CODEWITHSUNDEEP
c++typeid
G-s
G-s

Reputation: 143

Is it possible to attach a name to a runtime class in C++?

I have an hierarchy of classes, the base class having a function to print the class name:

#include <iostream>
using namespace std;

class base
{
public:
  virtual void print_name() { cout << typeid(*this).name() << endl; };
};

class derived1 : public base { };

class derived2 : public base { };

int main ()
{
  base Base;
  Base.print_name();

  derived1 Derived1;
  Derived1.print_name();

  derived2 Derived2;
  Derived2.print_name();
}

The output of the above is

class base
class derived1
class derived2

which is, in fact, platform dependent.

Is there a more or less standard way to "attach" some unique name to each class, so it could be used in printname() making the output the same for all platforms (and independent of any changes made to real class names)?

Upvotes: 0

Views: 237

Answers (3)

Matthieu M.
Matthieu M.

Reputation: 299730

You can effectively use type_info for this.

type_info supports a before method conforming to a Weak Order, which allows its use in a std::map (for example) as long as a user-supplied predicate is provided.

struct TypeInfoLess {
     bool operator()(std::type_info const* lhs, std::type_info const* rhs) const {
         return lhs->before(rhs);
     }
};

struct AdditionalTypeInfo {
    std::string name;
};

typedef std::map<std::type_info const*, AdditionalTypeInfo, TypeInfoLess> TypeInfoMap;

Then, you can just add/search types:

template <typename T>
void add(TypeInfoMap& map, T const& t, AdditionalTypeInfo const& ati) {
    map[&typeid(t)] = ati;
}

template <typename T>
AdditionalTypeInfo const* find(TypeInfoMap const& map, T const& t) {
    TypeInfoMap::const_iterator it = map.find(&typeid(t));
    if (it == map.end()) { return 0; }
    return &it->second;
}

int main() {
    TypeInfoMap timap;

    add(timap, timap, { "TypeInfoMap" });

    if (AdditionalTypeInfo const* const ati = find(timap, timap)) {
        std::cout << ati->name << "\n";
    }
}

Note: it is then your responsability to add to the map every type that you might want.

Upvotes: 1

Steed
Steed

Reputation: 1300

You can use template functions (or traits classes) to get names:

template<typename T> const char * ClassName(T const * objPtr);

template<> const char * ClassName<derived1>(derived1 const *objPtr) { return "derived1"; }
template<> const char * ClassName<derived2>(derived2 const *objPtr) { return "derived2"; }

(Here the function parameter is used for template parameter matching only). And in calling code you can cast your pointer to specific class and use ClassName

derived1 * pDerived = static_cast<derived1*>(pPointer);
cout << ClassName(pDerived);

However this is all compile-time, and you will need some mechanism (like GUIDs) to identify a class at runtime (to cast it to the right pointer type). If your classes have none and you can't modify them, than I can't help you;( But you can use traits technique to localize your platform-dependent type selection code.

Upvotes: 0

Fred Foo
Fred Foo

Reputation: 363487

Sure:

class base {
  public:
    virtual char const *name() const { return "base"; }
};

class derived1 : public base {
  public:
    virtual char const *name() const { return "derived1"; }
};

However, if you do not override name in a class, its name will be that of its superclass. That may be a bug or a feature, depending on your use case. If it's a bug, then you can add some runtime checks to make sure the method is overridden:

virtual char const *name() const {
    if (typeid(*this) != typeid(base))
        throw std::logic_error("name() not overridden");
    return "base";
}

But you'll have to repeat this check in every implementation of name that must be overridden.

Upvotes: 2

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