How can I convert a LazySeq of Characters to a String in Clojure?

Question

Let's say I have a LazySeq of java.lang.Character like

(\b \ \! \/ \b \ \% \1 \9 \/ \. \i \% \$ \i \space \^@) 

How can I convert this to a String? I've tried the obvious

(String. my-char-seq) 

but it throws

java.lang.IllegalArgumentException: No matching ctor found for class java.lang.String (NO_SOURCE_FILE:0) [Thrown class clojure.lang.Compiler$CompilerException] 

I think because the String constructor expects a primitive char[] instead of a LazySeq. So then I tried something like

(String. (into-array my-char-seq)) 

but it throws the same exception. The problem now is that into-array is returning a java.lang.Character[] instead of a primitive char[]. This is frustrating, because I actually generate my character sequence like this

(map #(char (Integer. %)) seq-of-ascii-ints) 

Basically I have a seq of ints representing ASCII characters; 65 = A, etc. You can see I explicitly use the primitive type coercion function (char x).

What this means is that my map function is returning a primitive char but the Clojure map function overall is returning the java.lang.Character object.

Answer 1

This works:

(apply str my-char-seq) 

Basically, str calls toString() on each of its args and then concatenates them. Here we are using apply to pass the characters in the sequence as args to str.