Why?? (prog question)

[Josh Carroll]

On Mon, Mar 30, 2009 at 10:57 PM, Gary Kline <kline at thought.org> wrote:
> people, i've been under the weather for days and will probably be for a few more.
> new  and TEMPORARY meds dont like me, ugh.
>
> can anybody clue me in why the followin joinline program fails to catch if argc == 1?
>
>
> /*
>  * simple prog to join all | very nearly all lines of a text file that
>  * make up one paragraph into one LONG line.
>  *
>  * paragraphs are delimiated by a single \n break.
>  */
>
> #include <stdio.h>
> #include <string.h>
> #include <stdlib.h>
>
> main(int argc, char argv[])
> {
>   char buf[65536];
>
>   if (argc == 1)
>   {
>        printf("Usage: %s < file > newfile\n", argv[0]);
>        exit (-1);
>   }
>   while (fgets(buf, sizeof buf, stdin) )
>   {
>     if (*buf == '\n')
>     {
>       fprintf(stdout, "\n\n");
>     }
>     else
>     {
>       buf[strlen(buf)-1] = ' ';
>       fputs(buf, stdout);
>     }
>   }
> }

main should be:

int main(int argc, char **argv)

or perhaps

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

As is, you're defining int as   char argv[]  (e.g. char *) instead of char **.

What will likely happen is you'll get a segmentation fault when you
try to run the program, since your printf format spec has %s, but
you're passing it a char.

In fact, if you compile it with -Wall, you'll see the two problems
I've mentioned:

t.c:13: warning: second argument of 'main' should be 'char **'
t.c: In function 'main':
t.c:20: warning: format '%s' expects type 'char *', but argument 2 has
type 'int'

Change char argv[] to char *argv[] or char **argv and it should work properly.

Note also that your main should have an int return type and should
return a value.

Regards,
Josh