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> Macros modify code structure at runtime so obviously that is fraught with danger. You may have inadvertently misspoke, but macros operate at compile time, at

> Macros modify code structure at runtime so obviously that is fraught with danger. You may have inadvertently misspoke, but macros operate at compile time, at least in the Lisps I have used.

lispm 6y ago
If one uses a Lisp interpreter, macro expansion may happen at runtime. CLISP example: [2]> (defmacro add2 (place) (print 'add2) `(incf ,place 2)) ADD2 [3]> (let ((a 1)) (dotimes (i 4) (add2 a))) ADD2 ADD2 ADD2 ADD2 NIL [4]> As one can see the macro form is expanded four times at runtime.
- dzsekijo 6y ago
  This is a confusing example, because in a REPL steps of compilation and evaluation are interleaved. Indeed, can you write a program for CLISP that works like this: - takes one command line argument (a file name) - reads in the given file, interprets it as Common Lisp code, expecting it to deliver a definition for the add2 macro - then runs (let ((a 1)) (dotimes (i 4) (add2 a)))
  - lispm 6y ago
    > This is a confusing example, because in a REPL steps of compilation and evaluation are interleaved. The CLISP REPL does not compile, thus it can't be interleaved. > then runs It will still be interpreted and the macro will still be expanded at runtime.
vram22 6y ago
I don't know Lisp well, but I remember PG saying in one of his books or essays, that Lisp is a language in which you can compile and run at read time, and the other two possibilities, too.
lmilcin 6y ago
In full Lisp "compile time" is part of application execution. Now, Clojure is kind of impaired Lisp because it is written for a VM that was not intended to be used this way and so this is not that much pronounced (but you still get REPL, etc.)
- thu2111 6y ago
  Is there anything lacking from the JVM beyond tail-call elimination that Lisp needs? I know Clojure is very slow to start up but my understanding is that this is because it uses the JVM very inefficiently, and they don't seem to care much.
ludston 6y ago
Macro's are evaluated inside of 'defun, however, 'defun is evaluated during runtime (when the .lisp file is being loaded into the implementation) so base698 is technically correct. Excepting that it is not really "fraught with danger" unless you are redefining macro's that have already been expanded and cached in function definitions.

> Macros modify code structure at runtime so obviously that is fraught with danger. You may have inadvertently misspoke, but macros operate at compile time, at least in the Lisps I have used.

lispm 6y ago

If one uses a Lisp interpreter, macro expansion may happen at runtime. CLISP example: [2]> (defmacro add2 (place) (print 'add2) `(incf ,place 2)) ADD2 [3]> (let ((a 1)) (dotimes (i 4) (add2 a))) ADD2 ADD2 ADD2 ADD2 NIL [4]> As one can see the macro form is expanded four times at runtime.

dzsekijo 6y ago
This is a confusing example, because in a REPL steps of compilation and evaluation are interleaved. Indeed, can you write a program for CLISP that works like this: - takes one command line argument (a file name) - reads in the given file, interprets it as Common Lisp code, expecting it to deliver a definition for the add2 macro - then runs (let ((a 1)) (dotimes (i 4) (add2 a)))
- lispm 6y ago
  > This is a confusing example, because in a REPL steps of compilation and evaluation are interleaved. The CLISP REPL does not compile, thus it can't be interleaved. > then runs It will still be interpreted and the macro will still be expanded at runtime.

vram22 6y ago

I don't know Lisp well, but I remember PG saying in one of his books or essays, that Lisp is a language in which you can compile and run at read time, and the other two possibilities, too.

lmilcin 6y ago

In full Lisp "compile time" is part of application execution. Now, Clojure is kind of impaired Lisp because it is written for a VM that was not intended to be used this way and so this is not that much pronounced (but you still get REPL, etc.)

thu2111 6y ago
Is there anything lacking from the JVM beyond tail-call elimination that Lisp needs? I know Clojure is very slow to start up but my understanding is that this is because it uses the JVM very inefficiently, and they don't seem to care much.

ludston 6y ago

Macro's are evaluated inside of 'defun, however, 'defun is evaluated during runtime (when the .lisp file is being loaded into the implementation) so base698 is technically correct. Excepting that it is not really "fraught with danger" unless you are redefining macro's that have already been expanded and cached in function definitions.