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Legal to overwrite std::string's null terminator?

Asked 2012-10-05T06:03:42.037
42

In C++11, we know that std::string is guaranteed to be both contiguous and null-terminated (or more pedantically, terminated by charT(), which in the case of char is the null character 0).

There is this C API I need to use that fills in a string by pointer. It writes the whole string + null terminator. In C++03, I was always forced to use a vector<char>, because I couldn't assume that string was contiguous or null-terminated. But in C++11 (assuming a properly conforming basic_string class, which is still iffy in some standard libraries), I can.

Or can I? When I do this:

std::string str(length);

The string will allocate length+1 bytes, with the last filled in by the null-terminator. That's good. But when I pass this off to the C API, it's going to write length+1 characters. It's going to overwrite the null-terminator.

Admittedly, it's going to overwrite the null-terminator with a null character. Odds are good that this will work (indeed, I can't imagine how it couldn't work).

But I don't care about what "works". I want to know, according to the spec, whether it's OK to overwrite the null-terminator with a null character?

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2 Answers

25

Unfortunately, this is UB, if I interpret the wording correct (in any case, it's not allowed):

§21.4.5 [string.access] p2

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.

(Editorial error that it says T not charT.)

.data() and .c_str() basically point back to operator[] (§21.4.7.1 [string.accessors] p1):

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

answered 2012-10-05T06:44:31.370
5

I suppose n3092 isn't current any more but that's what I have. Section 21.4.5 allows access to a single element. It requires pos <= size(). If pos < size() then you get the actual element, otherwise (i.e. if pos == size()) then you get a non-modifiable reference.

I think that as far as the programming language is concerned, a kind of access which could modify the value is considered a modification even if the new value is the same as the old value.

Does g++ have a pedantic library that you can link to?

answered 2012-10-05T06:33:16.643

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