Xcode - shallow copy in STL string assignment

Question

When running the following C++ code in Xcode:

std::wstring str1 = L"1111";
std::wstring str2 = str1;

void* ptr1 = (void*)str1.c_str();
void* ptr2 = (void*)str2.c_str();

Result is that both pointers are equal. Is this by standard? In Visual Studio it's not the case.

Answer 1

It looks like the implementation is using the copy-on-write (COW) optimization, in which a strings internal state is only really set when a write operation is performed^*. This was allowed in pre-C++11 implementations, but I don't think this is standard since C++11.

Note that you can check that the address of the underlying pointer changes when you access the string in a non-const manner, even without writing to it:

str2[0];

The evaluation of this expression should trigger a write operation which would change the address of the pointer. Here is a working example:

#include <string>
#include <iostream>

int main() 
{
  std::wstring str1 = L"1111";
  std::wstring str2 = str1;

  std::cout << (void*)str1.c_str() << " " << (void*)str2.c_str() << std::endl;

  str2[0]; // forces a write operation. c_str() changes.

  std::cout << (void*)str1.c_str() << " " << (void*)str2.c_str() << std::endl;
}

On a recent gcc, this yields

0x8a8e014 0x8a8e014
0x8a8e014 0x8a8e03c

---

^{* Some non-const accesses can trigger a write even if they don't semantically mutate the string, as shown in the example above.}