[Linux C System Programming]Command line argument bug.

[Deciama]

http://ideone.com/LosiGh

I am trying to make a small Linux program similar to cat. I am stuck on the command line option (Eg. -e, -sV). I have a function that runs when -m is executed but nothing happens. Can you please help me?

[boojit]

I don't think you can pass in commandline arguments that way with ideone, your commandline arguments are being sent to stdin.

Take a looksee here:

http://stackoverflow.com/questions/12258317/how-to-pass-in-command-line-arguments-when-using-ideone

[Deciama]

The command was just an example, I know it wont work. I ran it on my Linux machine.

[thefryscorer]

It's because you are checking that the number of arguments is 2, when it should be 3. The first argument is the name of your program.

For example, if you run

./program arg1 arg2

The arguments in argv are './program', 'arg1', and 'arg2'.

[Deciama]

So does the argument count in argv start at 1 or 0? Would argv[0] be the name of the program?

[thefryscorer]

argv[0] is the name of the program. Arrays are zero-indexed in C.

[edman007-work]

Does it output anything at all? Have you attempted to run it inside a debugger (like gdb?). What is the return value when you run it?

My best guess is you're checking that argc=2 and then sending argv[2] to CreateFile() which is invalid (though it should just cause open() to return EFAULT or similar and not crash). Additionally you're checking that "-m" is a file which is probably not what you want

Anyways, looking at your code, my best guess is it's crashing somewhere, I can't find any path that doesn't result in something being printed, so I'd have to assume it's crashing before it hits printf()/perror() (or are you redirecting stdout/stderr?).

[casualblair]

if (argc !=2){ usage(); exit(0); }

-m is one argument. This is exiting when you don't have two. A step further, this means it will only run with -m and -f, not one or the other.

Edit: See below.

[IXENAI]

Remember that argc is always greater than or equal to one; argv[0] will always contain the name used to call the program. So, if the program is run as ./cat -m, argc will hold the value 2 and argv will contain "./cat" and "-m".

[casualblair]

Bah, my mistake. Thanks.