string-join list with function as list member

Question

I want to string-join a list of strings, one of which is returned by a function call.

Like this:

(defun foobar () "foobar")
(string-join '("foo" (foobar) "bar") "|")

That results in Wrong type argument: characterp, foobar and not in "foo|foobar|bar" as expected.

What's the elegant way to do that?

Does this answer your question? How to evaluate the variables before adding them to a list? — Drew
– Drew, Commented Jan 17, 2021 at 19:04
There are multiple questions that deal with the same thing: quoting some sexp and expecting some part of the quoted sexp to be evaluate. It would be great if Someone (TM) created a general Q & A for that, as a Community question. Many, many questions like this one have been closed as dups of the one I cited, but that Q is not stated in as general a way as it could be. — Drew
– Drew, Commented Jan 17, 2021 at 19:06

Muihlinn · Accepted Answer · 2021-01-17 06:29:25Z

3

A quoted list doesn't evaluate its args, so it consists of a string, a list containing a symbol and another string. You can selectively evaluate it using backquote and unquote:

`("foo" ,(foobar) "bar") ;=> ("foo" "foobar" "bar")
(string-join `("foo" ,(foobar) "bar") "|") ;=> "foo|foobar|bar"

edited Jan 17, 2021 at 6:29

Muihlinn

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answered Jan 17, 2021 at 0:08

wasamasa

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