Inside a shell script, which special parameter expands to the number of positional parameters the script was given?
LPIC-1 Exam 102-500, objective 105. Shells and shell scripting easy
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The options
Not correct $@
Wrong. $@ expands to the positional parameters themselves. Quoted as "$@" it produces one word per argument.
Not correct $0
Wrong. $0 is the name the script was invoked as. It is deliberately not counted in $#.
Not correct $?
Wrong. $? is the exit status of the last completed foreground command, which has nothing to do with argument count.
Correct $#
Correct. $# is the count of positional parameters, excluding $0. It is the usual way to validate usage, for example `[ $# -ne 2 ] && exit 1`.
Why
Positional parameters are $1 through $9, then ${10} and beyond with braces. $# counts them, $@ and $* expand them, and `shift` drops $1 and renumbers the rest, decreasing $# by one. $0 is not a positional parameter, so a script called with two arguments reports $# as 2.
Where this comes from
- Cited
- LPI exam objective 105.2
- What it says
- Use command line arguments passed to a script.
Practise this
Reading one question is not practice. The trainer will draw a short set from objective 105 and space the ones you get wrong.
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