How to read lines from stdin (in) in clojure

Question

I am writing my first clojure program, and want to read lines from stdin.

When I try this:

(doall (map #(println %) (line-seq *in*)))

I get this exception:

Exception in thread "main" java.lang.ClassCastException: clojure.lang.LineNumberingPushbackReader cannot be cast to java.io.BufferedReader (test.clj:0)

I get the same results in version 1.0 and 1.1

So how do I convert *in* into a seq I can iterate over? I would have thought that this is common enough that *in* itself would be iterable, but that does not work either - if I try to use it directly I get:

java.lang.IllegalArgumentException: Don't know how to create ISeq from: clojure.lang.LineNumberingPushbackReader (NO_SOURCE_FILE:0)

Also, are there any examples of doing general file handling in clojure?

replacing (map #(println %) (line-seq.. )) with just (map println (line-seq..)) could be nice — GabiMe, Jan 10 '10 at 12:44

score 39 · Accepted Answer · edited May 20 '16 at 12:04

39

Try wrapping *in* in a java.io.BufferedReader. And also use doseq instead of doall, as devstopfix pointed out:

(doseq [ln (line-seq (java.io.BufferedReader. *in*))]
   (println ln))

Note that line-seq is documented to require a BufferedReader as its source.

edited May 20 '16 at 12:04

bfontaine

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answered Jan 09 '10 at 17:17

seh

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This code does not work. I get the error "doseq requires a vector for its binding" Try was @devstopfix wrote. It worked fro me. – Jason Dec 11 '14 at 17:35
1

@Jason Use dorun instead of doseq for this form. doseq requires a binding form. dorun forces a lazy seq. – user1009908 Jun 18 '15 at 01:42
1

How can use a escape for this loop? – Alexandre Nascimento Oct 27 '19 at 17:01

score 25 · Answer 2 · answered Dec 22 '11 at 19:52

25

Just a note that for anyone who wants to only read a single line, there's the read-line function.

answered Dec 22 '11 at 19:52

Mike Crittenden

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score 24 · Answer 3 · edited Sep 12 '13 at 06:48

24

You should probably use doseq instead of doall:

(doseq [line (line-seq (java.io.BufferedReader. *in*))] 
    (println line))

doall:

Walks through the successive nexts of the seq, retains the head and returns it, thus causing the entire seq to reside in memory at one time.

doseq:

Does not retain the head of the sequence. Returns nil.

edited Sep 12 '13 at 06:48

Evgeniy Berezovsky

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answered Jan 26 '11 at 20:56

devstopfix

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score 9 · Answer 4 · answered Mar 05 '18 at 14:53

9

For reasonably small inputs, the following would also work:

(let [input-string (slurp *in*)]
  (println input-string))

Or, splitting by lines:

(let [lines (clojure.string/split-lines (slurp *in*))]
  (println lines))

answered Mar 05 '18 at 14:53

mrucci

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What is "reasonably small"? Why is this only appropriate for "reasonably small" inputs? – Eric Ihli Feb 28 '20 at 13:17
3

@EricIhli, [`slurp`](https://clojuredocs.org/clojure.core/slurp) reads *all* of the input that's available (until EOF, i.e. end of file). So if there's a lot of input--at least many megabytes or several gigabytes, let's say--then you're loading all of into memory at the same time in a single variable `input-string`. And then *after* that, you're printing all of it out at once. That will work if you have enough RAM in your system and if you've started Clojure with Java options that tell Java to use enough RAM, but you still might have to wait awhile before you start seeing the printed output. – Mars Mar 01 '22 at 04:33

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