[SOLVED] bash arithmetic evaluation null variable strange behavior

doru · 01-11-2020, 12:55 AM

a is unset. Why would this happen?

Code:

$ echo $(($a+1))
1
$ echo $((1+$a))
bash: 1+: syntax error: operand expected (error token is "+")
$ bash --version 
GNU bash, version 4.2.46(2)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2011 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

doru · 01-11-2020, 12:58 AM

+ in +1 works like a unary operator, + in 1+ is wrong.

crts · 01-11-2020, 06:38 AM

Just as an FYI from the manpage (emphasis by me):

Code:

ARITHMETIC EXPRESSIONS
       ...

       Shell variables are allowed as operands; parameter  expansion  is  per‐
       formed before the expression is evaluated.  Within an expression, shell
       variables may also be referenced by name without  using  the  parameter
       expansion  syntax.  A shell variable that is null or unset evaluates to
       0 when referenced by name without using the parameter expansion syntax.
       The  value  of a variable is evaluated as an arithmetic expression when
       it is referenced, or when a variable which has been given  the  integer
       attribute using declare -i is assigned a value.  A null value evaluates
       to 0.  A shell variable need not have its integer attribute  turned  on
       to be used in an expression.

So the expression

Code:

$ unset a
$ echo $((1+a))
1

is allowed and does not lead to an error.

astrogeek · 01-11-2020, 07:44 PM

To expand on what has already been said...

Code:

cat null_math
#!/bin/bash
unset a
echo $((a+1))
echo $((1+a))
echo $(($a+1))
echo $((1+$a))

./null_math
1
1
1
./null_math: line 8: 1+: syntax error: operand expected (error token is "+")

You will see that only the last case fails.

The first one evaluates to 0+1 per the pasted man page snippet.

The second one evaluates to 1+0, again per the man page.

The third, which uses parameter expansion, evaluates to +1 which is just the value, +1.

The fourth, also using parameter expansion, fails because it evaluates to 1+ (nothing after the +), hence the message about the missing operand.

doru · 01-13-2020, 02:08 AM

Quote:

A null value evaluates to 0.

Apparently this refers to:

Code:

 ~/tmp]$ unset a
 ~/tmp]$ echo $(()) $((a)) $(($a))
0 0 0

They tell the story of the program, so to speak, in a concentrated way, and, of course, it can not be perfectly clear.

(@astrogeek: I fully disagree with your signature. Sources are information, just like software. Trade in information is subject to lemon dilemma and moral hazard.)