Emacs -- How to extract all elements of a list

Question

I am looking for some assistance, please, to extract all elements of a list of files and/or directories that have been marked in `dired-mode`. Essentially, if there were some way to just remove the parentheses from around the result of `(mapcar (lambda (filename) (file-name-nondirectory filename)) (dired-get-marked-files))`, then that would do the trick. ``` (start-process "name-of-process" "*output-buffer*" "/usr/bin/zip" "zip-file-name.zip" (mapcar (lambda (filename) (file-name-nondirectory filename)) (dired-get-marked-files)) ) ``` The result I am seeking will look like this: ``` (start-process "name-of-process" "*output-buffer*" "/usr/bin/zip" "zip-file-name.zip" "filename-number-one" "filename-number-two" "filename-number-three" ) ``` EDIT: The `start-process` function does not generally accept a single concatenated string of arguments. Instead, each argument must be separately spelled out (with quotation marks around each argument), or the argument can be a variable. Here is the…

Accepted Answer

What you want is to use `apply` with `(mapcar ...)` as its last argument:

```
(apply 'start-process
       "name-of-process"
       "*output-buffer*"
       "/usr/bin/zip"
       "zip-file-name.zip"
       (mapcar #'file-name-nondirectory (dired-get-marked-files)))
```

Note that `(mapcar #'function list)` is a shorter spelling of `(mapcar (lambda (arg) (function arg)) list)`.

Emacs -- How to extract all elements of a list

Solution

Problem