One-dimensional access to a multidimensional array: is it well-defined behaviour?

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All arrays (including multidimensional ones) are padding-free. Even if it’s never explicitly mentioned, it can be inferred from sizeof rules.

Now, array subscription is a special case of pointer arithmetics, and C99 section 6.5.6, §8 states clearly that behaviour is only defined if the pointer operand and the resulting pointer lie in the same array (or one element past), which makes bounds-checking implementations of the C language possible.

This means that your example is, in fact, undefined behaviour. However, as most C implementations do not check bounds, it will work as expected – most compilers treat undefined pointer expressions like

mtx[0] + 5

identically to well-defined counterparts like

(int *)((char *)mtx + 5 * sizeof (int))

which is well-defined because any object (including the whole two-dimensional array) can always be treated as a one-dimensinal array of type char.

On further meditation on the wording of section 6.5.6, splitting out-of-bounds access into seemingly well-defined subexpression like

(mtx[0] + 3) + 2

reasoning that mtx[0] + 3 is a pointer to one element past the end of mtx[0] (making the first addition well-defined) and as well as a pointer to the first element of mtx[1] (making the second addition well-defined) is incorrect:

Even though mtx[0] + 3 and mtx[1] + 0 are guaranteed to compare equal (see section 6.5.9, §6), they are semantically different. For example, the former can’t be dereferenced and thus does not point to an element of mtx[1].

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