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so -> and ->> behave the same as long as my functions only take one
parameter?

Am 02.11.2011 10:58, schrieb Matt Hoyt:
> First question answer:
> 
> The code isn't the same.  The ~x is the first argument for -> and
> is the last argument for ->>.  So if you do a long chain of
> expression the first argument is passed into the return value into
> the next function with -> and ->> pass that value as the last
> parameter.
> 
> Second question:
> 
> I think what you mean is (->> "hi" println) .   If you wanted to
> give the function multiple arguments you could do this:
> 
> (-> "hello" (println "world!")) and the other form it would be
> (->> "world!" (println "Hello"))
> 
> Matt Hoyt 
> ------------------------------------------------------------------------
>
> 
*From:* Dennis Haupt <d.haup...@googlemail.com>
> *To:* clojure@googlegroups.com *Sent:* Wednesday, November 2, 2011
> 4:37 AM *Subject:* -> vs ->> and names vs anonymous function
> 
> hi there,
> 
> i stumbled over two odd things 1) -> and ->> have the same source
> code. why is that? 2) why can't i use "(-> "hi" #(println %))"
> directly? why do i have to put my function into a symbol first? is
> there a way to avoid this?
> 
> (let [dummy #(println %)] (->> "hi" dummy))
> 
> 
> 
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