CODEWITHSUNDEEP
macroslispcommon-lispmetaprogrammingquote
HiroIshida
HiroIshida

Reputation: 1603

pass a list to macro in Common Lisp

I'm having a problem passing a list to macro, where the list will be used to generate function name. For example, the code below causes an error.

(defmacro gen (str-lst)
  `(defun ,(intern (string-upcase (car str-lst))) () (print "foo")))

(gen '("foo" "bar"))

The resulting error was:

*** - DEFUN/DEFMACRO: QUOTE is a special operator and may not be redefined. The following restarts are available: ABORT :R1
Abort main loop

How should I modify my code, and what is wrong with my code?

The thing makes me even more confused is that the code below, about which answer exits here, works fine.

(defmacro easy-one (str-lst)
  `(mapc #'(lambda (str) (print str)) ,str-lst))
(easy-one '("foo" "bar"))

Upvotes: 2

Views: 471

Answers (1)

Barmar
Barmar

Reputation: 780655

Don't quote the list. Macros don't evaluate their arguments, so you don't need to quote them to prevent them from being evaluated, like you do for ordinary functions.

(gen ("foo" "bar"))

When you quote it, you're executing

(get (quote ("foo" "bar")))

The value of str-list is the list (quote ("foo" "bar")), so (car str-list) is the symbol QUOTE. As a result, the macro expands to

(defun quote () (print "foo"))

That's why you get an error complaining that you're trying to redefine the built-in QUOTE.

The difference in your second example is that you're just substituting the parameter into the expansion, you're not using its value in the expansion code. So it expands to

(mapc #'(lambda (str) (print str)) '("foo" "bar")))

Here the list will be used when the expansion runs, not while the macro is being expanded. It needs to be quoted there, to prevent it from being evaluated as a function call.

You should use macroexpand to see how your macros are being expanded when debugging.

Upvotes: 7

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