Argument Processing

[Daniel Rudy]

Hello,

I'm trying to learn how command line arguments are handled in C. The
following code segment that I wrote as a test program compiles, but when
I try to run it, it core dumps. This is under a FreeBSD environment.
What am I doing wrong here?

/*

just echos the command line arguments onto the screen
tests the format of argument processing

*/

#include <stdio.h>
#include <string.h>
int main(int argc, char argv[])
{
int i; /* generic counter */

printf("argc = %d", argc);
for (i = 0; i <= argc; i++)
{
printf("argv[%d] = %s\n", i, argv[i]);
}

/* return to operating system */
return(0);
}
--
Daniel Rudy

Email address has been encoded to reduce spam.
Remove all numbers, then remove invalid, email, no, and spam to reply.

[Chris Dollin]

su******@yahoo. com wrote:

no problem with the declartion it is fine

Context would have been useful. The original post had:

int main(int argc, char argv[])

which *is* wrong - it's `char *argv[]` or `char **argv` (I prefer
the latter, myself, but the former means the same thing as a
function argument declaration).

--
Chris "electric hedgehog" Dollin

[Daniel Rudy]

At about the time of 11/4/2004 10:00 PM, Ravi Uday stated the following:

"Daniel Rudy"
<i0************ *************** *******@n0o1p2a 3c4b5e6l7l8s9p0 a1m2.3n4e5t6>
wrote in message news:fo******** *********@newss vr14.news.prodi gy.com...

Hello,

I'm trying to learn how command line arguments are handled in C. The
following code segment that I wrote as a test program compiles, but when
I try to run it, it core dumps. This is under a FreeBSD environment.
What am I doing wrong here?

/*

just echos the command line arguments onto the screen
tests the format of argument processing

*/

#include <stdio.h>
#include <string.h>
int main(int argc, char argv[])
{
int i; /* generic counter */

printf("argc = %d", argc);
for (i = 0; i <= argc; i++)
{
printf("argv[%d] = %s\n", i, argv[i]);
}

/* return to operating system */
return(0);
}

What are you passing to main from command line ?

Just test stuff to see how it works.
You might try adding this before the first printf.

if ( argc < 3 )
{
printf ("Insufficie nt arguments to main\n");
return 0;
}
I did as you suggested, now it comes up with argc = 4 and then core dumps.

- Ravi

--
Daniel Rudy

Email address has been encoded to reduce spam.
Remove all numbers, then remove invalid, email, no, and spam to reply.

--
Daniel Rudy

Email address has been encoded to reduce spam.
Remove all numbers, then remove invalid, email, no, and spam to reply.

[Daniel Rudy]

At about the time of 11/5/2004 12:28 AM, Mike Wahler stated the following:

"Daniel Rudy"
<i0************ *************** *******@n0o1p2a 3c4b5e6l7l8s9p0 a1m2.3n4e5t6>
wrote in message news:fo******** *********@newss vr14.news.prodi gy.com...

Hello,

I'm trying to learn how command line arguments are handled in C. The
following code segment that I wrote as a test program compiles, but when
I try to run it, it core dumps. This is under a FreeBSD environment.
What am I doing wrong here?

See below.

/*

just echos the command line arguments onto the screen
tests the format of argument processing

*/

#include <stdio.h>
#include <string.h>
int main(int argc, char argv[])

int main(int argc, char *argv[])

/* or */

int main(int argc, char **argv)

{
int i; /* generic counter */

printf("argc = %d", argc);
for (i = 0; i <= argc; i++)

for (i = 0; i < argc; i++)

{
printf("argv[%d] = %s\n", i, argv[i]);
}

/* return to operating system */
return(0);
}

-Mike

Thanks to everyone who replied. The code now looks like this:

strata:/home/dcrudy/c 1047 $$$ ->cat argtest.c
/*

just echos the command line arguments onto the screen
tests the format of argument processing

*/

#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i; /* generic counter */

if (argc < 3)
{
printf("error: need more arguments.\n");
return(0);
}
printf("argc = %d\n", argc);
for (i = 0; i < argc; i++)
{
printf("argv[%d] = %s\n", i, argv[i]);
}

/* return to operating system */
return(0);
}
So now when I run it, instead of core dumping, I get the following output:

strata:/home/dcrudy/c 1046 $$$ ->./argtest arg1 arg2 arg3
argc = 4
argv[0] = ./argtest
argv[1] = arg1
argv[2] = arg2
argv[3] = arg3

Now there is something about this that I do not understand. What
exactly is the nature of argv? I know it's an array, but is it an array
of char? A number of people pointed out that the * was required.
AFAIK, * is a pointer reference.

Thanks again.

--
Daniel Rudy

Email address has been encoded to reduce spam.
Remove all numbers, then remove invalid, email, no, and spam to reply.

[Nils O. Selåsdal]

Daniel Rudy wrote:

Now there is something about this that I do not understand. What
exactly is the nature of argv? I know it's an array, but is it an array
of char? A number of people pointed out that the * was required.
AFAIK, * is a pointer reference.

Thanks again.

argv is char *argv[] .A n array, of pointers to char.
The pointers actually points to the 1. element of a
nul terminated array of chars, but you can't tell from
that declaration though.

Think of it as an array of pointers to C strings.

[Mike Wahler]

"Daniel Rudy"
<i0************ *************** *******@n0o1p2a 3c4b5e6l7l8s9p0 a1m2.3n4e5t6>
wrote in message news:kG******** ***********@new ssvr21.news.pro digy.com...

Thanks to everyone who replied. The code now looks like this:

strata:/home/dcrudy/c 1047 $$$ ->cat argtest.c
/*

just echos the command line arguments onto the screen
tests the format of argument processing

*/

#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i; /* generic counter */

if (argc < 3)
{
printf("error: need more arguments.\n");
return(0);
}
printf("argc = %d\n", argc);
for (i = 0; i < argc; i++)
{
printf("argv[%d] = %s\n", i, argv[i]);
}

/* return to operating system */
return(0);
}
So now when I run it, instead of core dumping, I get the following output:

strata:/home/dcrudy/c 1046 $$$ ->./argtest arg1 arg2 arg3
argc = 4
argv[0] = ./argtest
argv[1] = arg1
argv[2] = arg2
argv[3] = arg3

Now there is something about this that I do not understand. What
exactly is the nature of argv? I know it's an array,
Actually, it's not. Arrays cannot be passed to or returned
from functions.
but is it an array
of char?
No. It's a pointer (to an array of pointers (to char)).
This allows for arguments of varying lengths.
A number of people pointed out that the * was required.
AFAIK, * is a pointer reference.

:-)

-Mike

[Mike Wahler]

"Nils O. Selåsdal" <NO*@Utel.no> wrote in message
news:8Y******** ***********@new s2.e.nsc.no...

Daniel Rudy wrote:

Now there is something about this that I do not understand. What
exactly is the nature of argv? I know it's an array, but is it an array
of char? A number of people pointed out that the * was required.
AFAIK, * is a pointer reference.

Thanks again. argv is char *argv[] .A n array, of pointers to char.

No, 'argv' is *not* an array, it's a pointer.
The pointers actually points to the 1. element of a
nul terminated array of chars, but you can't tell from
that declaration though.

Think of it as an array of pointers to C strings.

No. It's a pointer to such an array.

-Mike

[Tim Rentsch]

Jack Klein <ja*******@spam cop.net> writes:

Note that 'return' is a statement, not a function. Parentheses do no
harm, but have not been required since the first ANSI standard 15
years ago.

Just a question - were parentheses EVER required on a 'return'
statement in C?

[Chris Torek]

>Jack Klein <ja*******@spam cop.net> writes:

Note that 'return' is a statement, not a function. Parentheses do no
harm, but have not been required since the first ANSI standard 15
years ago.

In article <news:kf******* ******@alumnus. caltech.edu>
Tim Rentsch <tx*@alumnus.ca ltech.edu> wrote:Just a question - were parentheses EVER required on a 'return'
statement in C?

It is quite possible.

Note that the syntax for "struct" once used parentheses rather than
braces, too:

struct ( int a; int b; );

As late as Version 6 Unix, the op= operators were spelled the other
way around, initializers for static-duration objects were written
without an "=" sign, and there were other oddities:

int x 5;

int main(argc, argv) char **argv; {

printf(2, "this goes to stderr: argc = %d\n", argc);

argc =- 1; /* ignore program name */
argv =+ 1;

...

}

The "standard I/O" library was originally a separate option; to
use it you had to link with "-lS" (I think -- I never actually
*used* V6, much less anything earlier like V5).
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

[Tim Rentsch]

Chris Torek <no****@torek.n et> writes:

Jack Klein <ja*******@spam cop.net> writes:

Note that 'return' is a statement, not a function. Parentheses do no
harm, but have not been required since the first ANSI standard 15
years ago.

In article <news:kf******* ******@alumnus. caltech.edu>
Tim Rentsch <tx*@alumnus.ca ltech.edu> wrote:

Just a question - were parentheses EVER required on a 'return'
statement in C?

It is quite possible.

[numerous obsolete/obsolescent C-isms noted]

It might be good to ask the question this way - can any CLC-er
identify an old C compiler (or even remember using such a compiler)
where the syntax for 'return' required parentheses?

[Randy Howard]

In article <9w************ *******@newssvr 21.news.prodigy .com>,
i0************* *************** ******@n0o1p2a3 c4b5e6l7l8s9p0a 1m2.3n4e5t6 says...

for (i = 0; i <= argc; i++)

What's wrong with this picture?

--
Randy Howard (2reply remove FOOBAR)