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Re: Inconsistent arithmetic evaluation of parameters


From: Chet Ramey
Subject: Re: Inconsistent arithmetic evaluation of parameters
Date: Thu, 3 Sep 2015 09:04:44 -0400
User-agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.10; rv:38.0) Gecko/20100101 Thunderbird/38.2.0

On 9/2/15 11:48 AM, Greg Wooledge wrote:
> On Wed, Sep 02, 2015 at 11:24:42AM -0400, Chet Ramey wrote:
>> On 9/2/15 11:19 AM, Greg Wooledge wrote:
>>> On Wed, Sep 02, 2015 at 10:16:14AM -0500, Dennis Williamson wrote:
>>>> The $ is implied.
>>>
>>> That is completely absurd.  (And wrong.)
>>
>> Not exactly.  When the arithmetic evaluator encounters a token that is of
>> the form of a shell identifier (`bar'), in a context where an operand is
>> needed, it treats it as a shell variable and looks up the variable's
>> value.  In that sense, it's an expansion.
>>
>> The difference between bash and dash is what each shell does with that
>> value.
> 
> $foo and foo are not equivalent in dash, as we've already discussed:
> 
> $ dash
> $ foo=bar bar=5
> $ echo $((foo))
> dash: 4: Illegal number: bar
> $ echo $(($foo))
> 5

Yes.  I explained exactly why that is and what's happening.  You can
take that explanation and understand why Dennis uses the term
`indirection' to mean the subsequent expansion by the arithmetic
evaluator, and why it appears as if there is an implied `$' in the
specific case we're discussing.  It's not a general property, just
observed behavior in arithmetic evaluation.

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



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