Does “&s[0]” point to contiguous characters in a std::string?

A std::string’s allocation is not guaranteed to be contiguous under the C++98/03 standard, but C++11 forces it to be. In practice, neither I nor Herb Sutter know of an implementation that does not use contiguous storage.

Notice that the &s[0] thing is always guaranteed to work by the C++11 standard, even in the 0-length string case. It would not be guaranteed if you did str.begin() or &*str.begin(), but for &s[0] the standard defines operator[] as:

Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type T with value charT(); the referenced value shall not be modified

Continuing on, data() is defined as:

Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].

(notice the square brackets at both ends of the range)

Notice: pre-standardization C++0x did not guarantee &s[0] to work with zero-length strings (actually, it was explicitly undefined behavior), and an older revision of this answer explained this; this has been fixed in later standard drafts, so the answer has been updated accordingly.

Leave a Comment