How do I turn (seq "foo") back into a string?

I'm exploring clojure and am puzzled. What fills in the blank to make the following expression eval to true?

(= "foo"开发者_Go百科 ___ (str (seq "foo")))

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You need to use apply function:

user=> (apply str (seq "foo"))
"foo"

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Actually nothing can fill in the blank to make the expression (= "foo" _ (str (seq "foo"))) eval to true because (str (seq "foo")) => "(\f \o \o)" which is not equal to "foo" so we have an inequality already and a third item, no matter what value, to fill the blank cannot make the expression evaluate to true

If you meant to ask

(= "foo"
    (____ str (seq "foo")))

Then the answer would rightly be apply as answered by alex.

user> (doc apply)
-------------------------
clojure.core/apply
([f args* argseq])
  Applies fn f to the argument list formed by prepending args to argseq.

Apply takes a function (in this case str) and calls str with the args present in the seq

user> (seq "foo")
(\f \o \o)

user> (str \f \o \o)
"foo"

And oh, btw:

user> (= 1 1 1)
true

user> (= 1 2 1)
false

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Or you could use the convient clojure.string/join

(require 'clojure.string)
(clojure.string/join (seq "foo"))

Will print "foo" as well. You could off course define your own join using the previously mentioned technique (to avoid write (apply str seq) all the time).