Legality of pointer arithmetic

[houstorx]

I've been looking at the committee draft of the C99 specification,
specifically the one at this URI:
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n843.pdf. I don't have
a copy of the official international standard, but I assume that it is
similar.

The rules for pointer arithmetic (6.5.6.7-8) are appended below for
ease of reference. They seem astonishingly restrictive. As far as I can
see, there is no guarantee that the code:

char *p = malloc(10); /* assume this malloc succeeds */
p[2] = 'x';

will write an 'x' into the third byte of the allocated store; in fact,
the behaviour of that code is completely undefined.

(The expression p[2] is defined to be equivalent to *(p+2); p+2 is not
part of the same array object as p, hence the behaviour is undefined.)

Can that possibly be right? Am I misreading the spec, is the spec
misrepresenting the language (or is there another possiblity)?

In anticipation of your collective wisdom,
Robin

From section 6.5.6 "Additive operators":

7 For the purposes of these operators, a pointer to a nonarray object
behaves the same as a pointer to the first element of an array of
length one with the type of the object as its element type.

8 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.

 

[Robert Gamble]

I've been looking at the committee draft of the C99 specification,
specifically the one at this URI:
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n843.pdf. I don't have
a copy of the official international standard, but I assume that it is
similar.

The most recent draft is n869 available at:
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n869/

The rules for pointer arithmetic (6.5.6.7-8) are appended below for
ease of reference. They seem astonishingly restrictive. As far as I can
see, there is no guarantee that the code:

char *p = malloc(10); /* assume this malloc succeeds */
p[2] = 'x';

will write an 'x' into the third byte of the allocated store; in fact,
the behaviour of that code is completely undefined.

(The expression p[2] is defined to be equivalent to *(p+2); p+2 is not
part of the same array object as p, hence the behaviour is undefined.)

7.20.3p1 (malloc):
'The pointer returned if the allocation succeeds is suitably aligned so
that it may be assigned to a pointer to any type of object and then
used to access such an object or an array of such objects in the space
allocated'

Robert Gamble

 

[Eric Sosman]

I've been looking at the committee draft of the C99 specification,
specifically the one at this URI:
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n843.pdf. I don't have
a copy of the official international standard, but I assume that it is
similar.

The rules for pointer arithmetic (6.5.6.7-8) are appended below for
ease of reference. They seem astonishingly restrictive. As far as I can
see, there is no guarantee that the code:

char *p = malloc(10); /* assume this malloc succeeds */
p[2] = 'x';

will write an 'x' into the third byte of the allocated store; in fact,
the behaviour of that code is completely undefined.

(The expression p[2] is defined to be equivalent to *(p+2); p+2 is not
part of the same array object as p, hence the behaviour is undefined.)
[...]

You've overlooked 7.20.3: "[...] The pointer returned [...]
may be assigned to a pointer to any type of object and then used
to access such an object *or an array of such objects* [...]"
(Emphasis mine.)

 

[CBFalconer]

The most recent draft is n869 available at:
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n869/

try: <http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf>

which is the 2005 draft incorporating the corrigendae.
Unfortunately there is no text version.

 

[houstorx]

The most recent draft is n869 available at:
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n869/

That's useful to know. Thanks. (Also thanks to CBFalconer for the link
to the 2005 draft.)

7.20.3p1 (malloc):
'The pointer returned if the allocation succeeds is suitably aligned so
that it may be assigned to a pointer to any type of object and then
used to access such an object or an array of such objects in the space
allocated'

Ah yes, I overlooked that sentence. Thanks again, and to Eric Sosman
for pointing out the same thing.

Robin

 

[Tim Rentsch]

try: <http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf>

which is the 2005 draft incorporating the corrigendae.
Unfortunately there is no text version.

I believe the plural of corrigendum is corrigenda,
not corrigendae.

 

[CBFalconer]

I believe the plural of corrigendum is corrigenda, not corrigendae.

By getting abysmal marks in latin in grade 10, I managed to
persuade my parents to let me replace it with mathematics. The
effects still linger.

 

[Tim Rentsch]

By getting abysmal marks in latin in grade 10, I managed to
persuade my parents to let me replace it with mathematics. The
effects still linger.

Lucky you. I had to get poor marks in latin for two years
before I could stop taking it.

In this case, it wasn't my HS latin that led me to look up
corrigendum/corrigenda, but a parallel construction with
agendum/agenda. This small but interesting item is one of
many that came from reading "Quiddities", by W. V. Quine. I
thoroughly recommend it.

"I have been accused of denying consciousness, but I'm not
conscious of having done so." - W. V. Quine, in "Quiddities"

 

[lawrence.jones]

I believe the plural of corrigendum is corrigenda,
not corrigendae.

In English, the plural is corrigendums.

-Larry Jones

Oh, now don't YOU start on me. -- Calvin

 

[Old Wolf]

In English, the plural is corrigendums.

Which version of English are you referring to?

http://dictionary.reference.com/search?q=corrigendum

 

[Eric Sosman]

Which version of English are you referring to?

The corrigendified version, obviouslywise.