C++ 中有 __CLASS__ 宏吗？
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Is there a __CLASS__ macro in C++?
提问by mortal
Is there a __CLASS__macro in C++ which gives the class name similar to __FUNCTION__macro which gives the function name
__CLASS__C++中是否有一个宏，它给出的类名类似于__FUNCTION__给出函数名的宏
采纳答案by Aleksei Potov
The closest thing there's is to call typeid(your_class).name()- but this produces compiler specific mangled name.
最接近的是调用typeid(your_class).name()- 但这会产生编译器特定的损坏名称。
To use it inside class just typeid(*this).name()
在课堂上使用它只是 typeid(*this).name()
回答by Andrew Prock
The problem with using typeid(*this).name()is that there is no thispointer 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.
using 的问题typeid(*this).name()在于this静态方法调用中没有指针。该宏__PRETTY_FUNCTION__报告静态函数和方法调用中的类名。但是，这只适用于 gcc。
Here's an example of extracting the information through a macro style interface.
这是通过宏样式界面提取信息的示例。
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.
宏__METHOD_NAME__将返回一个形式为 的字符串，<class>::<method>()从__PRETTY_FUNCTION__给你的内容中修剪返回类型、修饰符和参数。
For something which extracts just the class name, some care must be taken to trap situations where there is no class:
对于只提取类名的东西，必须注意捕捉没有类的情况：
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__)
回答by Spacemoose
I would like to suggest boost::typeindex, which I learned about from Scott Meyer's "Effective Modern C++" Here's a basic example:
我想推荐boost::typeindex，这是我从 Scott Meyer 的“Effective Modern C++”中了解到的，这是一个基本示例：
Example
例子
#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
用“g++ --std=c++14”编译这会产生以下结果
Output
输出
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
如果您想打印模板类型，那很容易。
T = 双倍
要从对象实例中获取它，只需使用 decltype：
fb 的类型是： foo_bar
回答by Michael Krelin - hacker
Not yet. (I think __class__is proposed somewhere). You can also try to extract class part from __PRETTY_FUNCTION__.
还没有。（我认为__class__是在某处提出的）。您也可以尝试从__PRETTY_FUNCTION__.
回答by vaibhav
I think using __PRETTY_FUNCTION__is good enough though it includes namespace as well i.e. namespace::classname::functionnameuntil __CLASS__is available.
我认为 using__PRETTY_FUNCTION__已经足够好了，尽管它也包含命名空间，即namespace::classname::functionname直到__CLASS__可用。
回答by ndim
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).
如果您的编译器恰好是g++并且您要求是__CLASS__因为您想要一种获取当前方法名称（包括类）的方法，那么__PRETTY_FUNCTION__应该会有所帮助（根据5.43info gcc节函数名称作为字符串）。
回答by Charles.Lee
You can get the function name including class name. This can process C-type funcitons.
您可以获得包括类名的函数名。这可以处理 C 型函数。
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) + "()";
}
回答by Ruben Bartelink
If you're talking MS C++ (You should state, esp as __FUNCTION__is a non-standard extension), there are __FUNCDNAME__and __FUNCSIG__symbolswhich you could parse
如果你在谈论MS C ++（应该说明，尤指__FUNCTION__是一种非标准的扩展名），有__FUNCDNAME__和__FUNCSIG__符号，你可以解析
回答by Andrey Epifantsev
My solution:
我的解决方案：
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.
我为 Visual C++ 12 工作。
回答by Sven Vranckx
Here's a solution based on the __FUNCTION__macro and C++ templates:
这是基于__FUNCTION__宏和 C++ 模板的解决方案：
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
这是一个如何将其用于记录器类的示例
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::Runwill then be
的输出Example::Run将是
Example: 0
Logger<Example>: 0
