跟进这一问题的讨论，我不知道如何来使用本机C ++编程方式确定与否的标准::字符串实现他们使用的是开发前景的的副本上写（COW）的

Following up on the discussion from this question, I was wondering how does one using native C++ determine programmatically whether or not the std::string implementation they are using utilizes Copy-On-Write (COW)

我有以下功能：

#include <iostream>
#include <string>

bool stdstring_supports_cow()
{
   //make sure the string is longer than the size of potential
   //implementation of small-string.
   std::string s1 = "012345678901234567890123456789"
                    "012345678901234567890123456789"
                    "012345678901234567890123456789"
                    "012345678901234567890123456789"
                    "012345678901234567890123456789";
   std::string s2 = s1;
   std::string s3 = s2;

   bool result1 = (&s1[0]) == (&s2[0]);
   bool result2 = (&s1[0]) == (&s3[0]);

   s2[0] = 'X';

   bool result3 = (&s1[0]) != (&s2[0]);
   bool result4 = (&s1[0]) == (&s3[0]);

   s3[0] = 'X';

   bool result5 = (&s1[0]) != (&s3[0]);

   return result1 && result2 &&
          result3 && result4 &&
          result5;
}

int main()
{
  if (stdstring_supports_cow())
      std::cout << "std::string is COW." << std::endl;
   else
      std::cout << "std::string is NOT COW." << std::endl;
   return 0;
}

现在的问题是，我似乎无法找到一个C ++工具链中，则返回true。有没有在我的假设一个缺陷有关如何牛是性病:: string的实现？

The problem is I can't seem to find a C++ tool chain where it returns true. Is there a flaw in my assumption about how COW is implemented for std::string?

更新：的基础上kotlinski的意见，我已经改变了使用writeble引用数据（）中的功能，但现在看来，返回真的一些实施

Update: Based on kotlinski comments, I've changed the use of writeble references to data() in the function, it now seems to return "true" for some implementations.

bool stdstring_supports_cow()
{
   //make sure the string is longer than the size of potential
   //implementation of small-string.
   std::string s1 = "012345678901234567890123456789"
                    "012345678901234567890123456789"
                    "012345678901234567890123456789"
                    "012345678901234567890123456789"
                    "012345678901234567890123456789";
   std::string s2 = s1;
   std::string s3 = s2;

   bool result1 = s1.data() == s2.data();
   bool result2 = s1.data() == s3.data();

   s2[0] = 'X';

   bool result3 = s1.data() != s2.data();
   bool result4 = s1.data() == s3.data();

   s3[0] = 'X';

   bool result5 = s1.data() != s3.data();

   return result1 && result2 &&
          result3 && result4 &&
          result5;
}

注：根据 N2668：并发修改以基本字符串，在即将到来的C ++ 0x标准，COW选项将被从basic_string的删除。由于詹姆斯和Beldaz为使这件事。

Note: According N2668: "Concurrency Modifications to Basic String", in the upcoming C++0x standard, COW option will be removed from basic_string. thanks to James and Beldaz for bringing that up.

推荐答案

使用＆放大器; S1 [0] 带是不是你想要的联系地址， [0] 返回一个可写的参考，并创建一个副本。

Using &s1[0] to take the adress is not what you want, [0] returns a writable reference and will create a copy.

使用数据（）代替，它返回一个const char *，并且您的测试可以通过。

Use data() instead, it returns a const char*, and your tests may pass.