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AFAIK the following is implementation-defined behaviour, am I right?:


#include <stdio.h>


int main(void)
{
int n= 0;

printf("%d\n", n++);


return 0;
}

Ioannis Vranos wrote:

> AFAIK the following is implementation-defined behaviour, am I right?:
>
> #include <stdio.h>
>
> int main(void)
> {
> int n= 0;
> printf("%d\n", n++);
> return 0;
> }

I don't think so.

Ioannis Vranos wrote:
> AFAIK the following is implementation-defined behaviour, am I right?:
>
>
> #include <stdio.h>
>
>
> int main(void)
> {
> int n= 0;
>
> printf("%d\n", n++);
>
>
> return 0;
> }

Only by quibbles: the form of "successful termination"
returned to the host environment is implementation-defined,
the actual new-line representation written to stdout in
response to the '\n' is implementation-defined, and things
of that sort.

What aspect do you believe is not Standard-defined?

--
Eric Sosman
lid

santosh wrote:
> Ioannis Vranos wrote:
>
>> AFAIK the following is implementation-defined behaviour, am I right?:
>>
>> #include <stdio.h>
>>
>> int main(void)
>> {
>> int n= 0;
>> printf("%d\n", n++);
>> return 0;
>> }
>
> I don't think so.

There are similar examples at 2.12 of K&R2.


Two quotes from there:

"printf("%d %d\n", ++n, power(2, n)); /* WRONG */"


"One unhappy situation is typified by the statement

a[i]= i++;"

Eric Sosman wrote:
> Ioannis Vranos wrote:
>> AFAIK the following is implementation-defined behaviour, am I right?:
>>
>>
>> #include <stdio.h>
>>
>>
>> int main(void)
>> {
>> int n= 0;
>>
>> printf("%d\n", n++);
>>
>>
>> return 0;
>> }
>
> Only by quibbles: the form of "successful termination"
> returned to the host environment is implementation-defined,
> the actual new-line representation written to stdout in
> response to the '\n' is implementation-defined, and things
> of that sort.
>
> What aspect do you believe is not Standard-defined?


I am talking about the implementation-defined behaviour of the printf()
call described at 2.12 of K&R2.

Ioannis Vranos wrote:

> santosh wrote:
>> Ioannis Vranos wrote:
>>
>>> AFAIK the following is implementation-defined behaviour, am I
>>> right?:
>>>
>>> #include <stdio.h>
>>>
>>> int main(void)
>>> {
>>> int n= 0;
>>> printf("%d\n", n++);
>>> return 0;
>>> }
>>
>> I don't think so.
>
> There are similar examples at 2.12 of K&R2.
>
>
> Two quotes from there:
>
> "printf("%d %d\n", ++n, power(2, n)); /* WRONG */"
>
>
> "One unhappy situation is typified by the statement
>
> a[i]= i++;"

Well these examples are different from what you have shown above. The
first example above invokes unspecified behaviour while the second one
invokes undefined behaviour. The example in your first post does
neither as far as I can see.

Ioannis Vranos wrote:
> Eric Sosman wrote:
>> Ioannis Vranos wrote:
>>> AFAIK the following is implementation-defined behaviour, am I right?:
>>>
>>>
>>> #include <stdio.h>
>>>
>>>
>>> int main(void)
>>> {
>>> int n= 0;
>>>
>>> printf("%d\n", n++);
>>>
>>>
>>> return 0;
>>> }
>> Only by quibbles: the form of "successful termination"
>> returned to the host environment is implementation-defined,
>> the actual new-line representation written to stdout in
>> response to the '\n' is implementation-defined, and things
>> of that sort.
>>
>> What aspect do you believe is not Standard-defined?
>
>
> I am talking about the implementation-defined behaviour of the printf()
> call described at 2.12 of K&R2.
>
There's nothing wrong with your example, the one you cite is completely
different. If you had tried to use n again in the printf, you would hit UB.

--
Ian Collins.

Ian Collins wrote:
> Ioannis Vranos wrote:
>> Eric Sosman wrote:
>>> Ioannis Vranos wrote:
>>>> AFAIK the following is implementation-defined behaviour, am I right?:
>>>>
>>>>
>>>> #include <stdio.h>
>>>>
>>>>
>>>> int main(void)
>>>> {
>>>> int n= 0;
>>>>
>>>> printf("%d\n", n++);
>>>>
>>>>
>>>> return 0;
>>>> }
>>> Only by quibbles: the form of "successful termination"
>>> returned to the host environment is implementation-defined,
>>> the actual new-line representation written to stdout in
>>> response to the '\n' is implementation-defined, and things
>>> of that sort.
>>>
>>> What aspect do you believe is not Standard-defined?
>>
>> I am talking about the implementation-defined behaviour of the printf()
>> call described at 2.12 of K&R2.
>>
> There's nothing wrong with your example, the one you cite is completely
> different. If you had tried to use n again in the printf, you would hit UB.


So, printf("%d "%d\n", x++, x++); invokes implementation-defined
behaviour, while printf("%d\n", x++); doesn't invoke
implementation-defined behaviour?

"Ioannis Vranos" <> wrote in message
news:fshjmg$1avp$...
> AFAIK the following is implementation-defined behaviour, am I right?:
>
>
> #include <stdio.h>
>
>
> int main(void)
> {
> int n= 0;
>
> printf("%d\n", n++);
>
>
> return 0;
> }

C:\tmp>lin foo.c

C:\tmp>"C:\Lint\Lint-nt" +v -i"C:\Lint" std.lnt -os(_LINT.TMP) foo.c
PC-lint for C/C++ (NT) Vers. 8.00u, Copyright Gimpel Software 1985-2006

--- Module: foo.c (C)

C:\tmp>type _LINT.TMP | more

--- Module: foo.c (C)
_
printf("%d %d\n", n++, n);
foo.c( : Warning 564: variable 'n' depends on order of evaluation

---
output placed in _LINT.TMP

C:\tmp>splint foo.c
Splint 3.1.1 --- 12 Mar 2007

foo.c: (in function main)
foo.c(8,24): Argument 2 modifies n, used by argument 3 (order of evaluation
of
actual parameters is undefined): printf("%d %d\n", n++, n)
Code has unspecified behavior. Order of evaluation of function parameters
or
subexpressions is not defined, so if a value is used and modified in
different places not separated by a sequence point constraining evaluation
order, then the result of the expression is unspecified. (Use -evalorder
to
inhibit warning)

Finished checking --- 1 code warning

C:\tmp>type foo.c
#include <stdio.h>
int main(void)
{
int n = 0;
/* Not a problem, function call is a sequence point: */
printf("%d\n", n++);
/* Problem here, order unspecified */
printf("%d %d\n", n++, n);
return 0;
}



--
Posted via a free Usenet account from http://www.teranews.com

"Ioannis Vranos" <> wrote in message
news:fsho27$1o9d$...
> Ian Collins wrote:
>> Ioannis Vranos wrote:
>>> Eric Sosman wrote:
>>>> Ioannis Vranos wrote:
>>>>> AFAIK the following is implementation-defined behaviour, am I right?:
>>>>>
>>>>>
>>>>> #include <stdio.h>
>>>>>
>>>>>
>>>>> int main(void)
>>>>> {
>>>>> int n= 0;
>>>>>
>>>>> printf("%d\n", n++);
>>>>>
>>>>>
>>>>> return 0;
>>>>> }
>>>> Only by quibbles: the form of "successful termination"
>>>> returned to the host environment is implementation-defined,
>>>> the actual new-line representation written to stdout in
>>>> response to the '\n' is implementation-defined, and things
>>>> of that sort.
>>>>
>>>> What aspect do you believe is not Standard-defined?
>>>
>>> I am talking about the implementation-defined behaviour of the printf()
>>> call described at 2.12 of K&R2.
>>>
>> There's nothing wrong with your example, the one you cite is completely
>> different. If you had tried to use n again in the printf, you would hit
>> UB.
>
>
> So, printf("%d %d\n", x++, x++); invokes implementation-defined
so does:
printf("%d %d\n", x++, x);

> behaviour, while printf("%d\n", x++); doesn't invoke
> implementation-defined behaviour?

This one does not.



--
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