Parameter expansion without colon (default value to empty string) in a condition

Question

I know the difference between ${var:-word} and ${var-word}, but I don't see the purpose of using a default value in this condition:

[ -z "${NVM_PROFILE-}" ]

Why is an empty default value needed there?
Is it something related to compatibility with other shells (zsh)?

That code makes no sense to me. If the test succeeds, it goes ahead and uses a (potentially unset) `${NVM_PROFILE}` anyway. If the test doesn't succeed, it still uses `${NVM_PROFILE-}` in the `else` branch, even though `$NVM_PROFILE` must be set at that point. — melpomene, May 02 '18 at 05:53

score 1 · Accepted Answer · answered May 02 '18 at 10:05

1

I like set -u.

$ sh -c '[ -z "NVM_PROFILE" ] && echo empty'
empty
$ sh -c 'set -u; [ -z "$NVM_PROFILE" ] && echo empty'
sh: NVM_PROFILE: unbound variable
$ sh -c 'set -u; [ -z "${NVM_PROFILE-}" ] && echo empty'
empty

answered May 02 '18 at 10:05

fumiyas

367
3
11

score 1 · Answer 2 · answered May 02 '18 at 17:40

If the nounset option is enabled with set -u or set -o nounset, [ -z "$NVM_PROFILE" ] would result in an error if NVM_PROFILE isn't set. Using - or :- to explicitly expand the unset variable to an empty string avoids that error.

An alternative to using set -u is to check explicitly if the variable is set. (This works in zsh and bash.)

[[ ! -v NVM_PROFILE || -z $NVM_PROFILE ]]

Parameter expansion without colon (default value to empty string) in a condition

2 Answers2