Re: Scheme question on strict substitution

[Erik Sandberg]

On Thursday 30 November 2006 21:32, address@hidden wrote:
> Hello,
>
> > . . . one problem is that this [giving the syntax of each keyword] would
> > still just tell a small part of the full syntax.
>
> I think it would give a big part of the full syntax, even if not the whole
> picture.  It would enable a user to know
>    (1) what type of "arguments" have to follow the keyword;
>    (2) whether (s)he has accidentally omitted one of the required
>        arguments (when debugging);
>    (3) whether (s)he should write a macro definition or a function
>        definition to specify a "\whatever" construct involving the
>        keyword;
>    (4) what the types should be in the event that (s)he winds up writing a
>        function definition.
> "Full" or not, that's a lot of useful information.  And I'm talking about
> just a list of the syntax for each keyword, *not* the semantics.  A syntax
> specification can be short but still tremendously useful.  Specification
> of the semantics takes a lot of blah-blah, and nobody would have the time
> to write up all the semantics.
>
> > Another problem is that the
> > rule I gave really was too simple. For example, you cannot say
> > ^"some text" in a .ly file
> > without having a note in front of it, still you can define a macro
> > mytext = ^"some text"
> > and then use it as c \mytext
>
> That's a general problem in LilyPond, knowing what kind of expression
> constitutes a valid right-hand side of a macro definition.  (I'm amazed
> that the above definition works.)
>
> A different but related problem concerns something I don't know the right
> term for, something like "context-free equivalence" maybe:  for example,
> I can say
>
>    foo = \markup { \bold "Zanzibar" }
>    bar = \markup { "Stand on" \foo }
>
> but I can't say
>
>    bar = \markup { "Stand on" \markup { \bold "Zanzibar" } }
>
> You will object that I would have no *reason* to say that anyway,
> but it nevertheless constitutes a toy example of non-transferability:
>
>    \foo   is not equivalent to   \markup { \bold "Zanzibar" }
>
> despite the "=" sign in the macro definition, and that's surprising.

The reason for this is that foo is not a macro; it's a variable. Similarly, in 
C you can write 
int i = 5+9;
foo(&i);
which is _not_ the same as the (invalid) expression
foo(&(5+9));
(this example is not exactly analogous to what you're trying, but it 
illustrates that variables and macros are different things).

The special thing about 'markup' is that it isn't a function; it's rather a 
special token that marks that the following code should be interpreted as 
markups.

-- 
Erik