Is there a __CLASS__ macro in C++? - c++

Question

Is there a __CLASS__ macro in C++ which gives the class name similar to __FUNCTION__ macro which gives the function name

Answer 1

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__)

Answer 2

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()

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

I created a function using __PRETTY_FUNCTION__ and constexpr with C++17 constexpr std::string_view methods. I also updated the algorithm a bit to be more reliably (Thanks to #n. 'pronouns' m for your help in 64387023).
constexpr std::string_view method_name(const char* s)
{
std::string_view prettyFunction(s);
size_t bracket = prettyFunction.rfind("(");
size_t space = prettyFunction.rfind(" ", bracket) + 1;
return prettyFunction.substr(space, bracket-space);
}
#define __METHOD_NAME__ method_name(__PRETTY_FUNCTION__)
In C++20, one can declare the function as consteval forcing it to evaluate at compile-time. Furthermore, there is std::basic_fixed_string for use as template parameter.

Answer 8

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).

Answer 9

If you're talking MS C++ (You should state, esp as __FUNCTION__ is a non-standard extension), there are __FUNCDNAME__ and __FUNCSIG__ symbols which you could parse

Answer 10

You can get the function name including class name.
This can process C-type funcitons.
static std::string methodName(const std::string& prettyFunction)
{
size_t begin,end;
end = prettyFunction.find("(");
begin = prettyFunction.substr(0,end).rfind(" ") + 1;
end -= begin;
return prettyFunction.substr(begin,end) + "()";
}

Answer 11

My solution:
std::string getClassName(const char* fullFuncName)
{
std::string fullFuncNameStr(fullFuncName);
size_t pos = fullFuncNameStr.find_last_of("::");
if (pos == std::string::npos)
{
return "";
}
return fullFuncNameStr.substr(0, pos-1);
}
#define __CLASS__ getClassName(__FUNCTION__)
I works for Visual C++ 12.

Answer 12

Here's a solution based on the __FUNCTION__ macro and C++ templates:
template <class T>
class ClassName
{
public:
static std::string Get()
{
// Get function name, which is "ClassName<class T>::Get"
// The template parameter 'T' is the class name we're looking for
std::string name = __FUNCTION__;
// Remove "ClassName<class " ("<class " is 7 characters long)
size_t pos = name.find_first_of('<');
if (pos != std::string::npos)
name = name.substr(pos + 7);
// Remove ">::Get"
pos = name.find_last_of('>');
if (pos != std::string::npos)
name = name.substr(0, pos);
return name;
}
};
template <class T>
std::string GetClassName(const T* _this = NULL)
{
return ClassName<T>::Get();
}
Here's an example of how this could be used for a logger class
template <class T>
class Logger
{
public:
void Log(int value)
{
std::cout << GetClassName<T>() << ": " << value << std::endl;
std::cout << GetClassName(this) << ": " << value << std::endl;
}
};
class Example : protected Logger<Example>
{
public:
void Run()
{
Log(0);
}
}
The output of Example::Run will then be
Example: 0
Logger<Example>: 0

Answer 13

This works quite nicely if you are willing to pay the cost of a pointer.
class State
{
public:
State( const char* const stateName ) :mStateName( stateName ) {};
const char* const GetName( void ) { return mStateName; }
private:
const char * const mStateName;
};
class ClientStateConnected
: public State
{
public:
ClientStateConnected( void ) : State( __FUNCTION__ ) {};
};

Answer 14

Works with msvc and gcc too
#ifdef _MSC_VER
#define __class_func__ __FUNCTION__
#endif
#ifdef __GNUG__
#include <cxxabi.h>
#include <execinfo.h>
char *class_func(const char *c, const char *f)
{
int status;
static char buff[100];
char *demangled = abi::__cxa_demangle(c, NULL, NULL, &status);
snprintf(buff, sizeof(buff), "%s::%s", demangled, f);
free(demangled);
return buff;
}
#define __class_func__ class_func(typeid(*this).name(), __func__)
#endif

Answer 15

All the solutions posted above that rely on the __PRETTY_FUNCTION__ do have specific edge case(s) where they do not return the class name / class name only. For example, consider the following pretty function value:
static std::string PrettyFunctionHelper::Test::testMacro(std::string)
Using the last occurence of "::" as delimter won't work since the function parameter also contains a "::" (std::string).
You can find similar edge cases for "(" as delimiter and more.
The only solution I found takes both the __FUNCTION__ and __PRETTY_FUNCTION__ macros as parameters. Here is the full code:
namespace PrettyFunctionHelper{
static constexpr const auto UNKNOWN_CLASS_NAME="UnknownClassName";
/**
* #param prettyFunction as obtained by the macro __PRETTY_FUNCTION__
* #return a string containing the class name at the end, optionally prefixed by the namespace(s).
* Example return values: "MyNamespace1::MyNamespace2::MyClassName","MyNamespace1::MyClassName" "MyClassName"
*/
static std::string namespaceAndClassName(const std::string& function,const std::string& prettyFunction){
//AndroidLogger(ANDROID_LOG_DEBUG,"NoT")<<prettyFunction;
// Here I assume that the 'function name' does not appear multiple times. The opposite is highly unlikely
const size_t len1=prettyFunction.find(function);
if(len1 == std::string::npos)return UNKNOWN_CLASS_NAME;
// The substring of len-2 contains the function return type and the "namespaceAndClass" area
const std::string returnTypeAndNamespaceAndClassName=prettyFunction.substr(0,len1-2);
// find the last empty space in the substring. The values until the first empty space are the function return type
// for example "void ","std::optional<std::string> ", "static std::string "
// See how the 3rd example return type also contains a " ".
// However, it is guaranteed that the area NamespaceAndClassName does not contain an empty space
const size_t begin1 = returnTypeAndNamespaceAndClassName.rfind(" ");
if(begin1 == std::string::npos)return UNKNOWN_CLASS_NAME;
const std::string namespaceAndClassName=returnTypeAndNamespaceAndClassName.substr(begin1+1);
return namespaceAndClassName;
}
/**
* #param namespaceAndClassName value obtained by namespaceAndClassName()
* #return the class name only (without namespace prefix if existing)
*/
static std::string className(const std::string& namespaceAndClassName){
const size_t end=namespaceAndClassName.rfind("::");
if(end!=std::string::npos){
return namespaceAndClassName.substr(end+2);
}
return namespaceAndClassName;
}
class Test{
public:
static std::string testMacro(std::string exampleParam=""){
const auto namespaceAndClassName=PrettyFunctionHelper::namespaceAndClassName(__FUNCTION__,__PRETTY_FUNCTION__);
//AndroidLogger(ANDROID_LOG_DEBUG,"NoT2")<<namespaceAndClassName;
assert(namespaceAndClassName.compare("PrettyFunctionHelper::Test") == 0);
const auto className=PrettyFunctionHelper::className(namespaceAndClassName);
//AndroidLogger(ANDROID_LOG_DEBUG,"NoT2")<<className;
assert(className.compare("Test") == 0);
return "";
}
};
}
#ifndef __CLASS_NAME__
#define __CLASS_NAME__ PrettyFunctionHelper::namespaceAndClassName(__FUNCTION__,__PRETTY_FUNCTION__)
#endif

Answer 16

Following method (based on methodName() above) can also handle input like "int main(int argc, char** argv)":
string getMethodName(const string& prettyFunction)
{
size_t end = prettyFunction.find("(") - 1;
size_t begin = prettyFunction.substr(0, end).rfind(" ") + 1;
return prettyFunction.substr(begin, end - begin + 1) + "()";
}