From: Dave Thomas <dave@...>
Date: 2009-01-05T12:26:03+09:00
Subject: Re: VERY simple question about "?"


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