Asked  7 Months ago    Answers:  5   Viewed   50 times

I have seen it asserted several times now that the following code is not allowed by the C++ Standard:

int array[5];
int *array_begin = &array[0];
int *array_end = &array[5];

Is &array[5] legal C++ code in this context?

I would like an answer with a reference to the Standard if possible.

It would also be interesting to know if it meets the C standard. And if it isn't standard C++, why was the decision made to treat it differently from array + 5 or &array[4] + 1?

 Answers

32

Your example is legal, but only because you're not actually using an out of bounds pointer.

Let's deal with out of bounds pointers first (because that's how I originally interpreted your question, before I noticed that the example uses a one-past-the-end pointer instead):

In general, you're not even allowed to create an out-of-bounds pointer. A pointer must point to an element within the array, or one past the end. Nowhere else.

The pointer is not even allowed to exist, which means you're obviously not allowed to dereference it either.

Here's what the standard has to say on the subject:

5.7:5:

When an expression that has integral 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 integral 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 over?ow; otherwise, the behavior is unde?ned.

(emphasis mine)

Of course, this is for operator+. So just to be sure, here's what the standard says about array subscripting:

5.2.1:1:

The expression E1[E2] is identical (by de?nition) to *((E1)+(E2))

Of course, there's an obvious caveat: Your example doesn't actually show an out-of-bounds pointer. it uses a "one past the end" pointer, which is different. The pointer is allowed to exist (as the above says), but the standard, as far as I can see, says nothing about dereferencing it. The closest I can find is 3.9.2:3:

[Note: for instance, the address one past the end of an array (5.7) would be considered to point to an unrelated object of the array’s element type that might be located at that address. —end note ]

Which seems to me to imply that yes, you can legally dereference it, but the result of reading or writing to the location is unspecified.

Thanks to ilproxyil for correcting the last bit here, answering the last part of your question:

  • array + 5 doesn't actually dereference anything, it simply creates a pointer to one past the end of array.
  • &array[4] + 1 dereferences array+4 (which is perfectly safe), takes the address of that lvalue, and adds one to that address, which results in a one-past-the-end pointer (but that pointer never gets dereferenced.
  • &array[5] dereferences array+5 (which as far as I can see is legal, and results in "an unrelated object of the array’s element type", as the above said), and then takes the address of that element, which also seems legal enough.

So they don't do quite the same thing, although in this case, the end result is the same.

Tuesday, June 1, 2021
 
fhonics
answered 7 Months ago
75

Yes, you can take the address one beyond the end of an array, but you can't dereference it. For your array of 10 items, array+10 would work. It's been argued a few times (by the committee, among others) whether &array[10] really causes undefined behavior or not (and if it does, whether it really should). The bottom line with it is that at least according to the current standards (both C and C++) it officially causes undefined behavior, but if there's a single compiler for which it actually doesn't work, nobody in any of the arguments has been able to find or cite it.

Edit: For once my memory was half correct -- this was (part of) an official Defect Report to the committee, and at least some committee members (e.g., Tom Plum) thought the wording had been changed so it would not cause undefined behavior. OTOH, the DR dates from 2000, and the status is still "Drafting", so it's open to question whether it's really fixed, or ever likely to be (I haven't looked through N3090/3092 to figure out).

In C99, however, it's clearly not undefined behavior.

Wednesday, June 2, 2021
 
rlanvin
answered 7 Months ago
44

Yes, this is perfectly legal - of course you wouldn't return that address from the function, because by the time foo returns, it's meaningless.

Sunday, August 15, 2021
 
IcedAnt
answered 4 Months ago
85

12.6.2, [class.base.init], item 6, lists the steps of initialization, and this is the final one:

Finally, the compound-statement of the constructor body is executed.

So once the body has executed, initialization is complete.

Wednesday, September 1, 2021
 
etsous
answered 4 Months ago
96

This change was a result of CWG 1769, addressing the status of exception objects bound to unnamed catch handler parameters:

catch (std::exception&) // <==
{ 
}

That is now a variable. This simplifies the conceptual model around exception objects.

The first sentence, which remained unchanged, is still the complete definition of the term variable.

Monday, November 1, 2021
 
Sheakspear Zitouni
answered 1 Month ago
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