After executing the command
declare -i a=5, the command
a+=2 succeeds, but the command
a-=2 fails. Can someone explain this strange behavior of bash?
Bash doesn't have an
-= assignment operator in the main shell syntax (arithmetic context is different, see below). That is to say, while you can use
= to assign to variables, and
+= to append to non-integer variables, or add to integer variables, there's no
*= etc. to go with them. The situation is the same in Ksh, where Bash's syntax is borrowed from (in this case, as in many others); and in Zsh, which also has similar features.
The other combined assignment operators apart from
+= probably would make little sense for non-integers anyway, and since regular "string" variables are the most common ones, it's probably not worth it to have those operators in the main syntax. Especially since
var*=123 is also a glob, and
var/=123 looks like a path. But as said,
+= does work for non-integers though:
$ foo=123; foo+=456; echo $foo 123456
The manual, as usual is somewhat brief on this, documenting the absence of
-= only by omission. Section 3.4 Shell Parameters describes variable assignment and mentions
+=, but no others.
Of course, in an arithmetic context (
$(( .. )),
(( .. )) etc.), all of
*= etc. are available:
$ foo=456; (( foo -= 123 )); echo $foo 333
In Bash, arithmetic evaluation is done inside
(( )), e.g.
((i=i+3)). From Bash's man page (
The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION.
+= are documented in the ARITHMETIC EVALUATION section, along with
= *= /= %= <<= >>= &= ^= |=, and all work as you expect
if you use the arithmetic notation.
+= working without that notation is an exception described under PARAMETERS
section of the manual.
When += is applied to a variable for which the integer attribute has been set, value is evaluated as an arithmetic expression and added to the variable's current value, which is also evaluated.
All in all, to get the desired behavior,
#!/bin/bash declare -i a=5 ((a+=2)) echo $a ((a-=2)) echo $a
The output is 7 and 5.
Bash allows arithmetic evaluation implicitly when used with
+= operator for a variable whose attribute is set to be integer type with
declare -i. Without
-i, it tells the shell to perform the "append" instead of "add" operation. The
-= or other operators do not have a special meaning anywhere other than, when used inside arithmetic context.
See this excerpt from GNU bash man page
+=is applied to a variable for which the integer attribute has been set, value is evaluated as an arithmetic expression and added to the variable’s current value, which is also evaluated.
declare -i var=2 var+=2 printf '%d\n' "$var" 4
declare foo=zoo foo+=2 printf '%s\n' "$foo" zoo2
Now for the other operators
|= are all supported inside
foo=144; (( foo /= 12 )); printf '%d\n' "$foo" 12
One other behavior associated with
+= when used with arrays
arr+=foo appends the
foo string to the element at the first index, while
arr+=(foo) appends a new element
foo to the array at the next index available.