variable expansion in curly braces
There are two variants of the ${var-value}
notation, one without a colon, as shown, and one with a colon: ${var:-value}
.
The first version, without colon, means 'if $var
is set to any value (including an empty string), use it; otherwise, use value
instead'.
The second version, with colon, means 'if $var
is set to any value except the empty string, use it; otherwise, use value
instead'.
This pattern holds for other variable substitutions too, notably:
-
${var:=value}
- if
$var
is set to any non-empty string, leave it unchanged; otherwise, set$var
tovalue
.
- if
-
${var=value}
- if
$var
is set to any value (including an empty string), leave it unchanged; otherwise, set$var
tovalue
.
- if
-
${var:?message}
- if
$var
is set to any non-empty string, do nothing; otherwise, complain using the given message' (where a default message is supplied ifmessage
is itself empty).
- if
-
${var?message}
- if
$var
is set to any value (including an empty string), do nothing; otherwise, complain using the given message'.
- if
These notations all apply to any POSIX-compatible shell (Bourne, Korn, Bash, and others). You can find the manual for the bash
version online — in the section Shell Parameter Expansion. Bash also has a number of non-standard notations, many of which are extremely useful but not necessarily shared with other shells.
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user1929290
Updated on September 15, 2022Comments
-
user1929290 over 1 year
This is code
a='' b=john c=${a-$b} echo $c
And the output is empty line
And for similar code where first variable is not initialized
b1=doe c1=${a1-$b1} echo $c1
And the output is
doe
I do not understand how bash deals with expanding of variables leading to different results.