Implementation-defined behaviour

[Ioannis Vranos]

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;
}

[santosh]

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]

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]

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
es*****@ieee-dot-org.invalid

[Ioannis Vranos]

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.

[santosh]

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.

[Ian Collins]

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.

[Ioannis Vranos]

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?

[Dann Corbit]

"Ioannis Vranos" <iv*****@nospam .no.spamfreemai l.grwrote in message
news:fs******** ***@ulysses.noc .ntua.gr...

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.

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

[Andrey Tarasevich]

Ioannis Vranos wrote:

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

'printf("%d "%d\n", x++, x++)' invokes _undefined_ behavior. 'printf("%d
"%d\n", x++, x)' also invokes _undefined_ behavior.

'printf("%d\n", x++)' does not invoke undefined behavior.

"Implementa tion-defined behavior" is not immediately relevant to the
nature of your question.

--
Best regards,
Andrey Tarasevich