basic c i/o and EOF

[Dave]

Below is the code ive written just to count the characters typed in. I
assumed EOF is -1, so if i type -1 and then press enter shouldnt the
program end? It orks if i put something like 'q' in the while loop to
end the loop.

what is up?

<code>

#include <stdio.h>

void main() {

long nc;

nc = 0;
while (getchar() != 'EOF') {
++nc;
}
printf("%ld\n", nc);

}

</code>

thanks
Dave

[Tim Cambrant]

"Dave" <no*****@noemai l.com> skrev i meddelandet
news:bj******** **@titan.btinte rnet.com...

Below is the code ive written just to count the characters typed in. I
assumed EOF is -1, so if i type -1 and then press enter shouldnt the
program end?

No, because when you type -1 it isn't stored as a character, but two
characters '-' and '1'. '-1' however is a single character, and those two
values doesn't match.

However, you could do a test yourself, and check for a series of letters.
Should be quite easy for you to do. I believe the theory is to put all the
input characters into an array, and check if the last two characters in the
array is equal to '-' and '1'.

--
Tim Cambrant
<tim at cambrant dot com>

[Lew Pitcher]

On Thu, 11 Sep 2003 15:09:58 GMT, "Tim Cambrant" <ti*@cambrant.c om.net> wrote:

"Dave" <no*****@noemai l.com> skrev i meddelandet
news:bj******* ***@titan.btint ernet.com...

Below is the code ive written just to count the characters typed in. I
assumed EOF is -1, so if i type -1 and then press enter shouldnt the
program end?

No, because when you type -1 it isn't stored as a character, but two
characters '-' and '1'. '-1' however is a single character, and those two
values doesn't match.

However, you could do a test yourself, and check for a series of letters.
Should be quite easy for you to do. I believe the theory is to put all the
input characters into an array, and check if the last two characters in the
array is equal to '-' and '1'.

Or better yet, if your platform supports it, enter the key or key combination
that signals end-of-input-data.

On MSDOS-based systems, this would be the ^Z key combination
On Unix-based systems, this would be what ever combination generates the 'eof'
signal (typically ^D, but check your stty settings to be sure).
--
Lew Pitcher
IT Consultant, Enterprise Technology Solutions
Toronto Dominion Bank Financial Group

(Opinions expressed are my own, not my employers')

[Irrwahn Grausewitz]

Dave <no*****@noemai l.com> wrote:

Below is the code ive written just to count the characters typed in. I
assumed EOF is -1, so if i type -1 and then press enter shouldnt the
program end? Already addressed in the other replies.
It orks if i put something like 'q' in the while loop to
end the loop.

what is up?

<code>

#include <stdio.h>

void main() { int main( void ) {

long nc;

nc = 0;
while (getchar() != 'EOF') { ^^^^^
illegal character constant, just write EOF (it is defined in stdio.h)
++nc;
}
printf("%ld\n", nc);
return 0;}

</code> AFAIK it is implementation defined how to produce an EOF via the
keyboard (e.g. Ctr-Z, Ctrl-C, Ctrl-D, ... ).
thanks
Dave

Regards

Irrwahn
--
Close your eyes and press escape three times.

[Johan Aurér]

On Thu, 11 Sep 2003, Irrwahn Grausewitz wrote:

Dave <no*****@noemai l.com> wrote:

long nc;

nc = 0;
while (getchar() != 'EOF') {

^^^^^
illegal character constant, just write EOF (it is defined in stdio.h)

It's not "illegal," just implementation-defined. In fact, 'EOF' might
even equal EOF if you're lucky. :-)

--
au***@axis.com

[Dan Pop]

In <ef************ *************** *****@4ax.com> Irrwahn Grausewitz <ir*****@freene t.de> writes:

Dave <no*****@noemai l.com> wrote:

Below is the code ive written just to count the characters typed in. I
assumed EOF is -1, so if i type -1 and then press enter shouldnt the
program end?

Already addressed in the other replies.

It orks if i put something like 'q' in the while loop to
end the loop.

what is up?

<code>

#include <stdio.h>

void main() {

int main( void ) {

long nc;

nc = 0;
while (getchar() != 'EOF') {

^^^^^
illegal character constant, just write EOF (it is defined in stdio.h)

'EOF' is perfectly legal as a character constant, just not what the OP
needs in his program.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Irrwahn Grausewitz]

Johan Aurér <au***@axis.com > wrote:

On Thu, 11 Sep 2003, Irrwahn Grausewitz wrote:

Dave <no*****@noemai l.com> wrote:

>
> long nc;
>
> nc = 0;
> while (getchar() != 'EOF') { ^^^^^
illegal character constant, just write EOF (it is defined in stdio.h)

It's not "illegal," just implementation-defined.

Hmph, you're so right.
In fact, 'EOF' might
even equal EOF if you're lucky. :-)

I would call this bad luck, in that it would hide the fact that
one wrote unportable code. :-)

Irrwahn
--
Close your eyes and press escape three times.

[amanayin]

Do you people know that code is taken from page 18
of The C programming language by kernigan & ritchie.

[Andreas Kahari]

In article <bj**********@t itan.btinternet .com>, amanayin wrote:

Do you people know that code is taken from page 18
of The C programming language by kernigan & ritchie.

Yes. It is similar to, but not the same as the code on that
page in that book.

--
Andreas Kähäri

[Ben Pfaff]

Irrwahn Grausewitz <ir*****@freene t.de> writes:

Dave <no*****@noemai l.com> wrote:

while (getchar() != 'EOF') {

^^^^^
illegal character constant, just write EOF (it is defined in stdio.h)

Not illegal, but its value is implementation-defined. It is
certainly not what the OP wants.
--
"It wouldn't be a new C standard if it didn't give a
new meaning to the word `static'."
--Peter Seebach on C99