如何从 C + + 对象获得类名？

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You could try using "typeid".

This doesn't work for "object" name but YOU know the object name so you'll just have to store it somewhere. The Compiler doesn't care what you namned an object.

Its worth bearing in mind, though, that the output of typeid is a compiler specific thing so even if it produces what you are after on the current platform it may not on another. This may or may not be a problem for you.

The other solution is to create some kind of template wrapper that you store the class name in. Then you need to use partial specialisation to get it to return the correct class name for you. This has the advantage of working compile time but is significantly more complex.

Edit: Being more explicit

template< typename Type > class ClassName
{
public:
static std::string name()
{
return "Unknown";
}
};

Then for each class somethign liek the following:

template<> class ClassName<MyClass>
{
public:
static std::string name()
{
return "MyClass";
}
};

Which could even be macro'd as follows:

#define DefineClassName( className ) \
\
template<> class ClassName<className> \
{ \
public: \
static std::string name() \
{ \
return #className; \
} \
}; \

Allowing you to, simply, do

DefineClassName( MyClass );

Finally to Get the class name you'd do the following:

ClassName< MyClass >::name();

Edit2: Elaborating further you'd then need to put this "DefineClassName" macro in each class you make and define a "classname" function that would call the static template function.

Edit3: And thinking about it ... Its obviously bad posting first thing in the morning as you may as well just define a member function "classname()" as follows:

std::string classname()
{
return "MyClass";
}

which can be macro'd as follows:

DefineClassName( className ) \
std::string classname()  \
{ \
return #className; \
}

Then you can simply just drop

DefineClassName( MyClass );

into the class as you define it ...

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use typeid(class).name

// illustratory code assuming all includes/namespaces etc

#include <iostream>
#include <typeinfo>
using namespace std;


struct A{};
int main(){
cout << typeid(A).name();
}

It is important to remember that this gives an implementation defined names.

As far as I know, there is no way to get the name of the object at run time reliably e.g. 'A' in your code.

EDIT 2:

#include <typeinfo>
#include <iostream>
#include <map>
using namespace std;


struct A{
};
struct B{
};


map<const type_info*, string> m;


int main(){
m[&typeid(A)] = "A";         // Registration here
m[&typeid(B)] = "B";         // Registration here


A a;
cout << m[&typeid(a)];
}

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Do you want [classname] to be 'one' and [objectname] to be 'A'?

If so, this is not possible. These names are only abstractions for the programmer, and aren't actually used in the binary code that is generated. You could give the class a static variable classname, which you set to 'one' and a normal variable objectname which you would assign either directly, through a method or the constructor. You can then query these methods for the class and object names.

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最佳答案

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.

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Just write simple template:

template<typename T>
const char* getClassName(T) {
return typeid(T).name();
}


struct A {} a;


void main() {
std::cout << getClassName(a);
}

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An improvement for @Chubsdad answer,

//main.cpp


using namespace std;


int main(){
A a;
a.run();
}


//A.h
class A{
public:
A(){};
void run();
}


//A.cpp
#include <iostream>
#include <typeinfo>
void A::run(){
cout << (string)typeid(this).name();
}

Which will print:

class A*

小开

To get class name without mangling stuff you can use func macro in constructor:

class MyClass {
const char* name;
MyClass() {
name = __func__;
}
}

小开

You can try this:

template<typename T>
inline const char* getTypeName() {
return typeid(T).name();
}


#define DEFINE_TYPE_NAME(type, type_name)  \
template<>                               \
inline const char* getTypeName<type>() { \
return type_name;                      \
}


DEFINE_TYPE_NAME(int, "int")
DEFINE_TYPE_NAME(float, "float")
DEFINE_TYPE_NAME(double, "double")
DEFINE_TYPE_NAME(std::string, "string")
DEFINE_TYPE_NAME(bool, "bool")
DEFINE_TYPE_NAME(uint32_t, "uint")
DEFINE_TYPE_NAME(uint64_t, "uint")
// add your custom types' definitions

And call it like that:

void main() {
std::cout << getTypeName<int>();
}