CODEWITHSUNDEEP
bashshellparameter-expansion
Glitch
Glitch

Reputation: 406

Why does ${##parameter} always return 0?

Suppose that parameter=value is an internal variable in bash. When I try ${#parameter} in bash it returns a 5. But when I try, for instance, ${##parameter} or ${###parameter} it always give back 0 in return. Why it does not say that it is a substitution error as in other cases?

Upvotes: 2

Views: 283

Answers (2)

Ondrej K.
Ondrej K.

Reputation: 9664

Short version

You could express ${##parameter} as:

x=$#
${x#parameter}

and ${###parameter} as:

x=$#
${x##parameter}

You are performing a prefix removal on $# (number of arguments passed to a script / function).

Longer version

If parameter has any special meaning, it would be likely specific to your environment.

$ echo \""${parameter}"\"
""

or even:

$ set -u
$ echo \""${parameter}"\"
bash: parameter: unbound variable

Now with that out of the way, quick look into docs:

${#parameter}

The length in characters of the expanded value of parameter is substituted.

So with variable parameter not being set:

$ echo ${#parameter}
0

or with set -u:

$ echo ${#parameter}
bash: parameter: unbound variable

Additional # changed meaning to Remove matching prefix pattern. using: ${parameter#word} or ${parameter##word} syntax on ${#}, i.e. number of argument passed to a script/function. Good way to see and understand that behavior would be:

$ f() { echo -n "$# " ; echo ${##parameter}; }
$ f
0 0
$ f a b c
3 3

As you see in first call ${#} was 0 and (and attempting to strip prefix "parameter" does nothing to it really) and the second call value of ${#} is 3.

Prefix removal, the different between # and ## is, when matching patterns whether shorted or longest match is stripped.

Upvotes: 3

Hellmar Becker
Hellmar Becker

Reputation: 2972

Because ## is a special operator that removes a prefix. See for instance here: What is the meaning of the ${0##...} syntax with variable, braces and hash character in bash?

Upvotes: 0

Related Questions