Say you are typing in a termainl running the bash shell and write "ls -la /usr/bin"
What is considered the argument, option and flag?
Also, how does the interpreter know where the end of one is? For example are options deliminated by spaces; so if there's a space the interpreter knows it's a different option?
The way I understand it is these three things are arguments:
1. ls
2. -la
3. /usr/bin
The "-la" are two options, but sometimes people call them flags.
The basic parsing a terminal can do is separate all of the strings delimited by spaces and send them to the program as an array, which is received in the implementation of the program by language-dependent means. For any other advanced parsing, like understanding that -la means two different things, you need to program it.
As mentioned spaces delimit arguments so if you need to include a space in the argument you normally need to quote it in your shell, for exampleOriginally Posted by c_weed
ls -l "Program Files"
If an argument begins with a "-" and it is not a flag you cannot quote it because the program itself sees the - and might think it is a flag. In that case you can usually do something like this instead:
ls -l "./-file-beginning-with-a-dash"
If the tool you are using does GNU style options processing then the following also works (the "--" marks the end of the options, so that all the remaining arguments are considered non-option arguments)
ls -l -- "-argument-beginning-with-a-dash"
