4

Let's say, If I have a directory variable named a-directory with its value. Also I have another variable named b-directory which gets its value from the previous variable. 

(setq a-directory "C:/Users/Paper")
(setq b-directory (expand-file-name "org-mode.org" a-directory))

Every time when I change the value of a-directory, I need to eval the b-directory again to get its new value. So here comes the question: How can I get the updated value without evaling b-directory when changing the value of a-directory?

Improve this question
4
  • Please add the missing parenthesis for your code. (And consider getting rid of the backquote and comma.) – Drew May 22 '16 at 3:15
  • 4
    Instead of being a variable, b-directory could be a function: (defun b-directory () (expand-file-name "org-mode.org" a-directory)) - in this case it will evaluate the expression every time (slight overhead), but will be always up to date. – wvxvw May 22 '16 at 5:30
  • This is a good idea. – Leu_Grady May 22 '16 at 5:47
  • 3
    You could also never change a-directory directly through setq but from a change-a-directory function that would update both a-directory and b-directory. – JeanPierre May 22 '16 at 7:38
-1

Described behavior similar to reactive programming, you may found some implementations for elisp.

But I have no experience with them yet. Please, feel free to add usage examples bellow.

Improve this answer
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.