On Thursday, March 8, 2012 2:19:37 PM UTC-8, mefesto wrote:
>
> If `x` is a list then is the call to `(apply list x)` necessary?
>
Yes, meant to take that out.
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Not sure what kind of input you can get, but apply-ing list only on strings
would decompose them, so your function can be written as:
(defn ls [x]
(cond
(string? x)
(apply list (list x))
:else
(apply list x
as '(apply list nil)' will yield '().
Or, you can write i
> So I've ended up writing the function with a conditional, like so. Is there
> a tidier way?
>
> (defn ls [x] (cond (list? x) (apply list x)
> (nil? x) '()
> :else (list x)))
If `x` is a list then is the call to `(apply list x)` necessary?
(defn ls [x]
(
Hi,
I'm seeking a small & idiomatic function that will take input and ensure
it's wrapped in a list. If nil, I'd like the list to be empty. If the
input is already a list, I don't want to wrap it in another list layer. So:
"hi" => ("hi")
nil=> ()
("hi") => ("hi")
(list '("hi"))
