I have a variable with some common values between dap-mode debug templates and I want to use that variable along with other cons elements for :environment-variable in languages like Golang or Typescript you would do something like this:

variable = [1, 2, 3]
another_variable = [...variable, 4, 5, 6]

then another_variable will carry the value of [1, 2, 3, 4, 5, 6]

How can I do this in Emacs Lisp?

EDIT: I know how you can evaluate something inside a list, this is using the backquote, my problem was that I didn't know how Elisp processes lists in a fashion that keep the result plain because, for instance in Python, you can end with something like [1, [2, 3, 4]] if x is [1] and you do x.append([2, 3, 4]); I didn't know that ,@ was a thing and that neither append keeps a plain list.


2 Answers 2


You can use the backquote mechanism in combination with ,@ to splice the value of a variable into a list; or equivalently, use append:

(setq variable '(1 2 3))
(1 2 3)

(setq another-variable `(-2 -1 0 ,@variable 4 5 6))
(-2 -1 0 1 2 3 4 5 6)

;; or...
(setq another-variable (append '(-2 -1 0) variable '(4 5 6)))
(-2 -1 0 1 2 3 4 5 6)

Note: A Lisp program should not try to modify self-evaluating forms or constant lists (bug#40671). These are internal data structures [Message 42].

Be aware that the final argument for append is not copied.

ELISP> (setq var1 (vector 'a 'b))
ELISP> (symbol-value 'var1)
[a b]
ELISP> (setq lst (append var1 (list 1)))
ELISP> (symbol-value 'lst)
(a b 1)
  • Not sure exactly what you are objecting to, since in my example I do not modify existing list structure at all. I did not read all of the bug you referred to, but AFAICT all of the examples that are discussed involve modifying a "constant" list, which (I think) I'm not doing. How is the fact that append does not copy its final argument make any difference to the example in my answer?
    – NickD
    Commented Jul 16, 2020 at 14:37
  • @NickD: These objects are not explicitly modified, but neither should they be implicitly modifiable. You can modify the constant list through the variable: (setcar another-variable 0 'a).
    – user19761
    Commented Jul 16, 2020 at 15:30
  • So what? Are you proposing outlawing setcar and its ilk?
    – NickD
    Commented Jul 16, 2020 at 16:01
  • @NickD: no, I'm just saying it's better to assign a new list than a literal list. For instance, it is better to do this (setq lst (list 'a 'b)) rather than (setq lst '(a b)).
    – user19761
    Commented Jul 16, 2020 at 16:45
  • @NickD: The issue is that Elisp may "merge" literal objects so IMO it is better to not store literal objects in "variables".
    – user19761
    Commented Jul 16, 2020 at 16:55

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