Re: macros and macroexpand

[Heime]

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------- Original Message -------
On Tuesday, August 8th, 2023 at 2:22 AM, Philip Kaludercic <philipk@posteo.net> 
wrote:


> Heime heimeborgia@protonmail.com writes:
> 
> > Sent with Proton Mail secure email.
> > 
> > ------- Original Message -------
> > On Monday, August 7th, 2023 at 11:46 PM, Yuri Khan yuri.v.khan@gmail.com 
> > wrote:
> > 
> > > On Mon, 7 Aug 2023 at 18:04, Heime heimeborgia@protonmail.com wrote:
> > > 
> > > > I have made a macro and know that they are supposed to return
> > > > expanded code for use. Still I cannot understand the need to
> > > > call "macroexpand". Should't the macro already perform the
> > > > expansion ?
> > > 
> > > You should be posting small examples of code that you’re trying,
> > > otherwise, there is high chance people will either misunderstand you
> > > or just disregard your questions as ill-posed.
> > > 
> > > ----
> > > 
> > > When you define a macro, you indeed write the definition similarly to
> > > a function that returns expanded code.
> > > 
> > > (defmacro foo (&rest body)
> > > `(bar ,@body)) When you evaluate a form that references a macro, Elisp 
> > > will (1) expand the macro, and (2) evaluate the result of the expansion: 
> > > (foo 'quux) ⇒ Debugger entered--Lisp error: (void-function bar) On the 
> > > other hand, calling ‘macroexpand’ on a data representation of that form 
> > > will just return the expansion result: (macroexpand '(foo 'quux)) ⇒ (bar 
> > > 'quux) In this example, I did not bother to define ‘bar’, so Elisp 
> > > assumes it would be a function and complains at evaluation time. But I 
> > > could further define ‘bar’ as a macro: (defmacro bar (&rest body)` (baz 
> > > ,@body))
> > > 
> > > In this case, evaluating the original form shows that Elisp expanded
> > > both macros ‘foo’ and ‘bar’, and then tried to call the undefined
> > > function ‘baz’:
> > > 
> > > (foo 'quux)
> > > ⇒ Debugger entered--Lisp error: (void-function baz)
> > > 
> > > Meanwhile, ‘macroexpand’ still just expands a single level of macros:
> > > 
> > > (macroexpand '(foo 'quux))
> > > ⇒ (bar 'quux)
> > > 
> > > and you can invoke it repeatedly until you get to the fixed point:
> > > 
> > > (macroexpand (macroexpand '(foo 'quux)))
> > > ⇒ (baz 'quux)
> > > 
> > > (macroexpand (macroexpand (macroexpand '(foo 'quux))))
> > > ⇒ (baz 'quux)
> > 
> > Then macroexpand is useful for diagnostics to expand at one level only
> > at a time. Thusly, if I just want to get the expanded code produced by
> > a macro, I can just do pp-to-string upon the object made by a macro.
> > 
> > (defmacro adder (mopi mopj)
> > `(+ ,(cl-second mopi) ,(cl-third mopj)))
> > 
> > (princ (pp-to-string '(adder (* 3 5) (* 5 7)) ))
> 
> ^
> don't do this
> 
> If you quote an expression, it won't be evaluated or macro-expanded any
> further. You can sort-of think of a macro like a kind of inline
> function call. The evaluation would go along these lines:
> 
> (princ (pp-to-string (adder (* 3 5) (* 5 7))))
> 
> will be transformed into this at macro-expansion time, and evaluation
> would do this:
> 
> (princ (pp-to-string (+ (cl-second '(* 3 5)) (cl-third '(* 5 7)))))
> (princ (pp-to-string (+ 3 7)))
> (princ (pp-to-string 10))
> (princ "10\n")
> "10\n"

What I want to do is print the code made by adder of its final expansion code.
Rather than the last evaluation of 10, I want to print (+ 3 7).  

Can my print command be modified in such a way that the message shows (+ 3 7) ?
IT seems that I would need to use macroexpand-all, to get to the final 
unevaluated
sexp. 
 
> > I would not do
> > 
> > (princ (pp-to-string (macroexpand '(adder (* 3 5) (* 5 7))) ))