From: Tom Cloyd <tomcloyd@...>
Date: 2009-01-05T15:42:41+09:00
Subject: Re: VERY simple question about "?"

Dave Thomas wrote:
>
> On Jan 4, 2009, at 8:14 PM, Tom Cloyd wrote:
>
>> I'm still struggling here. When syntax is productive of something, I 
>> see an operation occurring - active functionality. I'm perfectly 
>> content to see operators as methods. Syntactically there's no problem 
>> that I can detect in that construct. But to call "?<char>" a 
>> literal...well, I must go study up a bit on this. To me, "abcdef" is 
>> a literal. '\n' is too. "\n" is a literal if you allow meaning to be 
>> contexturalized, which should be no problem. As input to various 
>> processes, '\n' and "\n" have different outcomes, but by themselves 
>> produce nothing. "?<char>" produces something. Irb makes this clear 
>> (to me):
>>
>> irb(main):006:0> puts '\n'
>> \n
>> => nil
>> irb(main):007:0> puts "\n"
>>
>> => nil
>> irb(main):008:0> puts ?\n
>> 10
>> => nil
>> irb(main):009:0>
>>
>> If in "-1", we consider the "-" to be an operator, then I see an 
>> exact parallel to the function of "?a".
>>
>> It's not the same as what quote marks do - THEY provide context, and 
>> change meaning. Operators PRODUCE something. They "operate". It seems 
>> very simple. If I'm making a fundamental error, I'd be very grateful 
>> to have it point out to me. I highly value arriving at a place of 
>> clear understanding, when it's possible
>
> I think you're over thinking it. For example, in
>
>   0x10
>
> which is 16 in decimal, is the '0x' part an operator? If you say no 
> (which I hope you do) then why would you say that ? is an operator in 
> ?a (which is 97). They are simply characters that are used by the 
> lexical analysis phase to help interpret what follows. No executable 
> code is generated: there's no operation.
>
> Similarly, in your example of -1, the - sign is NOT an operator. It 
> simply means that when parsing the constant, a negative value is 
> required. Again, so executable code is generated. That's the 
> difference between
>
>    a = -1      # -1 is a constant, no operator
>    b = -a      # the unary minus operator (-@) is invoked on the 
> object referenced by a
>
> Cheers
>
>
> Dave
>
>
>
>
Thanks, Dave. This helps. You have a genuine knack for simple clear 
statement and example. Thanks for the dialog!

t.

-- 

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Tom Cloyd, MS MA, LMHC - Private practice Psychotherapist
Bellingham, Washington, U.S.A: (360) 920-1226
<< tc@tomcloyd.com >> (email)
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