bash is interpreting * and passing it as the list of files in the
current directory, you could try passing escape sequence with * and
see what is the behavior.
-sujay

On Wed, Jun 10, 2009 at 8:02 PM, prabhjot
singh<prabhjotsinghengineer@...> wrote:
>
>
> Hi,
>
> I write a very simple program to print command line arguments.
> As and when program hits '*' in received arguments It converts it to list of
> files in current directory.
>
> My questions/doubts:
> 1) What is the reason behind this?
> 2) Can't I pass * as command line argument?
>
> #include <stdio.h>
> int main(int argc, char *argv[])
> {
> int i = 0;
> printf("argc: [%d]", argc);
> while (i < 10 && i < argc)
> printf("%s\n", argv[i++]);
> }
> output:-
>
>>./a.out *
> argc: [38]
> /*here goes list of files*/
>
> --
> Platform used: RH Linux - Bash shell | compiler: gcc [egcs-2.91.66 ]
>
> --
> Thanks
> PS
>
> [Non-text portions of this message have been removed]
>
>