Re: arithmetic evaluation : no assignment

[Chet Ramey]

On 11/8/24 5:32 AM, watael rocketmail via wrote:
> hi,
>
> a=1; b=2 #no c
> echo $(( c=$b-$a , $c>$a+$b ? $c-$a : $c+$b ))
> bash: c=2-1 , >1+2 ? -1 : +2 : syntax error: operand expected (error token 
> is ">1+2 ? -1 : +2 ")
>
> I don't remember this didn't work.

The (( compound command performs word expansions on the individual words
between the (( and closing )) before evaluating the result as an
expression. This is from the man page:

"The expression undergoes the  same  ex-
 pansions  as  if  it were within double quotes, but double quote
 characters in expression are not treated specially and  are  re-
 moved."

`(( expression ))' is just about equivalent to `let "expression"'.


--
``The lyf so short, the craft so long to lerne.'' - Chaucer
                 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU    chet@case.edu    http://tiswww.cwru.edu/~chet/

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