Why does peek on a seq of vector fail?

[Jason Gilman]

Why does (peek (seq [1])) result in: 

ClassCastException clojure.lang.PersistentVector$ChunkedSeq cannot be cast to clojure.lang.IPersistentStack  clojure.lang.RT.peek (RT.java:634)

Peek documentation "For a list or queue, same as first, for a vector, same as, but much more efficient than, last. If the collection is empty, returns nil." implies that there shouldn't be a problem working with vectors.

This works without issue:

(peek [1]) 

This is on Clojure 1.5.1

[Jason Gilman]

Also, I really appreciate anyone who takes the time to answer. My company is currently evaluating Clojure. I'm trying to determine if this is non-idiomatic use of the language or some other issue. Doing something like this can cause the bug to occur without directly calling seq on a vector:

(defn bar [my-list n]

  (if (= n 0)

      (peek my-list)

      (bar (rest my-list) (dec n))))

(bar [1 2 3] 1)

Both rest and peek work on a vector without a problem but the combination of the two causes problems.

[David Nolen]

Calling seq on vector returns a sequence which is no longer a vector. The docstring is pretty specific about which types it's supposed to work on.

clojure.core/peek

([coll])

  For a list or queue, same as first, for a vector, same as, but much

  more efficient than, last. If the collection is empty, returns nil.

David

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[László Török]

Hi, 

I think it's by design. 

It only works with concrete data structure (list, vector, queue) types that implement 

clojure.lang.IPersistentStack

Las

2013/6/20 Jason Gilman <jason....@gmail.com>

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László Török

[John D. Hume]

On Jun 20, 2013 3:11 PM, "Jason Gilman" <jason....@gmail.com> wrote:
>
> (defn bar [my-list n]
>   (if (= n 0)
>       (peek my-list)
>       (bar (rest my-list) (dec n))))
> (bar [1 2 3] 1)
>

It seems likely you want either first and rest* (to work from the front of any seqable) or peek and pop (to work from the back of a vector or the front of a list or queue). Combining peek and rest on a vector as the bar function does makes no sense, since they work at opposite ends.

*or next in place of rest.

[Jason Gilman]

I think I get the difference now between a sequence, list, and a vector. Thanks for the quick answers. 

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