The fine man page, bash(1), says
((expression))
The expression is evaluated according to the rules described
below under ARITHMETIC EVALUATION. If the value of the
expression is non-zero, the return status is 0; otherwise the
return status is 1. This is exactly equivalent to let
"expression".
IOW, the behaviour you're seeing is correct.
** Changed in: bash (Ubuntu)
Status: New => Invalid
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https://bugs.launchpad.net/bugs/536129
Title:
++ operator exits with error when the variable equals to zero
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