[boost] Re: missing ## ... juxtaposition?

From: Dan W. (danw_at_[hidden])
Date: 2003-12-03 02:24:32

Paul Mensonides wrote:

 >No.  The preprocessor does not "insert spaces" *ever*.  At this point in
 >translation, the preprocessor is operating on preprocessing tokens, not
 >characters.  There is a big difference between a lack of whitespace and
 >concatenation.

Hi again, Paul. I'm sorry to bother so much. I forwarded your reply to 
Walter, of Digital Mars, and he still doesn't seem convinced. I am 99% 
convinced but I need to understand this fully if I'll ever be able to 
convince him :-)

Let me try to put what you said in my own words, and then I'll ask you a 
pointed question:

I always thought, myself, that the preprocessor worked at a textual level. 
Now I'm getting the idea that the first pass in preprocessing is 
'tokenization' which I take it to mean tagging tokens, like "abracadabra" 
or "." or ")". So, if we have,

#define a(x) x
  a(something).else

outputting

  something.else

is NOT 'concatenation', as no tokens are being merged.

Could we say, then, that ## was invented as a special power tool for 
'violating initial tokenization' and forcing a merger of two tokens?
If so, we could say that in the example above, the tokens are not being 
violated, and therefore ## is not needed.

Now the pointed question: What then if I write,

  a(something)else

without a "."?  Should or shouldn't the preprocessor add a space in this 
particular case?

Yours
dan