Thanks Pedro.
Can u explain the IF_CONDITION below?
Thanks in advance
asma
Pedro Izecksohn <[EMAIL PROTECTED]> wrote:
asma sabir wrote:
I'm doing OS course in Linux where I m not understanding the command line. is
it like DOS cmd line? And I am too confused in cammand_line_arguments like argc
and pointer argv. I want to know about each and every thing about these,
because of this i could not understand the below code and similar of it.There
are 2 files below.Can anybody explain this code specially the main() function
with its arguments?Pleaseeee........
My reply:
> the command line. is it like DOS cmd line?
Are you referring to bash? It is much better than Windows' cmd. DOS used
command.com .
> There are 2 files below.Can anybody explain this code specially the main()
> function with its arguments?
argv points to an array of pointers to strings. Try to printf ("%p %p\n", argv,
argv[0]);
I decided to modify the code to be easier to understand. If you do not
understand, ask what does a specific line.
# include
# include
# include
# include
int main(int argc,char *argv[])
{
FILE *in;
int count=0;
int ch;
what is the purpose of this IF_condition......
if(argc<2)
{
fprintf(stderr, "I %s require a filename.\n", argv[0]);
exit(1);
}
if((in=fopen(argv[1],"r"))== NULL)
{
fprintf (stderr, "%s : %s\n", argv[1], strerror(errno));
exit(2);
}
while ((ch=getc(in))!=EOF)
if(ch=='\n')
count++;
fclose(in);
printf("%d\n",count);
return 0;
}
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