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Drac
#1 / 7
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 command line argument /* could anyone tell me why this doesn't work? */
#include <stdio.h>
main(int argc, int * argv[])
{
if(* argv[1] >= 100)
printf("argv is bigger or equal to 100!\n");
else
printf("argv is less than 100!\n");
return 0;
Quote: }
/* I cant see why this shouldn't work! */
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| Sun, 27 Jun 2004 19:09:05 GMT |
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Drac
#2 / 7
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command line argument /* could anyone tell me why this doesn't work? */
#include <stdio.h>
main(int argc, int * argv[])
{
if(* argv[1] >= 100)
printf("argv is bigger or equal to 100!\n");
else
printf("argv is less than 100!\n");
return 0;
Quote: }
/* I cant see why this shouldn't work! */
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| Sun, 27 Jun 2004 19:12:07 GMT |
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k..
#3 / 7
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command line argument
Quote: > /* could anyone tell me why this doesn't work? */ > #include <stdio.h>
> main(int argc, int * argv[])
be a char**, not an int**.
Quote: > { > if(* argv[1] >= 100)
> printf("argv is bigger or equal to 100!\n");
> else
> printf("argv is less than 100!\n");
> return 0;
> }
> /* I cant see why this shouldn't work! */
--
Chris "C being C, and none of CPL, Ada, PL/1, Lisp, Smalltalk, or Pop" Dollin
C FAQs at: http://www.faqs.org/faqs/by-newsgroup/comp/comp.lang.c.html
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| Sun, 27 Jun 2004 19:59:56 GMT |
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Jens.Toerr..
#4 / 7
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command line argument Quote:
> /* could anyone tell me why this doesn't work? */
> #include <stdio.h>
> main(int argc, int * argv[])
You got this wrong, argv is an array of char pointers (and main() always
returns an int):
int main( int argc, char *argv[ ] )
Quote: > {
> if(* argv[1] >= 100)
> printf("argv is bigger or equal to 100!\n");
> else
> printf("argv is less than 100!\n");
Now it should be obvious why this won't work - you must convert the
string argv[ 1 ] to an integer before comparing. But before you do so
you should first check if argv[ 1 ] isn't NULL! And you should also be
prepared for situations where you don't get passed a string that can
be interpreted as an integer, e.g. if the user typed "lO0" instead of
"100".
Regards, Jens
--
_ _____ _____
_ | | | | | | AG Moebius, Institut fuer Molekuelphysik
| |_| | | | | | Fachbereich Physik, Freie Universitaet Berlin
\___/ens|_|homs|_|oerring Tel: ++49 (0)30 838 - 53394 / FAX: - 56046
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| Sun, 27 Jun 2004 20:28:14 GMT |
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Lawrence Kir
#5 / 7
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command line argument On Wednesday, in article
Quote: >/* could anyone tell me why this doesn't work? */ >#include <stdio.h>
>main(int argc, int * argv[])
int main(int argc, char *argv[])
argv points to the first element of an array of pointers to strings.
Quote: argv[1] is a pointer to a string. If that string has the form of an integer
you can convert it to an integer type using a fucntion like strtol().
E.g.
{
long number;
number = strtol(argv[1], NULL, 10);
if (number >= 100)
Quote: > printf("argv is bigger or equal to 100!\n");
>else
> printf("argv is less than 100!\n");
That isn't argv but the argument corresponding to argv[1]. Note that
the code assumes that argv[1] exists, you should test that argc>1 before
attempting toa ccess argv[1]. Also to use strtol() you should
#include <stdlib.h> at the top.
Quote: --
-----------------------------------------
-----------------------------------------
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| Sun, 27 Jun 2004 20:26:31 GMT |
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Roger Faus
#6 / 7
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command line argument Quote:
> > /* could anyone tell me why this doesn't work? */ > > #include <stdio.h>
> > main(int argc, int * argv[])
> You got this wrong, argv is an array of char pointers (and main() always
> returns an int):
> int main( int argc, char *argv[ ] )
> > {
> > if(* argv[1] >= 100)
> > printf("argv is bigger or equal to 100!\n");
> > else
> > printf("argv is less than 100!\n");
> Now it should be obvious why this won't work - you must convert the
> string argv[ 1 ] to an integer before comparing. But before you do so
i'm not very sure, but i think, *argv[1] >= 100 will work (if: char
*argv[]), but it will compaire the Ascii number, surely not what it is
expected to do :-)
Quote: > you should first check if argv[ 1 ] isn't NULL! And you should also be
> prepared for situations where you don't get passed a string that can
> be interpreted as an integer, e.g. if the user typed "lO0" instead of
> "100".
> Regards, Jens
--
#!/usr/bin/perl
use Inline C=>q{char trans (int n, char c){
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| Mon, 28 Jun 2004 02:02:13 GMT |
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Jens.Toerr..
#7 / 7
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command line argument Quote:
>> Now it should be obvious why this won't work - you must convert the
>> string argv[ 1 ] to an integer before comparing. But before you do so
> i'm not very sure, but i think, *argv[1] >= 100 will work (if: char
> *argv[]), but it will compaire the Ascii number, surely not what it is
> expected to do :-)
That's of course correct. I was just assuming from the context that the
OP was expecting an integer as input to the program ;-)
Regards, Jens
--
_ _____ _____
_ | | | | | | AG Moebius, Institut fuer Molekuelphysik
| |_| | | | | | Fachbereich Physik, Freie Universitaet Berlin
\___/ens|_|homs|_|oerring Tel: ++49 (0)30 838 - 53394 / FAX: - 56046
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| Mon, 28 Jun 2004 03:26:09 GMT |
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