如何从C ++对象获取类名？ [英] How can I get the class name from a C++ object?

问题描述

是否也可以获取对象名称？

 ＃include< cstdio> 
 
 class one {
 public：
 int no_of_students; 
 one（）{no_of_students = 0; } 
 void new_admission（）{no_of_students ++; } 
}; 
 
 int main（）{
 one A; 
 for（int i = 0; i <99; i ++）{
 A.new_admission（）; 
} 
 cout <<class<< [classname]<<<< [objectname]<<has
< ; A.no_of_students<<students; 
} 
  

其中我可以获取名称，例如

  [classname] = A.classname（）= one 
 [objectname] = A.objectname（）= A 
  
 
 
 
  C ++是否提供了任何机制来实现这一点？ 
解决方案
可以使用预处理器显示变量的名称。例如
  #include< iostream> 
 #define quote（x）#x 
 class one {}; 
 int main（）{
 one A; 
 std :: cout<< typeid（A）.name（）<<\t< quote（A）<<\\\
; 
 return 0; 
} 
  
输出
  3one A 
  
 ＃在预处理行之后将令牌更改为字符串
  std :: cout<< typeid（A）.name（）<<\t< A<\ n; 
  
当然，如果你做了
  void foo（one B）{
 std :: cout<< typeid（B）.name（）<<\t< quote（B）<<\\\
; 
} 
 int main（）{
 one A; 
 foo（A）; 
 return 0; 
} 
  
您将获得
  3one B 
  
 。
 
 
 
正如在gcc中发生的一样，typeid（）的结果name（）是被改变的类名，得到 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& >一个; 
 int status; 
 char * demangled = abi :: __ cxa_demangle（typeid（A）.name（），0,0，& status）; 
 std :: cout<< demangled<<\t<< quote（A）<<\\\
; 
免费（demangled）; 
 return 0; 
} 
  
 
  one< int，one< double，int> > A 
  
其他编译器可能使用不同的命名方案。
 
Is it possible to get the object name too?
#include<cstdio>

class one {
public:
    int no_of_students;
    one() { no_of_students = 0; }
    void new_admission() { no_of_students++; }
};

int main() {
    one A;
    for(int i = 0; i < 99; i++) {
        A.new_admission();
    }
    cout<<"class"<<[classname]<<" "<<[objectname]<<"has "
        <<A.no_of_students<<" students";
}
where I can fetch the names, something like 
[classname] = A.classname() = one
[objectname] = A.objectname() = A
Does C++ provide any mechanism to achieve this?
解决方案
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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