Question

在登记物体必须具有独特名称的系统中,我想使用/列入标的<代码>。我想创造<代码>的最简单方式?

std:string name = ???

Answer 1

您可以利用地址的严格表述:

#include <sstream> //for std::stringstream 
#include <string>  //for std::string

const void * address = static_cast<const void*>(this);
std::stringstream ss;
ss << address;  
std::string name = ss.str();

Answer 2

你们指的是把点本身作为扼杀手段的形式?

std::ostringstream address;
address << (void const *)this;
std:string name = address.str();

或......在我照此作时,所有其他类似答案都是一样的!

Answer 3

#include <sstream>
#include <iostream>
struct T
{
    T()
    {
        std::ostringstream oss;
        oss << (void*)this;
        std::string s(oss.str());

        std::cout << s << std::endl;
    }
};

int main()
{
    T t;
}

Answer 4

你们可以利用这个要点的地址,把这种描述作为描述的价值?

Answer 5

A simpler one-liner that doesn t require a whole stringstream:

std::string address = std::to_string((unsigned long long)(void**)this);

Also implied, works with pointers of any time, not just this. Also works if you just need it as an integer:

unsigned long long address = (unsigned long long)(void**)this;

长期以来,你可以以该系统提供的种类取代未签字。视窗:uintptr_t。

Answer 6

在一个登记物体必须具有独特名称的系统中,我想以这个名字使用/列入物体。

物体地址不一定独特。例: 您积极分配这一物体,同时使用该物体,删除该物体,然后再分配另一个此类物体。新分配的物体可能与前一个物体相同。

找到一个独一无二的名称是更好的办法。例如:

// Base class for objects with a unique, autogenerated name.
class Named {
public:
  Named() : unique_id(gensym()) {}
  Named(const std::string & prefix) : unique_id(gensym(prefix)) {}

  const std::string & get_unique_id () { return unique_id; }

private:
  static std::string gensym (const std::string & prefix = "gensym");
  const std::string unique_id;
};  

inline std::string Named::gensym (const std::string & prefix) {
  static std::map<std::string, int> counter_map;
  int & entry = counter_map[prefix];
  std::stringstream sstream;
  sstream << prefix << std::setfill( 0 ) << std::setw(7) << ++entry;
  return sstream.str();
}   

// Derived classes can have their own prefix. For example,
class DerivedNamed : public Named {
public:
  DerivedNamed() : Named("Derived") {}
};

友情链接