Question about arithmetic logic.

[Steven W. Orr]

I happen to be running

GNU bash, version 4.0.35(1)-release (x86_64-redhat-linux-gnu)

I create an integer variable and assign it either a 0 or a 1. The arithmetic 
test always returns success regardless of value. For example:

typeset -i ss=0

(( ss ))
echo $?         # Returns 1. Expected because it should be
                # the same as (( ss != 0 )) No?

ss=1
(( ss ))
echo $?         # Also says 1. Should this be 0 because it should be the
                # success result same as (( ss != 0 ))

But if I use an operator...
ss=0
(( ! ss ))      # Says 0.

ss=1
(( ! ss ))      # Says 1.

So it seems to me that if I do not use the logical not operator, then the 
arithmetic test is doing a test to see if the string value of ss is not null 
(or something like that). Is this a bug? A feature? Or am I doing it wrong?

TIA

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