Shell Programming and Scripting

Hey all,

I not able to find it what -n stand operator stands for.

Code:

 
[[ -n $* ]] && print -u2 "ERROR: $*"

could you please let me know why we are using -n in the code.

Check out man test

Hope that helps
Regards
Peasant.

thanks a lot.

It checks whether the length of the string s1 is non-zero or not.

The test man page says absolutely nothing about how the [[ expr ]] compound command behaves in various shells. In some shells, [[ is an unknown command; in others it is a keyword in the syntax for that shell. There is a huge difference between the way a shell handles an unquoted $* when arguments are part of the shell's syntax versus when arguments are being parsed as arguments for a utility to be invoked (even if it is a built-in utility as in the case of test and [).

Thank you Don for further explanation, but the OP asked about -n operator specifically which is covered in man test.

He did not specify shell or operating system to make further assumptions.

Regards
Peasant.

Yes, but it is a different command altogether, in fact [[..]] is part of the shell syntax, whereas [ .. ] is not and using a regular test command with single brackets would not work properly in this case. Even if the -n operators happen to have a similar meaning, better advise would be to use the man page of the shell..

IMHO one should quote $*, to avoid substitution by file names and eventual "too many tokens"

Code:

[[ -n "$*" ]]