Is it Safe to strncpy Into a string That Doesn't Have Room for the Null Terminator?

Question

Consider the following code:

const char foo[] = "lorem ipsum"; // foo is an array of 12 characters
const auto length = strlen(foo); // length is 11
string bar(length, '\0'); // bar was constructed with string(11, '\0')

strncpy(data(bar), foo, length);
cout << data(bar) << endl;

My understanding is that strings are always allocated with a hidden null element. If this is the case then bar really allocates 12 characters, with the 12^th being a hidden '\0' and this is perfectly safe... If I'm wrong on that then the cout will result in undefined behavior because there isn't a null terminator.

Can someone confirm for me? Is this legal?

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There have been a lot of questions about why to use strncpy instead of just using the string(const char*, const size_t) constructor. My intent has been to make my toy code close to my actual code which contains a vsnprintf. Unfortunately even after getting excellent answers here I've found that vsnprintf doesn't behave the same as strncpy, and I've asked a follow up question here: Why is vsnprintf Not Writing the Same Number of Characters as strncpy Would?

Answer 1

This is safe, as long as you copy [0, size()) characters into the string . Per [basic.string]/3

In all cases, [data(), data() + size()] is a valid range, data() + size() points at an object with value charT() (a “null terminator”), and size() <= capacity() is true.

So string bar(length, '\0') gives you a string with a size() of 11, with an immutable null terminator at the end (for a total of 12 characters in actual size). As long as you do not overwrite that null terminator, or try to write past it, you're okay.