40

I wrote a Haskell program that preforms a binary search on a list. At least that's what I thought it does. When I compiled the program with ghc v7.6.3 and ran the program I got the following output:

progname: <<loop>>

What on earth does this output mean? Does it mean I had an infinite loop that ghc optimized away? How am I supposed to debug this?

Improve this question
7
  • Are you aware of the ghci debugger? You can also re-purpose HPC to find out which code is NOT being executed as a way to narrow down a loop. Commented Feb 1, 2014 at 23:58
  • @ThomasM.DuBuisson GHCI raises: Exception: <<loop>>. I assume the compiled output progname: <<loop>> is and STDERR message. Does this mean I have an infinite loop? Commented Feb 2, 2014 at 0:02
  • 4
    Yes, it's the RTS (runtime system) detecting an infinite loop (which it can do in certain cases). Commented Feb 2, 2014 at 0:16
  • 8
    It's specifically when it detects the infinite loop that results when evaluating a specific constructor requires evaluating that constructor. Commented Feb 2, 2014 at 1:41
  • 5
    @awashburn You can have self-referential values that are fully defined. fibs = 0 : scanl (+) 1 fibs, for instance. That's why it's allowed. Commented Feb 2, 2014 at 5:19

1 Answer 1

Reset to default
40

As several of the comments have said, this is the Haskell RTS detecting an infinite loop at run-time. It cannot always detect such loops, but in simple cases it can.

For example,

x = x + 1

will compile just fine, but provoke an exception at run-time. (Incidentally, this is an exception - in particular, you can catch it if you want. But you probably don't "want".)

So why does GHC even let this compile? Well, because if I replace + with, say, :, then the expression now terminates just fine. (It represents a 1-element circular list.) The compiler can't tell at compile-time what is and is not sensible recursion. The RTS can't always tell at run-time; but when it can tell something's wrong, it'll let you know by throwing an exception at you.

Improve this answer
Sign up to request clarification or add additional context in comments.

5 Comments

Is there any way to force Haskell to also output which infinite loop it found?
@Bakuriu Sadly no. About the best you could do is try to catch the exception and print out where you caught it from - but that would require you to have some idea where to start looking in the first place. Usually loop bugs are something as silly as a typo (e.g., you meant x = foo y but accidentally wrote x = foo x).
@Bakuriu If you compile with profiling enabled, the following will tell you where the exception was throw: ./progName +RTS -xc -RTS
@MathematicalOrchid Could you add your example ( x = foo x) to your answer? I know this is an old question but I was just sent here from google with the same error and it was a simple issue like what you said. That may provide value to other haskell noobs like myself (and comments aren't forever).
The comment from @recursion.ninja is what those who land on this question are looking for.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.