Is writing to &str[0] buffer (of a std:string) well-defined behaviour in C++11?

Question

char hello[] = "hello world";
std::string str;
str.resize(sizeof(hello)-1);
memcpy(&str[0], hello, sizeof(hello)-1);

This code is undefined behaviour in C++98. Is it legal in C++11?

Answer 1

Yes, the code is legal in C++11 because the storage for std::string is guaranteed to be contiguous and your code avoids overwriting the terminating NULL character (or value initialized CharT ).

From N3337, §21.4.5 [string.access]

  const_reference operator[](size_type pos) const; reference operator[](size_type pos); 

1 Requires: pos <= size() .
2 Returns: *(begin() + pos) if pos < size() . Otherwise, returns a reference to an object of type charT with value charT() , where modifying the object leads to undefined behavior.

Your example satisfies the requirements stated above, so the behavior is well defined.