* 2021-07-30 07:41:23-0400, Greg Wooledge wrote:
> On Fri, Jul 30, 2021 at 08:48:28AM +0300, Teemu Likonen wrote:
>> You have already got answers but here is another. Bash has a special
>> arithmetic evaluation mode which happens in:
>>
>> let ... # returns true (0) or false (1)
>> (( ... )) # returns true (0) or false (1)
>> $(( ... )) # returns the value of the expression
>
> The word "returns" is a bit misleading in the last case. It doesn't
> set the exit status (as seen in the $? special parameter) to the
> value of the expression the way the first two set it to 0 or 1.
>
> The last one is a substitution, meaning it's *replaced* inline by the
> value of the expression.
Indeed, that is very good correction. In normal programming languages an
expression is evaluated and it returns something. In Bash there are two
different things: commands' return values (exit status) and many
different expansions. So let us correct:
let ... # returns true (0) or false (1)
(( ... )) # returns true (0) or false (1)
$(( ... )) # expands to the value of the expression
--
/// Teemu Likonen - .-.. https://www.iki.fi/tlikonen/
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