How can I get the class name from a C++ object?

You can display the name of a variable by using the preprocessor. For instance

#include <iostream>
#define quote(x) #x
class one {};
int main(){
    one A;
    std::cout<<typeid(A).name()<<"\t"<< quote(A) <<"\n";
    return 0;
}

outputs

3one    A

on my machine. The # changes a token into a string, after preprocessing the line is

std::cout<<typeid(A).name()<<"\t"<< "A" <<"\n";

Of course if you do something like

void foo(one B){
    std::cout<<typeid(B).name()<<"\t"<< quote(B) <<"\n";
}
int main(){
    one A;
    foo(A);
    return 0;
}

you will get

3one B

as the compiler doesn't keep track of all of the variable's names.

As it happens in gcc the result of typeid().name() is the mangled class name, to get the demangled version use

#include <iostream>
#include <cxxabi.h>
#define quote(x) #x
template <typename foo,typename bar> class one{ };
int main(){
    one<int,one<double, int> > A;
    int status;
    char * demangled = abi::__cxa_demangle(typeid(A).name(),0,0,&status);
    std::cout<<demangled<<"\t"<< quote(A) <<"\n";
    free(demangled);
    return 0;
}

which gives me

one<int, one<double, int> > A

Other compilers may use different naming schemes.

use typeid(class).name

// illustratory code assuming all includes/namespaces etc

#include <iostream>
#include <typeinfo>
using namespace std;

struct A{};
int main(){
   cout << typeid(A).name();
}

It is important to remember that this gives an implementation defined names.

As far as I know, there is no way to get the name of the object at run time reliably e.g. 'A' in your code.

EDIT 2:

#include <typeinfo>
#include <iostream>
#include <map>
using namespace std; 

struct A{
};
struct B{
};

map<const type_info*, string> m;

int main(){
    m[&typeid(A)] = "A";         // Registration here
    m[&typeid(B)] = "B";         // Registration here

    A a;
    cout << m[&typeid(a)];
}