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

cubuspl42 picture cubuspl42 · Aug 6, 2014 · Viewed 18.1k times · Source
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

Praetorian picture Praetorian · Aug 7, 2014

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.