How to call a macro in a function in LISP?

Question

Let's assume that I got the following macro:

(defmacro my-check (number)
  `(> 3  ,(apply #'+ number)))

How can I call this macro in a function?

I tried,for example, the following function:

(defun do-test (my-object)
  (my-check my-object)))

but I get the following error when compiling:

during macroexpansion of (MY-CHECK MY-OBJECT). Use *BREAK-ON-SIGNALS* to
intercept.
The value MY-OBJECT is not of type LIST.

jkiiski · Accepted Answer · 2016-02-29 13:55:49Z

4

The comma in your macro is in the wrong place. It is trying to evaluate the entire (apply ...) during macroexpansion, which of course fails since number is a symbol rather than a list. Remember that macros are expanded during compilation, not at run-time.

The correct version would be:

(defmacro my-check (number)
  `(> 3  (apply #'+ ,number)))

answered Feb 29, 2016 at 13:55

jkiiski

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4 Comments

Xaving Over a year ago

thank you. ´Suppose I could not modify the macro. How could I do a function call instead of re-working the macro?

jkiiski Over a year ago

The only way you can use the original macro is by giving it a literal list as argument, for example (my-check (1 1)). It can't be used with variables.

Xaving Over a year ago

Tank you. Could you elaborate a little bit on how you came to this conclusion?

jkiiski Over a year ago

I'm not sure what you mean. That the original macro won't work with variables? That follows from the fact that macros are expanded once when you first compile or load the code. At that point the variable is just a symbol; it will only have a value attached to it at run-time (probably a different value every time you call the function).

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