (char* realName, Base* pbase) {
    printf ("\n%s ==> %s\n\n", realName, typeid(*pbase).name());
}

int
main(){
    printType("base", new Base());
    printType("drv", new Drv());
    return 0;
}
--------8<--------
0
 
rstaveleyCommented:
> #include <stdio.h>

Strictly speaking you should use...

#include <cstdio>

...with C++ too.

If you are relatively new to C++, note that typeid and dynamic_cast<> (the two aspects of RTTI) are seldom used and therefore not worth mastering initially. That's why Microsoft doesn't enable RTTI by default. Good OOP design and getting polymorphism to work for you will make them unnecessary 99.999% of the time.
0
 
mxjijoAuthor Commented:
great. that would do.
thanks
0
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rstaveleyCommented:
Hit the accept button to accept the answer, mxjijo. Otherwise, the question remains open.
0
 
mxjijoAuthor Commented:

I'm sorry I thought I did :)
0
 
rstaveleyCommented:
Thanks :-)
0

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