HP/UX Itanium C comiler & std C
Sion Roberts
this situation where the compiler acts differently to any other UNIX
C/C++ compiler that I have come across in the last 10 years (including
other HP-UX platforms).
Consider the following code:
#include <stdio.h>
int main()
{
char *ptr = "AB";
char *ans;
ans = (*ptr++) ? ptr : "";
printf ("Answer is %s\n", ans);
return 0;
}
When running this, I would usually expect to see:
Answer is B
On Itanium it displays:
Answer is AB
There are a few HPUX compilers to choose from on this platform, but
they all behave in the same way. Compiler switches don't make any
difference. I am now using GCC, which works as I expect.
aCC on HPUX 11 PA-RISC also displays "AB".
Has anyone come across this before or confirm that the results I've
seen are invalid for ANSI C?
Sion.
Mark A. Odell
news:ae5dbf8a.04111...@posting.google.com:
> While porting our application to HP/UX 11.23 Itanium, I came across
> this situation where the compiler acts differently to any other UNIX
> C/C++ compiler that I have come across in the last 10 years (including
> other HP-UX platforms).
>
> Consider the following code:
>
> #include <stdio.h>
>
> int main()
> {
> char *ptr = "AB";
> char *ans;
>
> ans = (*ptr++) ? ptr : "";
>
> printf ("Answer is %s\n", ans);
>
> return 0;
> }
>
> When running this, I would usually expect to see:
>
> Answer is B
No, not unless you used %c instead of %s.
>
> On Itanium it displays:
>
> Answer is AB
This is correct. Strings are null terminated. The char 'B' is not '\0'
thus it gets printed along with 'A'.
--
- Mark ->
--
Kenny McCormack
Mark A. Odell <odel...@hotmail.com> wrote:
>srob...@transoft.com (Sion Roberts) wrote in
>news:ae5dbf8a.04111...@posting.google.com:
>
>> While porting our application to HP/UX 11.23 Itanium, I came across
>> this situation where the compiler acts differently to any other UNIX
>> C/C++ compiler that I have come across in the last 10 years (including
>> other HP-UX platforms).
>>
>> Consider the following code:
>>
>> #include <stdio.h>
>>
>> int main()
>> {
>> char *ptr = "AB";
>> char *ans;
>>
>> ans = (*ptr++) ? ptr : "";
>>
>> printf ("Answer is %s\n", ans);
>>
>> return 0;
>> }
>>
>> When running this, I would usually expect to see:
>>
>> Answer is B
>
>No, not unless you used %c instead of %s.
Um, %c would print the lowest 8 bits of the pointer value stored in "ans",
as an ASCII character (on most machines - obviously O/T, of course...)
>> On Itanium it displays:
>>
>> Answer is AB
>
>This is correct. Strings are null terminated. The char 'B' is not '\0'
>thus it gets printed along with 'A'.
The idea of the above is that in:
ans = (*ptr++) ? ptr : "";
"ptr" is "supposed" to be incremented after it is tested but before its
value is fetched and assigned to "ans". The pedants in this group will
probably point out that this behavior is not required by "the standard".
Lawrence Kirby
> While porting our application to HP/UX 11.23 Itanium, I came across this
> situation where the compiler acts differently to any other UNIX C/C++
> compiler that I have come across in the last 10 years (including other
> HP-UX platforms).
>
> Consider the following code:
>
> #include <stdio.h>
>
> int main()
> {
> char *ptr = "AB";
> char *ans;
>
> ans = (*ptr++) ? ptr : "";
>
> printf ("Answer is %s\n", ans);
>
> return 0;
> }
>
> When running this, I would usually expect to see:
>
> Answer is B
Correct, the ?: operator defines a sequence point after its first operand
is evaluated. Therefore the side-effect of incrementing ptr must be
complete before the value of ptr is read for the second operand (which it
is because *ptr++ is 'A' which is not null). So a pointer to the 2nd
character in the string should be assigned to ans.
> On Itanium it displays:
>
> Answer is AB
Looks like you've found a bona fide compiler bug.
Lawrence
pete
>
> srob...@transoft.com (Sion Roberts) wrote in
> news:ae5dbf8a.04111...@posting.google.com:
>
> > While porting our application to HP/UX 11.23 Itanium, I came across
> > this situation where the compiler acts differently to any other UNIX
> > C/C++ compiler that I have come across in the last 10 years (including
> > other HP-UX platforms).
> >
> > Consider the following code:
> >
> > #include <stdio.h>
> >
> > int main()
> > {
> > char *ptr = "AB";
> > char *ans;
> >
> > ans = (*ptr++) ? ptr : "";
> >
> > printf ("Answer is %s\n", ans);
> >
> > return 0;
> > }
> >
> > When running this, I would usually expect to see:
> >
> > Answer is B
>
> No, not unless you used %c instead of %s.
Answer is B.
ans points to a string regardless of whether it points
to the first or second element of "AB".
--
pete
Bjørn Augestad
> On Mon, 15 Nov 2004 05:16:00 -0800, Sion Roberts wrote:
>
>
>>While porting our application to HP/UX 11.23 Itanium, I came across this
>>situation where the compiler acts differently to any other UNIX C/C++
>>compiler that I have come across in the last 10 years (including other
>>HP-UX platforms).
>>
>>Consider the following code:
>>
>> #include <stdio.h>
>>
>> int main()
>> {
>> char *ptr = "AB";
>> char *ans;
>>
>> ans = (*ptr++) ? ptr : "";
>>
>> printf ("Answer is %s\n", ans);
>>
>> return 0;
>> }
>>
>>When running this, I would usually expect to see:
>>
>> Answer is B
>
>
> Correct, the ?: operator defines a sequence point after its first operand
> is evaluated. Therefore the side-effect of incrementing ptr must be
> complete before the value of ptr is read for the second operand (which it
> is because *ptr++ is 'A' which is not null). So a pointer to the 2nd
> character in the string should be assigned to ans.
Can someone please clarify what is meant by this text, from C99 §6.5.15
Conditional operator, #4.
"If an attempt is made to modify the result of a conditional operator or
to access it after the next sequence point, the behavior is undefined".
Does the "access it" part apply to the OP's code or am I just
misinterpreting the standard?
TIA
Bjørn
>
[snip]
pete
I think it means that (a ? b : c) is an rvalue
and that you can't do something like this: ((a ? b : c)++)
> or to access it after the next sequence point,
> the behavior is undefined".
> Does the "access it" part apply to the OP's code or am I just
> misinterpreting the standard?
I don't know what that means.
--
pete
Richard Tobin
Kenny McCormack <gaz...@interaccess.com> wrote:
>The idea of the above is that in:
>
> ans = (*ptr++) ? ptr : "";
>
>"ptr" is "supposed" to be incremented after it is tested but before its
>value is fetched and assigned to "ans". The pedants in this group will
>probably point out that this behavior is not required by "the standard".
No. The standard says there is a sequence point after the first operand
of the ? operator.
-- Richard
Lawrence Kirby
(rearranged to bring the standard text together)
>> "If an attempt is made to modify the result of a conditional
>> operator or to access it after the next sequence point,
>> the behavior is undefined".
>
> I think it means that (a ? b : c) is an rvalue
> and that you can't do something like this: ((a ? b : c)++)
That is true but I don't think it is what the text is for, because that is
caught by a constraint violation for ++ i.e. its operand must be a
modifiable lvalue. A constraint violation is the more significant error,
if you have a CV then further undefined behaviour is not an issue.
I'm wondering if this has something to do with "rvalue" structures e.g.
struct foo { int bar; } s1 = { 1 }, s2 = { 2 };
(expr ? s1 : s2).bar
but I haven't come up with an example that triggers undefined behaviour
from 6.5.15p4 and doesn't trip up elsewhere in the standard. That sentence
is an addition from C90, so either they spotted a case that hadn't been
spotted at the time of C90 or this has something to do with new C99
features.
>> Does the "access it" part apply to the OP's code or am I just
>> misinterpreting the standard?
>
> I don't know what that means.
The standard talks about modifying the RESULT of the conditional
operator or accessing after the next sequence point, which the OPs code
does not do.
Lawrence
Mark A. Odell
news:Xns95A25B2C33A1FC...@130.133.1.4:
>> Consider the following code:
>>
>> #include <stdio.h>
>>
>> int main()
>> {
>> char *ptr = "AB";
>> char *ans;
>>
>> ans = (*ptr++) ? ptr : "";
>>
>> printf ("Answer is %s\n", ans);
>>
>> return 0;
>> }
>>
>> When running this, I would usually expect to see:
>>
>> Answer is B
>
> No, not unless you used %c instead of %s.
Ignore me. I missed the increment on 'ptr' and the lack of dereference on
'ans'.
Dennis Handly
: In article <cnadav$b4g$1...@yin.interaccess.com>,
Right, we finally saw that too. This is CR JAGae79312:
Incorrect order of evaluation for: (x++) ? x ...
This is fixed in A.05.57:
http://h21007.www2.hp.com/dspp/tech/tech_TechSoftwareDetailPage_IDX/1,1703,1743,00.html
Christian Bau
gaz...@yin.interaccess.com (Kenny McCormack) wrote:
> The idea of the above is that in:
>
> ans = (*ptr++) ? ptr : "";
>
> "ptr" is "supposed" to be incremented after it is tested but before its
> value is fetched and assigned to "ans". The pedants in this group will
> probably point out that this behavior is not required by "the standard".
Only pedants who don't know C. Those who know C will replace "supposed"
with "required" and remove the quotes:
"ptr" is absolutely, one hundred percent required to be incremented
Kenny McCormack
Well, that is good to know. Always nice to hear of instances where
sensible behavior is required by the standard.
pete
> >> >>Consider the following code:
> >> >>
> >> >> #include <stdio.h>
> >> >>
> >> >> int main()
> >> >> {
> >> >> char *ptr = "AB";
> >> >> char *ans;
> >> >>
> >> >> ans = (*ptr++) ? ptr : "";
> >> >>
> >> >> printf ("Answer is %s\n", ans);
> >> >>
> >> >> return 0;
> >> >> }
> >> "If an attempt is made to modify the result of a conditional
> >> operator or to access it after the next sequence point,
> >> the behavior is undefined".
> >> Does the "access it" part apply to the OP's code or am I just
> >> misinterpreting the standard?
> >
> > I don't know what that means.
>
> The standard talks about modifying the RESULT of the conditional
> operator or accessing after the next sequence point,
> which the OPs code does not do.
I still don't understand.
Would you please change the code in the above program to something
which does attempt to access the result of the conditional
operator after the next sequence point,
to better illustrate what you and the standard are talking about?
--
pete
pete
> > ans = (*ptr++) ? ptr : "";
> "ptr" is absolutely, one hundred percent required to be incremented
> after it is tested but before its value is fetched and assigned to
> "ans".
I don't see ptr being tested.
*ptr++ is tested.
ptr may be incremented before the test instead of after.
It is the result of the postfix increment operation
which is the operand of the indirection operator. *(ptr++)
--
pete
Christian Bau
pete <pfi...@mindspring.com> wrote:
It is a very obscure thing, and difficult to write code that actually
creates the situation. Something like this:
typedef struct { int x; int a [2]; } mystruct;
static mystruct r, s;
static int i;
(i > 0 ? r : s).a [1] = 0;
The result of the ?: is some object, but not one of the objects r, s: It
is an unnamed object which contains either a copy of r or a copy of s.
Modifying that unnamed object invokes undefined behavior. You may have
to do some more work to make this compile.
Sion Roberts
Dennis,
Thank you for this information. This is exactly what I wanted to know.
Sion.
Lawrence Kirby
> Christian Bau wrote:
>
>> > ans = (*ptr++) ? ptr : "";
>
>> "ptr" is absolutely, one hundred percent required to be incremented
>> after it is tested but before its value is fetched and assigned to
>> "ans".
>
> I don't see ptr being tested.
True, although its value is used in the text expression.
> *ptr++ is tested.
True
> ptr may be incremented before the test instead of after.
True
> It is the result of the postfix increment operation
> which is the operand of the indirection operator. *(ptr++)
True, and the result of that operation is the unincremented value of ptr.
But you are correct that this doesn't stop the ptr object having an
incremented value written to it first.
Lawrence
Charlie Gordon
news:christian.bau-2BD...@slb-newsm1.svr.pol.co.uk...
> It is a very obscure thing, and difficult to write code that actually
> creates the situation. Something like this:
>
> typedef struct { int x; int a [2]; } mystruct;
> static mystruct r, s;
> static int i;
>
> (i > 0 ? r : s).a [1] = 0;
>
> The result of the ?: is some object, but not one of the objects r, s: It
> is an unnamed object which contains either a copy of r or a copy of s.
> Modifying that unnamed object invokes undefined behavior. You may have
> to do some more work to make this compile.
I guess the "proper" way to achieve this would be :
(i > 0 ? &r : &s)->a[1] = 0;
Are we dealing with the same issue in the code :
(ignore_case ? strcasecmp : strcmp) (s1, s2);
assuming proper declaration of strcasecmp.
How should it be written then ?
Chqrlie.
PS: I put emphasis around proper to express that standard compliant does not
necessarily mean readable and advisable.
pete
Thank you.
--
pete
Dan Pop
>"Christian Bau" <christ...@cbau.freeserve.co.uk> wrote in message
>news:christian.bau-2BD...@slb-newsm1.svr.pol.co.uk...
>> It is a very obscure thing, and difficult to write code that actually
>> creates the situation. Something like this:
>>
>> typedef struct { int x; int a [2]; } mystruct;
>> static mystruct r, s;
>> static int i;
>>
>> (i > 0 ? r : s).a [1] = 0;
>>
>> The result of the ?: is some object, but not one of the objects r, s: It
>> is an unnamed object which contains either a copy of r or a copy of s.
>> Modifying that unnamed object invokes undefined behavior. You may have
>> to do some more work to make this compile.
>
>I guess the "proper" way to achieve this would be :
>
>(i > 0 ? &r : &s)->a[1] = 0;
>
>Are we dealing with the same issue in the code :
>
>(ignore_case ? strcasecmp : strcmp) (s1, s2);
>
>assuming proper declaration of strcasecmp.
Nope. You're not attempting to modify the result of the conditional
operator.
>How should it be written then ?
Apart from stylistic issues, your version is correct. In real code,
you may want to either use a function pointer (if ignore_case doesn't
change its value in "random" places during normal program execution)
or to hide the conditional operator behind a macro (if it does).
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Dan...@ifh.de
Currently looking for a job in the European Union