Assign default value to variable if one not set, in Bash

Question

I know I can do:

echo ${var:-4}

to print 4 if var is null,

but how to assign -n in this case..?

echo ${var:--n} does not work.

although this does work..

echo ${var:---n}

but in that case it prints out --n and I need it to print -n.

echo is literally allowed to do anything when passed -n as an argument. See the POSIX specification: "If the first operand is -n, or if any of the operands contain a backslash ( '\' ) character, the results are implementation-defined". When passed -n, or any content with literal backslashes, echo could turn a backflip or kill your cat and still be POSIX-compliant. Follow the advice in the APPLICATION USAGE section of the spec, and use printf instead. — Charles Duffy
– Charles Duffy, Commented Oct 11, 2016 at 23:08
Possible duplicate of Assigning default values to shell variables with a single command in bash — Elmar Peise
– Elmar Peise, Commented Oct 11, 2016 at 23:09
@ElmarPeise, ...I'm not sure it is -- the immediate cause is certainly different, at least. — Charles Duffy
– Charles Duffy, Commented Oct 11, 2016 at 23:09

choroba · Accepted Answer · 2016-10-11 22:57:54Z

8

The problem is not in the default, but in echo: -n has a special meaning for it (check help echo). Use printf instead:

printf '%s\n' "${var:--n}"

answered Oct 11, 2016 at 22:57

choroba

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1 Comment

branquito Over a year ago

Yes, I knew about -n in echo.. but it didn't occur to me at the moment.. Interestingly this happened as a side effect.. if I happened to try with ${var:--a}, it would work ;)

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