Question

Is it possible to get the object name too?

#include<cstdio>

class one {
public:
    int no_of_students;
    one() { no_of_students = 0; }
    void new_admission() { no_of_students++; }
};

int main() {
    one A;
    for(int i = 0; i < 99; i++) {
        A.new_admission();
    }
    cout<<"class"<<[classname]<<" "<<[objectname]<<"has "
        <<A.no_of_students<<" students";
}

where I can fetch the names, something like

[classname] = A.classname() = one
[objectname] = A.objectname() = A

Does C++ provide any mechanism to achieve this?

Answer 1

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.