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JulioHerrlein

Argument Question (simple problem)

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Dear friends,

 

I'm still learning. This is a simple thing.

How can I put inside a function the name of a variable, like the example below.

 

I want to do this:

 

(gen-binary-row 12 '(0 2 5 7 8 11))

>> (1 0 1 0 0 1 0 1 1 0 0 1)

 

But in this way:

 

(setf binrow '(0 2 5 7 8 11))
(gen-binary-row 12 '(binrow))

 

What am I missing ?

 

Thanks for help.

Best,

Julio

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You can define local variables within and outside functions with let.

 

In principle, you can also use setf within a function, but then you are overwriting the value of a global variable. In most cases you want to avoid that. Functions that do not change anything outside them (so called side effects) are much safer. 

 

Both let and setf are explained in detail in the very informative online textbook Practical Common Lisp, chapter 6: http://www.gigamonkeys.com/book/variables.html. There you will also learn that Lisp actually distinguishes between lexical and dynamic variables; I was talking here about local and global variables instead for simplicity. In case this chapter covers something you don't follow, just go back a chapter or two and you will learn a lot :)

 

Simple example:


(defun my-function (x)
  (let ((y (+ x 1)))
    y)) 

(my-function 42)
; => 43

 

Best,

Torsten 

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But in this way:

(setf binrow '(0 2 5 7 8 11))
(gen-binary-row 12 '(binrow))

What am I missing ?

 

just do it like that:


(setf binrow '(0 2 5 7 8 11))
(gen-binary-row 12 binrow)

;; your "binrow" is now a variable with a LIST as value 
;; when you are writing '(binrow) it will be a LIST with the VALUE binrow (and not the values of "binrow")

 

 

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