Haskell程序输出`<< loop>>` [英] Haskell program outputs `<<loop>>`

问题描述

我写了一个Haskell程序，在列表上进行二进制搜索。至少这是我以为它是。当我用 ghc v7.6.3 编译程序并运行程序时，我得到以下输出：

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>>

这个输出是什么意思？是否意味着我有一个无限循环 ghc 优化掉了？

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?

推荐答案

正如几个意见所说，这是Haskell RTS检测到一个无限循环在运行时。它无法始终检测到此类循环，但在简单的情况下它可以。

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.

例如，

x = x + 1

将编译正常，但在运行时引发异常。 （顺便说一下，这是一个例外 - 特别是你可以捕捉它，如果你想要的，但你可能不是想要。）

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".)

那么为什么GHC甚至让这个编译？好吧，因为如果我用 + 替换，例如：，那么表达式现在终止了。 （它代表一个1元素的循环列表。）编译器不能告诉编译时什么是和不是明智的递归。 RTS不能总是在运行时告诉;但是当它可以说出错误时，它会通过向您抛出异常来通知您。

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.

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