[Maxima-discuss] A question about parsing: ":lisp #$expr$" is not low-level enough

[Eduardo O. <edu...@gm...>]

Hi Richard and Stavros,

I made some progress on the questions that I asked here,

  https://sourceforge.net/p/maxima/mailman/message/37415334/
  https://sourceforge.net/p/maxima/mailman/message/37415708/

but I need to ask more questions, and I felt that it was better to
create a new thread.

I am reading about expressions, simplification, evaluation, and the
internal representation mainly in these places:

  (info "(maxima)Expressions")
  (info "(maxima)Evaluation")
  (info "(maxima)Simplification")
  (info "(maxima)Lisp and Maxima")
  https://people.eecs.berkeley.edu/~fateman/papers/intro5.txt

The first two sentences of

  (info "(maxima)Introduction to Simplification")

are:

  Maxima performs a cycle of actions in response to each new
  user-typed command. This consists of four steps: reading or
  "parsing" the input, evaluation, simplification and output.

I am still in a stage in which the rules for evaluation and
simplification feel overwhelming, so I'm trying to understand parsing,
evaluation, simplification, and output separately...

The second part of

  https://people.eecs.berkeley.edu/~fateman/papers/intro5.txt

has a table with "Input String"s in the left column and "Parser
Output"s in the right column. The most obvious - and naïve - way to
transform that table into a series of executable examples/tests is to
wrap each "Input String" in a ":lisp #$___$"; I found that if I run a
"simp:false;" before the tests then all these tests yield essentially
the same results as in the paper (obs: I had to fix some typos - or
scanos - in the .txt... Richard, do you want a list of the
typos/scanos?):

  simp:false;
  :lisp #$a$
  :lisp #$?a$
  :lisp #$"a"$
  :lisp #$x+y$
  :lisp #$x-y$
  :lisp #$x*y$
  :lisp #$a(x)$
  :lisp #$a[1,2]$
  :lisp #$a[1,2](x)$
  :lisp #$sin(x)$
  :lisp #$x/y$
  :lisp #$x.y$
  :lisp #$x^2$
  :lisp #$x^^2$
  :lisp #$[a,b,c]$
  :lisp #$if a then b$
  :lisp #$if a then b else c$
  :lisp #$'diff(y,x)$
  :lisp #$'integrate(a,b,c,d)$
  :lisp #$not a$
  :lisp #$a or b$
  :lisp #$a and b$
  :lisp #$a=b$
  :lisp #$a>b$
  :lisp #$a>=b$
  :lisp #$a<b$
  :lisp #$a<=b$
  :lisp #$a#b$

but these entries give outputs that are very wrong:

  :lisp #$'a$
  :lisp #$(a,b,c)$
  :lisp #$for i:a thru b step c unless q do f(i)$
  :lisp #$for i:a next n unless q do f(i)$
  :lisp #$for i in L do f(i)$
  :lisp #$diff(y,x)$
  :lisp #$diff(y,x,2,z,1)$
  :lisp #$integrate(a,b,c,d)$
  :lisp #$block([l1,l2], s1,s2)$
  :lisp #$block(s1,s2)$
  :lisp #$a:b$
  :lisp #$a::b$
  :lisp #$a(x):f$

it seems that there is some kind of evaluation that is still going on.
How can I run just the parser? I guess that it would be something like
this,

  to_lisp();
    (mread-raw "a<b")

but using another function, as mread-raw expects a stream...

  Thanks in advance =),
    Eduardo Ochs
    http://angg.twu.net/eev-maxima.html
    http://angg.twu.net/e/maxima.e.html#input-string-parser-output