CODEWITHSUNDEEP
functionintlisp
hunterge
hunterge

Reputation: 653

x++ equivalent in Lisp?

I want an integer x, that increases by one every time a function is called.

I was told to use (defvar x) to declare the variable, but does this declare it as an int?

And also, is there a Lisp equivalent to x++ (x = x + 1) that I can use at the end of the function?

Thanks a lot.

Upvotes: 0

Views: 190

Answers (2)

Sylwester
Sylwester

Reputation: 48745

You don't need to declare what type it is. Just set it to a number.

Use the macro incf to change the binding to the successor of the current binding.

You don't need to have it as a global. You can have a closed over variable to update. Eg.

(let ((call-count 0))
  ;; create my-func in the scope of call-count
  (defun my-func (arg)
    (if (eq arg 'get)        ;; if the argument is the symbol get
        call-count           ;; return call-count
        (progn               ;; else we do first 
           (incf call-count) ;; update the binding to a hight value 
           arg)))))          ;; and then the job, as example it's #'identity


call-count     ; ==> ERROR call-count has not value
(my-func 'a)   ; ==> a
(my-func 'b)   ; ==> b
(my-func 'get) ; ==> 2

Upvotes: 2

sds
sds

Reputation: 60014

Try incf for incrementing:

(defvar *call-count* 0)

(defun my-func (...) 
  (incf *call-count*)
  ...)

Read up on defvar; it creates a dynamic variable but does not declare its type (use declaim for that).

There is no int in Lisp, just integer, whose size is not limited by the standard (there is a smaller fixnum size too, but there is no reason to declare this variable's type).

Upvotes: 3

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