Is there a __CLASS__ macro in C++?

Question

The problem with using typeid(*this).name() is that there is no this pointer in a static method call. The macro __PRETTY_FUNCTION__ reports a class name in static functions as well as method calls. However, this will only work with gcc.

Here's an example of extracting the information through a macro style interface.

inline std::string methodName(const std::string& prettyFunction)
{
    size_t colons = prettyFunction.find("::");
    size_t begin = prettyFunction.substr(0,colons).rfind(" ") + 1;
    size_t end = prettyFunction.rfind("(") - begin;

    return prettyFunction.substr(begin,end) + "()";
}

#define __METHOD_NAME__ methodName(__PRETTY_FUNCTION__)

The macro __METHOD_NAME__ will return a string of the form <class>::<method>(), trimming the return type, modifiers and arguments from what __PRETTY_FUNCTION__ gives you.

For something which extracts just the class name, some care must be taken to trap situations where there is no class:

inline std::string className(const std::string& prettyFunction)
{
    size_t colons = prettyFunction.find("::");
    if (colons == std::string::npos)
        return "::";
    size_t begin = prettyFunction.substr(0,colons).rfind(" ") + 1;
    size_t end = colons - begin;

    return prettyFunction.substr(begin,end);
}

#define __CLASS_NAME__ className(__PRETTY_FUNCTION__)

The closest thing there's is to call typeid(your_class).name() - but this produces compiler specific mangled name.

To use it inside class just typeid(*this).name()

I would like to suggest boost::typeindex, which I learned about from Scott Meyer's "Effective Modern C++" Here's a basic example:

Example

#include <boost/type_index.hpp>

class foo_bar
{
    int whatever;
};

namespace bti =  boost::typeindex;

template <typename T>
void from_type(T t)
{
    std::cout << "\tT = " << bti::type_id_with_cvr<T>().pretty_name() << "\n";
}

int main()
{
    std::cout << "If you want to print a template type, that's easy.\n";
    from_type(1.0);
    std::cout << "To get it from an object instance, just use decltype:\n";
    foo_bar fb;
    std::cout << "\tfb's type is : "
              << bti::type_id_with_cvr<decltype(fb)>().pretty_name() << "\n";
}

Compiled with "g++ --std=c++14" this produces the following

Output

If you want to print a template type, that's easy.

T = double

To get it from an object instance, just use decltype:

fb's type is : foo_bar

Not yet. (I think __class__ is proposed somewhere). You can also try to extract class part from __PRETTY_FUNCTION__.

I think using __PRETTY_FUNCTION__ is good enough though it includes namespace as well i.e. namespace::classname::functionname until __CLASS__ is available.

If you need something that will actually produce the class name at compile time, you can use C++11 to do this:

#define __CLASS__ std::remove_reference<decltype(classMacroImpl(this))>::type

template<class T> T& classMacroImpl(const T* t);

I recognize that this is not the same thing as __FUNCTION__ but I found this post while looking for an answer like this. :D

If your compiler happens to be g++ and you are asking for __CLASS__ because you want a way to get the current method name including the class, __PRETTY_FUNCTION__ should help (according to info gcc, section 5.43 Function Names as Strings).