HP/UX Itanium C comiler & std C

[Sion Roberts]

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

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]

sr******@transo ft.com (Sion Roberts) wrote in
news:ae******** *************** ***@posting.goo gle.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]

In article <Xn************ *************** *****@130.133.1 .4>,
Mark A. Odell <od*******@hotm ail.com> wrote:

sr******@trans oft.com (Sion Roberts) wrote in
news:ae******* *************** ****@posting.go ogle.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]

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.
On Itanium it displays:

Answer is AB

Looks like you've found a bona fide compiler bug.

Lawrence

[pete]

Mark A. Odell wrote:

sr******@transo ft.com (Sion Roberts) wrote in
news:ae******** *************** ***@posting.goo gle.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]

Lawrence Kirby wrote:

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]

Bjørn Augestad wrote:

Lawrence Kirby wrote:

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

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]

In article <cn**********@y in.interaccess. com>,
Kenny McCormack <ga*****@intera ccess.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]

On Mon, 15 Nov 2004 14:41:45 +0000, pete wrote:

Bjørn Augestad wrote:

Lawrence Kirby wrote:

> 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.

(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]

"Mark A. Odell" <od*******@hotm ail.com> wrote in
news:Xn******** *************** *********@130.1 33.1.4:

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'.

--
- Mark ->
--