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supercat@casperkitty.com writes:
> On Wednesday, February 21, 2018 at 5:51:37 PM UTC-6, Keith Thompson wrote:
>> supercat@casperkitty.com writes:
>> > The code above would create an array object sized to hold a total of 25
>> > values of type "int". Assume "int" is 4 bytes; the size of the object
>> > would be 100 bytes.
>>
>> Yes, it would create an array object that contains 25 int objects, but
>> not an array object of type int[25]. Each of the 5 elements of a is of
>> type int[5].
>
> There would be a 100-byte array object.
Yes, you already said that, and I already agreed.
>> > According to C89 3.3.6, the subexpression a[1] would multiply the value in
>> > the brackets (i.e. 1) by the size of *a [i.e. 20], and add that result to
>> > the address of the aforementioned array object. Then applying [7] to that
>> > would multiply the value 7 by the size of **a [i.e. 4] and that to the
>> > just-computed address, yielding an address 48 bytes past the start of the
>> > 100-byte array object.
>>
>> Yes, the C89 definition of pointer arithmetic refers to multiplication.
>> (In my opinion the current definition, which does not, is better.)
>
> So you acknowledge that there was a change.
Of course, the wording was changed. (In my opinion it was an
improvement; C89's reference to multiplication was imprecise.)
I'm not aware that the changed wording changed the semantics, but I
haven't studied it closely enough to be sure. Do you think the
semantics changed?
>> C89 3.3.6:
>> When an expression that has integral type is added to
>> or subtracted from a pointer, the integral value is first
>> multiplied by the size of the object pointed to. The result
>> has the type of the pointer operand. If the pointer operand
>> points to a member of an array object, and the array object is
>> large enough, the result points to a member of the same array
>> object, appropriately offset from the original member.
>>
>> For a[1][7] to work the way you want it to, there would have be an array
>> object of type int[N], with N > 7. (The most straightforward way to
>> have this would be to treat all of a as an array object of type
>> int[25].) The elements of a are not int objects; they're array objects
>> of type int[5].
>
> The Standard could have been clearer if it didn't overload the term
> "object" so much. The Standard clearly intends that the code:
>
> int arr[5], p=a+5, q=p-1;
>
> should have defined behavior even though "p" doesn't point to any object;
> thus "the size of the object pointed to" presumably means "the size of
> the lvalue yielded by dereferencing a pointer of that type".
C11 (N1570 6.5.6) is clearer on this point, though the wording IMHO
could still be fixed up a bit. It discusses pointer arithmetic where
the pointer operand points just past the end of an array.
> I would suggest that the Standard also clearly intends that the code:
>
> int *p = malloc(3 * sizeof (int));
> if (p) p[2] = 5;
>
> should have defined behavior even though there is no int[]. The malloc()
> function would be pretty well useless if its return pointer couldn't be
> treated, for purposes of pointer arithmetic, as though it pointed to an
> array object.
I think the effective type rules cover this. (C89/C90 didn't discuss
effective types.)
>> I didn't ask whether a such an implementation would be useful. I asked
>> what (normative) clause of C90 it would violate.
>
> The one which describes how pointer arithmetic works. There exists an
> array object. The lvalue a[1] points to four bytes within it.
lvalues don't point. The lvalue a[1] is an expression that designates
an object of type int[5]. Since it's an array expression, it's
converted to a pointer expression which is no longer an lvalue. The
value of that pointer expression is the address of an int object.
> Adding
> 28 to the address of a[1] would yield the address of different four
> bytes within that same array object.
What array object are referring to? You seem to be assuming the
existence of an object of type int[25], but you haven't demonstrated
where the C89 standard says that such an object exists, or that a can be
accessed as if it were such an object.
"If the pointer operand points to a member of an array object"
The array object in question is a[1]. The member in question is
a[1][0].
"and the array object is large enough,"
The array object a[1] has only 5 members, so it isn't large enough.
--
Keith Thompson (The_Other_Keith) kst-u@mib.org <http://www.ghoti.net/~kst>
Working, but not speaking, for JetHead Development, Inc.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
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Re: Opinion sought: Is this strictly conforming C? supercat@casperkitty.com - 2018-02-28 09:03 -0800
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Re: Opinion sought: Is this strictly conforming C? supercat@casperkitty.com - 2018-03-01 12:31 -0800
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Re: Opinion sought: Is this strictly conforming C? supercat@casperkitty.com - 2018-02-26 10:17 -0800
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