[amelius]

I think the authors of the spec really meant something else: reading/writing a memory location past the end of the array is illegal. But here "*" is used only in an address computation, not to actually access memory.

Shows how difficult it is to get a spec right.

So, IMO, you are right, the code in the article is illegal (strictly speaking).

But I think it is likely that most compilers would still allow it, because that clause in the spec essentially exempts the compiler from adding an explicit bounds check.

[bonzini]

I don't think this is illegal. What is the clause in the spec that allows &arr[1]? I would try and see if it also applies to (&arr)[1].

[clarry]

The clause that allows pointing one past the last element of an array is the same clause that explicitly forbids the latter (keeping in mind that E1[E2] is identical to ( * ((E1)+(E2)))).

N1256 6.5.6p8

When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i-n-th elements of the array object, provided they exist. Moreover, if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q)-1 points to the last element of the array object. If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined. If the result points one past the last element of the array object, it shall not be used as the operand of a unary * operator that is evaluated.

The last sentence right there forbids what we're doing here.

6.5.3.2p3 allows dereference with address-of (&a[1]):

The unary & operator yields the address of its operand. If the operand has type ''type'', the result has type ''pointer to type''. If the operand is the result of a unary * operator, neither that operator nor the & operator is evaluated and the result is as if both were omitted, except that the constraints on the operators still apply and the result is not an lvalue. Similarly, if the operand is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator were removed and the [] operator were changed to a + operator. Otherwise, the result is a pointer to the object or function designated by its operand.

The exception in this clause clearly does not apply to (&arr)[1] because the operand of & is not a result of the * (or []) operator.