Need help understanding a piece of code
Hi,
I came across an example program for command line arguments in C and it contains a piece of code I can't get my head around. The following is the program:
Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
main( int argc, char *argv[] )
{
int m, n, /* Loop counters. */
l, /* String length. */
x, /* Exit code. */
ch; /* Character buffer. */
char s[256]; /* String buffer. */
for( n = 1; n < argc; n++ ) /* Scan through args. */
{
switch( (int)argv[n][0] ) /* Check for option character. */
{
case '-':
case '/': x = 0; /* Bail out if 1. */
l = strlen( argv[n] );
for( m = 1; m < l; ++m ) /* Scan through options. */
{
ch = (int)argv[n][m];
switch( ch )
{
case 'a': /* Legal options. */
case 'A':
case 'b':
case 'B':
case 'C':
case 'd':
case 'D': printf( "Option code = %c\n", ch );
break;
case 's': /* String parameter. */
case 'S': if( m + 1 >= l )
{
puts( "Illegal syntax -- no string!" );
exit( 1 );
}
else
{
strcpy( s, &argv[n][m+1] );
printf( "String = %s\n", s );
}
x = 1;
break;
default: printf( "Illegal option code = %c\n", ch );
x = 1; /* Not legal option. */
exit( 1 );
break;
}
if( x == 1 )
{
break;
}
}
break;
default: printf( "Text = %s\n", argv[n] ); /* Not option -- text. */
break;
}
}
puts( "DONE!" );
}
What I don't understand is the code in the second switch statement marked with string parameter. What do I have to input to get to the printf statement: string = %s ? what does the -s option do?
All in all I am confused and I would really appreciate any help that can help me crawl out of the dark state of noobness I'm in now. Thanks.