How would you ask for the user to input files as arguments to be used (as many as they would like)? Also How would you print to a file?
scanf("%s", user_filename);
FILE *fp;
fp = fopen (user_filename, "r");
I have tried doing various things to it but I can only get it to take one file.
The easiest way to pass some file names to your C program is to pass them as arguments to your C program.
Arguments are passed to a C program using the parameters to main:
int main( int argc, char *argv[] )
{
...
}
The value argc indicates how many parameters there are, and argv[] is an array of string pointers containing the arguments. Note that the first argument at index 0 (the string pointed to by argv[0]) is the command name itself. The rest of the arguments passed to the command are in argv[1], argv[2], and so on.
If you compile your program and call it like this:
my_prog foo.txt bar.txt bah.txt
Then the value of argc will be 4 (remember it includes the command) and the argv values will be:
argv[0] points to "my_prog"
argv[1] points to "foo.txt"
argv[2] points to "bar.txt"
argv[3] points to "bah.txt"
In your program then, you only need to check argc for how many parameters there are. If argv > 1, then you have at least one parameter starting at argv[1]:
int main( int argc, char *argv[] )
{
int i;
FILE *fp;
// If there are no command line parameters, complain and exit
//
if ( argc < 2 )
{
fprintf( stderr, "Usage: %s some_file_names\n", argv[0] );
return 1; // This exits the program, but you might choose to continue processing
// the other files.
}
for ( i = 1; i < argc; i++ )
{
if ( (fp = fopen(argv[i], "r")) == NULL )
{
fprintf( stderr, "%s: Gah! I could not open file named %s!\n", argv[0], argv[i] );
return 2;
}
// Do some stuff with file named argv[i], file pointer fp
...
fclose( fp );
}
return 0;
}
This is just one of several different ways to do it (functionally and stylistically), depending upon how you need to process the files.
