What is the exact semantic of reduce in clojure?

Question

The Clojure docs regarding (reduce f val coll) state:

If val is supplied, returns the result of applying f to val and the first item in coll, then applying f to that result and the 2nd item, etc.

So I tried:

(reduce str ["s1" "s2"] "s3")

Which leads to:

"[\"s1\" \"s2\"]s3"

However, when I try

(apply str ["s1" "s2"])

I get:

"s1s2"

So why does the vector get converted into its string representation?

score 1 · Answer 1 · answered Sep 17 '13 at 11:11

1

Ok, I misread the docs. So reduce actually does this:

(str ["s1" "s2"] "s3")

Which evaluates to:

"[\"s1\" \"s2\"]s3"

Because str needs to convert its arguments to string!

answered Sep 17 '13 at 11:11

Arne

score 1 · Accepted Answer · answered Sep 17 '13 at 16:56

For a "picture" view for others who may see this question. Clojure's reduce is a left fold. For Haskell programmers foldl.

It creates the following structure

(reduce - 0 (range 1 3))
(- (- (- 0 1) 2) 3)

or infix

(((0 - 1) - 2) - 3)

The intuition being it's a left fold because the parens drift towards the left

2 Answers2