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I have a variable say some-var, its value is a string. I want to use its value in a quote expression.

'(some-var "some string")

The function who is using the above expression reports error because it gets some-var literally...not the value.

Any trick to eval some-var inside a quote expression?

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1 Answer 1

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You have two options:

1) Don't use quotes at all, as in:

(list some-var "some-string")

2) Use a backquote. They work like quotes, but part of an expression can be evaluated by using , and ,@. For example:

`(,some-var "some-string")
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    Why I did not think about list at the first place...lol! Thanks.
    – David S.
    Jul 1, 2016 at 0:54
  • In my case, I was trying to accomplish '((".*" . "PATH")), where "PATH" should be an evaluated expression. I ended up using: (list (cons ".*" (EXPR))).
    – trxgnyp1
    Nov 4, 2023 at 20:10

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