printf("%d", INT_MAX);

[Yevgen Muntyan]

Hey,

It was mentioned elsewhere that printf("%d", INT_MAX);
is implementation-defined. Why? Is it because INT_MAX
value is implementation-defined so output depends
on implementation, or is it something more subtle so
that output of the following is also implementation-defined:

if (INT_MAX == 65535)
printf("%d", INT_MAX);
else
printf("%d", 65535);

Thanks,
Yevgen

[Yevgen Muntyan]

Yevgen Muntyan wrote:

Hey,

It was mentioned elsewhere that printf("%d", INT_MAX);
is implementation-defined. Why? Is it because INT_MAX
value is implementation-defined so output depends
on implementation, or is it something more subtle so
that output of the following is also implementation-defined:

if (INT_MAX == 65535)
printf("%d", INT_MAX);
else
printf("%d", 65535);

Um, I again made it overcomplicated . Is this implementation-defined:

printf("%d\n", 65535);

(if \n matters, then we print \n after the first piece of code too,
it's there in main()).

Yevgen

[Yevgen Muntyan]

Yevgen Muntyan wrote:

Yevgen Muntyan wrote:

>Hey,

It was mentioned elsewhere that printf("%d", INT_MAX);
is implementation-defined. Why? Is it because INT_MAX
value is implementation-defined so output depends
on implementation, or is it something more subtle so
that output of the following is also implementation-defined:

if (INT_MAX == 65535)
printf("%d", INT_MAX);
else
printf("%d", 65535);

Um, I again made it overcomplicated . Is this implementation-defined:

printf("%d\n", 65535);

After consulting the standard, the above should be

printf("%d\n", 32767);

Oh well.

Yevgen

[Eric Sosman]

Yevgen Muntyan wrote On 03/14/07 13:29,:

Yevgen Muntyan wrote:

>>Yevgen Muntyan wrote:

>>>Hey,

It was mentioned elsewhere that printf("%d", INT_MAX);
is implementation-defined. Why? Is it because INT_MAX
value is implementation-defined so output depends
on implementation, or is it something more subtle so
that output of the following is also implementation-defined:

if (INT_MAX == 65535)
printf("%d", INT_MAX);
else
printf("%d", 65535);

Um, I again made it overcomplicated . Is this implementation-defined:

printf("%d\n" , 65535);

After consulting the standard, the above should be

printf("%d\n", 32767);

It seems there are several questions here.

The value of INT_MAX is implementation-defined, so the
output produced by printf("%d\n", INT_MAX) is implementation-
defined. There is nothing "wrong" with the operation, but
a strictly-conforming program must not perform it.

The effect of printf("%d\n", 65535) is implementation-
UNdefined (if I may coin a term). If the implementation-
defined value of INT_MAX is >= 65535, then the output is
well defined. If INT_MAX < 65535, the constant 65535 is
of type unsigned int and the behavior is undefined because
"%d" requires a signed int. Hence, it is implementation-
defined whether the behavior is well defined or undefined.

The effect of printf("%d\n", 32767) is the same on all
hosted implementations (barring I/O errors).

--
Er*********@sun .com

[Yevgen Muntyan]

Eric Sosman wrote:

Yevgen Muntyan wrote On 03/14/07 13:29,:

>Yevgen Muntyan wrote:

>>Yevgen Muntyan wrote:

Hey,

It was mentioned elsewhere that printf("%d", INT_MAX);
is implementation-defined. Why? Is it because INT_MAX
value is implementation-defined so output depends
on implementation, or is it something more subtle so
that output of the following is also implementation-defined:

if (INT_MAX == 65535)
printf("%d", INT_MAX);
else
printf("%d", 65535);
Um, I again made it overcomplicated . Is this implementation-defined:

printf("%d\n" , 65535);

After consulting the standard, the above should be

printf("%d\n ", 32767);

It seems there are several questions here.

The value of INT_MAX is implementation-defined, so the
output produced by printf("%d\n", INT_MAX) is implementation-
defined. There is nothing "wrong" with the operation, but
a strictly-conforming program must not perform it.

Sure.

The effect of printf("%d\n", 65535) is implementation-
UNdefined (if I may coin a term). ...

I thought (wrongly) 65535 was minimal value for INT_MAX.

[snip]

The effect of printf("%d\n", 32767) is the same on all
hosted implementations (barring I/O errors).

And this is what I asked about. I wanted to make sure there
is nothing special about passing INT_MAX as int to a variadic
function or something.
This printf thing was mentioned in connection to another instance
of implementation-defined behavior, where any behavior other
than the "right" one would break code in a serious way, so I wanted
to clarify if this INT_MAX is a usual pedant's story or something
serious.

Thanks,
Yevgen

[Eric Sosman]

Yevgen Muntyan wrote On 03/14/07 13:59,:

Eric Sosman wrote:

> The effect of printf("%d\n", 32767) is the same on all
hosted implementations (barring I/O errors).

And this is what I asked about. I wanted to make sure there
is nothing special about passing INT_MAX as int to a variadic
function or something.

No, nothing special. INT_MAX must expand to an expression
whose type is `int'; an expansion with the right value but
some other type would be wrong. Since INT_MAX is already an
`int', no promotions occur. Similarly, `32767' is already an
`int' and nothing happens to it.

This printf thing was mentioned in connection to another instance
of implementation-defined behavior, where any behavior other
than the "right" one would break code in a serious way, so I wanted
to clarify if this INT_MAX is a usual pedant's story or something
serious.

Neither, I think, since there's nothing wrong with it.
My point about printf("%d\n", 65535) is purest pedantry, but
printf("%d\n", INT_MAX) is beyond reproach (despite not being
strictly conforming).

--
Er*********@sun .com

[CBFalconer]

Yevgen Muntyan wrote:

>

.... snip ...

>
Um, I again made it overcomplicated . Is this implementation-defined:

printf("%d\n", 65535);

Yes, because there is no guarantee that 65535 is an int. It may be
a long int. It is outside the guaranteed range of an int.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net>

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

[Yevgen Muntyan]

CBFalconer wrote:

Yevgen Muntyan wrote:
... snip ...

>Um, I again made it overcomplicated . Is this implementation-defined:

printf("%d\n ", 65535);

Yes, because there is no guarantee that 65535 is an int. It may be
a long int. It is outside the guaranteed range of an int.

It's not implementation-defined if INT_MAX is less than 65535, is it?
Anyway, as you probably know, it was a wrong question.

Yevgen

[Jack Klein]

On Wed, 14 Mar 2007 17:24:02 GMT, Yevgen Muntyan
<mu************ ****@tamu.eduwr ote in comp.lang.c:

Hey,

It was mentioned elsewhere that printf("%d", INT_MAX);
is implementation-defined. Why? Is it because INT_MAX
value is implementation-defined so output depends
on implementation, or is it something more subtle so
that output of the following is also implementation-defined:

if (INT_MAX == 65535)
printf("%d", INT_MAX);
else
printf("%d", 65535);

Thanks,
Yevgen

Aside from newline issue, that you posted a correction to, the snippet
above, when packaged within an appropriate program with inclusion of
<stdio.hand <limits.his not strictly conforming, as pointed out in
4 p5 of the C99 standard:

"A strictly conforming program shall use only those features of the
language and library specified in this International Standard. It
shall not produce output dependent on any unspecified, undefined, or
implementation-defined behavior, and shall not exceed any minimum
implementation limit."

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Kenneth Brody]

Eric Sosman wrote:
[...]

The value of INT_MAX is implementation-defined, so the
output produced by printf("%d\n", INT_MAX) is implementation-
defined. There is nothing "wrong" with the operation, but
a strictly-conforming program must not perform it.

[...]

So, technically speaking, the following is non-conforming?

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(void)
{
printf("INT_MAX is %d\n",INT_MAX);
return(EXIT_SUC CESS);
}

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer .h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:Th***** ********@gmail. com>